A COURSE OF INSTRUCTION IN THE GENERAL PRINCIPLES OF CHEMISTRY BY ARTHUR A. NOYES AND MILES S. SHERRILL PRINTED IN PRELIMINARY FORM FOR THE CLASSES OF THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY BOSTON : THOMAS TODD Co., PRINTERS 1917 COPYRIGHT, 191? BY ARTHUR A. NOYES AND MILES S. SHERRILL CHAPTER I THE COMPOSITION OF SUBSTANCES 1. Definition of the Field of Chemistry. Its General Principles the Subject of this Course. Chemistry treats of the composition of substances, of their properties in relation to their composition, of changes in their composition, and of the effects attending such changes. General chemistry, often called also theoretical or physical chem- istry, treats of the general principles which have been found to ex- press certain common characteristics of the numerous phenomena of chemistry. To a discussion of the more important of these general principles this Course will be devoted. The divisions of the subject will be taken up in the order in which they were named in the above- given definition of the field of chemistry. 2. Pure Substances and Mixtures, and the Law of Definite Propor- tions. Out of the materials occurring in nature there can be prepared substances which, when subjected to suitable processes of fractionation (that is, to operations which resolve the materials into parts or frac- tions), always yield fractions whose properties are identical when measured at the same pressure and temperature. Such substances are called pure substances; other substances which can be resolved by such processes into fractions with different properties being called mixtures. For example, whether a solid material is a pure substance or mixture may be determined by partially melting or vaporizing it or by partially dissolving it in solvents, and by comparing the value of the density,, melting-point, or some other sensitive property, of the unmelted, un- vaporized, or undissolved part with that of the original material. The fundamental idea involved in the preceding considerations is that there exists an order of substances, called pure substances, of rela- tively great stability toward resolving agencies, each one of which has a perfectly definite set of properties, sharply differentiated from those of other pure substances; so that there is not a continuous series of pure substances whose properties pass over into one another by insensi- ble gradations. 1 I CMfO&ITION OF SUBSTANCES This principle of definiteness of properties in general applies also to the elementary composition of pure substances. This fact is ex- pressed by the law of definite proportions, which states that a pure sub- stance, however it be prepared, always contains its elements in exactly the same proportions by weight. 3. The Law of Combining Weights. To the various elements definite numerical values can be assigned which accurately express the weights of them, or small multiples of the weights of them, which are combined with one another in all pure substances. Such numerical values are called the combining weights of the elements. They are essentially relative quantities. Adopting 16 as the combining weight of oxygen, the combining weight of any other element may be defined to be that weight of it which combines with 16 parts of oxygen, or with some small multiple or submultiple of 16 parts of oxygen ; and the above principle, known as the law of combining weights f can be expressed as follows : Elements are present in pure substances only in the proportions of their combining weights or of small multiples of them. Prod. 1. The oxide of a certain element contains 30.06% of oxygen. and the sulphide of the same element contains 53.46% of sulphur. What does the law of combining weights show as to the relative weights of oxygen and sulphur that may be present in the pure compounds of these two elements? 4. Determination of Combining Weights. The way in which some important combining weights have been determined is illustrated by the following problem. Pro&. 2. Determine the exact combining weights of silver, potassium, and chlorine from the following data : In a series of eight experiments 801.48 g. of pure potassium chlorate were ignited or treated with hydro- chloric acid, yielding 485.66 g. of potassium chloride. In another series of five experiments 24.452 g. of pure potassium chloride were dissolved in water and precipitated with silver nitrate, whereby 47.013 g. of silver chloride were obtained. In a third series of ten experiments S2.669 g. of pure silver were dissolved in nitric acid and precipitated with hydro- chloric acid, yielding 109.840 g. of silver chloride. The combining weights adopted by the International Committee on Atomic Weights for the more important elements are presented in the following table, those multiples being given which have been shown to be the atomic weights, as described in Art. 17. The table is therefore also one of atomic weights. THE COMPOSITION OF SUBSTANCES 3 Aluminum . . . . Al 27.1 Antimony . . . . Sb 120.2 Argoii A 39.88 Arsenic As 74.96 Barium Ba 137.37 Beryllium .... Be 9.1 Bismuth Bi 208.0 Boron B 11.0 Bromine Br 79.92 Cadmium . . . . Cd 112.40 Calcium Ca 40.07 Carbon C 12.005 Chlorine 01 35.46 Chromium . . . . Cr 52.0 Cobalt Co 58.97 Copper Cu 63.57 Fluorine F 19.0 Gold Au 197.2 Helium He 4.00 Hydrogen . . . . H 1.008 Iodine . I 126.92 Iron Fe 55.84 Lead Pb 207.20 Lithium Li 6.94 Magnesium .... Mg 24.32 Manganese .... Mu 54.93 Mercury Hg 200.6 Nickel Ni 58.68 Nitrogen N 14.01 Oxygen ..... O 16.00 Phosphorus . . . . P 31.04 Platinum Pt 195.2 Potassium ....... K 39.10 Radium . . , .. . Ra 226.0 Silicon ..... Si 28.3 Silver Ag 107.88 Sodium Na 23.00 Strontium . . . . Sr 87.63 Sulphur S 32.06 Thallium Tl 204.0 Tin Sn 118.7 Zinc , Zn 65.37 5. The Atomic and Molecular Theories account for the fact that elements combine with one another only in the proportions of their combining weights or small multiples of them by assuming that any mass of each element is made up of a very large number of extremely small particles called atoms; that these are exactly alike in every respect; that they are not subdivided by chemical processes; that there are as many kinds of atoms as there are ele- ments; that the atoms associate with one another, usually in small numbers, forming a new order of distinct particles called molecules; that pure chemical substances are made up of only one kind of mole- cules, while mixtures contain two or more kinds; and that the molecules of elementary substances consist of atoms of the same kind, those of compound substances of atoms of different kinds. These assumptions in regard to atoms and molecules have now been confirmed in so many ways that they are no longer hypothetical. By the above statement chemical substances are implicitly defined from the molecular standpoint as pure substances which contain only a single kind of molecule. Thus, the pure substance liquid water contains the two chemical substances whose molecules are H 2 O and H 4 O 2 , these being always present in definite proportions at any definite temperature 4 THE COMPOSITION OF SUBSTANCES and pressure, since equilibrium is instantaneously established between them ; but the pure substance water-vapor, which contains only mole- cules of the form H 2 O, is a single chemical substance. Pure substances which, like water, water-vapor, and ice, are converted into each other by changes of pressure and temperature, are commonly spoken of as the same substance; but they may consist of different chemical sub- stances, as has been just illustrated. The relative weights of the atoms of the various elements and of the molecules of the various substances are called their atomic weights and molecular weights, respectively; and as a standard of reference, the weight of the oxygen atom taken as 16 is adopted. The atomic theory evidently requires that the weights of elements that combine with one another be proportional to the weights of their atoms or to small multiples of those weights; in other words, that the atomic weights be equal to the combining weights or to small multiples of them. Which multiple of the combining weight is the atomic weight cannot be derived from the elementary composition of substances. It can, however, be derived from other properties, with the aid of certain other principles. 6. Chemical Formulas, Formula- Weights, and Equivalent- Weights. In order to express the gravimetric composition of compounds, the symbols of the elements are considered to represent their atomic weights and are written in sequence with such integers as subscripts as will make the resulting formula express the proportions by weight of the elements in the compound. The formula is commonly so written as to represent also the number of atoms of each element present in the molecule, when this has been determined (by any of the methods described in later articles). The formula represents, in addition, a definite weight of tho substance, namely, the weight in grams which is equal to the sum of the numbers represented by the symbols of the elements in the formulas. This weight is called the formula-weight of the substance. Thus the formula HNO 3 denotes 1.008 -f- 14.01 -f (3 X 16.00) or 63.02 grams of nitric acid. Those weights of various substances which enter into chemical reactions with one another are called equivalent weights. Adopting one formula-weight or 1.008 grams of the element hydrogen as the THE COMPOSITION OF SUBSTANCES 5 standard of reference, the equivalent-weight or one equivalent of any substance is denned to be that weight of it which reacts with this standard weight of hydrogen, or with that weight of any other substance which itself reacts with this standard weight of hydrogen. Thus the equivalent-weight of each of the following substances is that fraction of its formula-weight which is indicated by the coeffi- cient preceding the formula: C1 2 ; |O 2 ; lAg; Zn; Bi; INaOH; Ba(OH) 2 ; H 2 SO 4 ; H 3 P0 4 ; JAlCi,; K 4 Fe(CN) 6 . The equiva- lent-weight of a substance may have different values depending on whether it is considered with reference to a reaction of metathesis or to one of oxidation and reduction. Thus, the metathetical equiva- lent of ferric chloride is ^FeCL,, but its oxidation-equivalent (with respect to its conversion to ferrous chloride) is !FeCl 3 ; the metathet- ical equivalent of potassium chlorate is 1KC1O 3 , but its oxidation- equivalent (with reference to its reduction to KOI) is CHAPTER II PROPERTIES OF GASES RELATED TO MOLECULAR COMPOSITION PRESSURE-VOLUME RELATIONS OF GASES 7. The Laws of Perfect Gases are the limiting laws to which gases conform more and more closely as their pressure approaches zero. These laws express fairly closely (within less than one percent) the behavior of gases up to pressures not much greater than one atmosphere, provided they are far removed from their temperatures of condensation. 8. Boyle's Law. At any definite temperature the pressure p of a definite weight m of any perfect gas is inversely proportional to its volume v. Or, since density d is defined to be the ratio m/v of the weight of a substance to its volume, the pressure of any perfect gas at any definite temperature is directly proportional to its density. 9. Gay-Lussac's Law of Temperature-Effect. When different perfect gases are heated or cooled from a definite initial to a definite final temperature, their pressure-volume products p v change by the same fractional amount. This principle and its degree of accuracy when applied to gases at moderate pressures are illustrated by the following data: When 1000 ccm. of gas at and a pressure of 1000 mm. of mercury are heated to 100 at constant pressure, the volume increases by 367 ccm. with nitrogen, 366 ccm. with hydrogen, and 374 ccm. with carbon dioxide; when heated at constant volume, the pressure in- creases by 367, 366, and 373 mm., with the three gases, respectively. 10. Definition of Absolute Temperature. Absolute temperature T is so defined that the pressure-volume product p v of a perfect gas is directly proportional to it. Since the pressure-volume product decreases -^ of its value at for each degree that the tempera- ture decreases, the absolute zero is at 273.1 ; and the absolute temperature T is equal to t -f- 273.1 (approximately 273*), where i is the ordinary centigrade temperature. *Approximate values of numerical constants which it is well to remember are printed in black-face type. These values will be given so as to be accurate to 0.1%. 6 PRESSURE-VOLUME RELATIONS OF GASES 7 11. Expression of the Physical Laws of Perfect Gases. The laws of Boyle and Gay-Lussac, the definition of absolute temperature, and the obvious proportionality between the value of the pressure-volume product and the weight ra of the gas are all expressed by the equation p i} = m R T, in which R is a constant, evidently representing the value of pv/T for one gram of the gas, which has different values for different gases. Pro 6. 1. From the fact that the density of oxygen is 0.001429 g. per ccm. at and 1 atm. calculate the volume in liters of 32 g. of it at 20 and 1 atm. 12. Law of Combining Volumes and the Principle of Avogairo. The foregoing physical laws acquire an important chemical significance by reason of the law of combining volumes, which may be stated as follows : Those quantities of perfect gases that are involved in chem- ical reactions with one another have at the same temperature and pressure volumes which are equal or small multiples of one another. Thus the quantities of hydrogen and of oxygen which unite with each other to form water have volumes whose ratio approaches the exact value 2 :1 as the pressure of the gases approaches zero. In other words, those quantities of different substances which as perfect gases have the same value of the product pv/T are equal to, or are small multiples of, the quantities which are involved in chemical reactions with one an- other. Now, since according to the molecular theory substances react by molecules, these quantities of different perfect gases which have the same value of pv/T must contain either an equal number of molecules or small multiples of an equal number. These considerations suggest a simpler hypothesis, known as the principle of Avogadro, which may be stated as follows: Those quanti- ties of any perfect gases which have the same volume at the same tem- perature and pressure, and therefore the same value of pv/T at any temperature and pressure, contain an equal number of molecules. This principle, originally hypothetical, has now been so fully veri- fied that it has become one of the fundamental laws of chemistry. Prob. 2. From the principle of Avogadro show that the densities of perfect gases at the same temperature and pressure are proportional to their molecular weights. 13. Empirical Definition of Molecular Weight and of Mol. The principle of Avogadro evidently enables the relative weights of the 8 MOLAL PROPERTIES OF GASES molecules of different gaseous substances to be determined. To express these relative weights more definitely, the ratio of the weight of the molecule of any substance to the weight of the molecule of oxygen taken as 32 is commonly considered, this ratio being called the molecu- lar weight M of the substance. The number 32 is adopted as the refer- ence quantity of oxygen, since (as shown in Prob. 13) it corresponds to the adoption of 16 as the atomic weight of oxygen. The so-defined molecular weight of a substance may evidently be experimentally determined by finding the number of grams of it which have that value of pv/T which 32 grams of oxygen have, when both substances are in the state of perfect gases. This number of grams is called one mol of the substance a unit-quantity* which has great im- portance in chemical considerations, both because it is directly related to molecular weight and because it makes possible a general expression of the laws of perfect gases, as shown in the following Article. 14. General Expression of the Laws of Perfect Gases. Represent- ing by R the constant value of pv/T for one mol and by N the number of mols of the gas present, the laws of perfect gases may be expressed in a general form by the following equation, hereafter called the perfect-gas equation: p v = N R T. The numerical value of the gas-constant R depends on the units in which the pressure, volume, and temperature are expressed. In sci- entific work, temperature is always expressed in centigrade degrees (here on the absolute scale) ; volume is ordinarily expressed in cubic centimeters or liters; and pressure in dynes per square centimeter, millimeters of mercury, or atmospheres. One dyne is a force of such magnitude that when it acts on a freely moving mass of one gram it increases its velocity each second by one centimeter per second; and the pressure of one dyne per square centimeter is called one ~bar, 10 8 bars being called one megdbar.^ One atmosphere is a pressure equal to that exerted by a column of mercury 76 cm. in height at at the sea-level in a latitude of 45. The value of the gas-constant R when *This unit-quantity is called by some authors one gram-molecular weight or one gram-molecule of the substance. tThe megabar is a unit of the same general magnitude as the atmosphere, and should replace it in scientific work ; but unfortunately it is not yet generally employed. PRESSURE-VOLUME RELATIONS OF GASES 9 the pressure is in atmospheres and the volume is in cubic centimeters is 82.07 (approximately 82). Prob. 3. Calculate the value of the atmosphere in dynes per sq. cm. and in megabars. The density of mercury at is 13.60. The force of gravity acting on any freely moving body increases its velocity each sec- ond by g centimeters per second. The value of g at the sea-level in a latitude of 45 is 980.6 (approximately 9'80) centimeters per second. Prob. 4. Calculate precisely the value of the gas-constant R when the pressure is in atmospheres and the volume in cubic centimeters, from the fact that one gram of oxygen has at and 0.1 atm. a volume of 7005 ccin. Assume in this problem and the following ones that the sub- stance behaves as a perfect gas. Pro 6. 5. Seven grams of a certain gas have a volume of 6.35 1. at 20 and 720 mm. How many grams make one mol? Pro 6. 6. What is the volume occupied by 12 g. of ether vapor (C 4 H 10 O) at 80 and 600 mm.? What is the density? What is the ratio of the density to that of oxygen at the same temperature and pressure? Pro&. 7. How many grams of iron must be taken to produce by its action on sulphuric acid one liter of hydrogen (H 2 ) at 27 and 1 atm.? 15. Dalton's Law of Partial Pressures. In a mixture of perfect gases each chemical substance (denned as in Art. 5 to be a substance consisting of a particular kind of molecule) has the same pressure as it would if it were alone present in the volume occupied by the mixture. The pressures of the separate substances are called partial pressures. Partial pressures have been experimentally determined in certain cases. Thus, when a platinum vessel containing a mixture of hydrogen and nitrogen at a high temperature is immersed in an atmosphere of hydrogen, hydrogen passes through the platinum walls until its pres- sures within and without the vessel become equal ; the difference then observed between the total pressure of the mixture within the vessel and the pressure of the hydrogen outside is due to the nitrogen, which does not pass through the wall ; and this difference is equal to its partial pressure. Walls of this kind, permeable for only one of the substances present in a mixture, are called semipermeable walls. The principal evidence in support of Dalton's law is, however, the fact that the total pressure of a mixture of perfect gases is actually equal to the sum of the partial pressures calculated by the law. By the molal composition of a mixture is meant its composition expressed in terms of the number of mols of each of the substances. 10 MOLAL PROPERTIES OF GASES The ratio of the number of mols of any one substance to the total number of mols in the mixture is called the mol- fraction x of that substance; that is, a? .= N 1 /(N 1 + N 2 + . .) The partial pressure of each substance in a mixture of perfect gases is evidently equal to the product of its mol-f raction by the total pressure of the mixture. Pure dry air contains 21.0 mol-percent of oxygen (O 2 ), 78.1 mol-per- cent of nitrogen (N 2 ), and 0.9 mol-percent of argon (A). The corre- sponding value of one mol of air is 29.00 grams, in agreement with the value directly derived from density measurements. With the aid of this value the perfect-gas equation may be applied directly to air. Prol. 8. a. From the composition of air given above, calculate the partial pressures of the separate substances in millimeters of mercury when the barometer stands at 1 atm. 6. Calculate the number of grams which make one mol of air. c. Calculate the percent by weight of oxygen in air. Prob. 9. A Bessemer converter is charged with 10,000 kilos of iron containing 3% carbon. How many cubic meters of air at 27 and 1 atm. are needed for the combustion of all the carbon, assuming one-third to burn to CO 2 and two-thirds to CO? What are the partial pressures of the gases evolved? Prob. 10. At 50 and 500 mm., nitrogen tetroxide has a density 2.15 times that of air under the same conditions. What percentage dissociation according to the equation N 2 O 4 = 2NO 2 must be assumed to explain this value? What is the partial pressure of each gas (NO, and N 2 O 4 ) in the mixture? 16. Determination of Molecular Weights. The experimental deter- mination of the molecular weight of a gaseous substance consists in measuring the volume of a weighed quantity of the substance at an observed pressure and temperature. From these quantities (v, m, p, and T) the molecular weight of the substance is calculated with the aid of the perfect-gas equation. Prob. 11. Determination of Vapor-Density by Hofmann's Method. 0.1035 g. of a volatile liquid is introduced into the vacuum above a mer- cury column in a graduated tube standing in an open vessel of mercury. The tube is entirely surrounded by a jacket through which steam at 100 is passed. The mercury column falls till it stands 260 mm. above the mercury-level in the vessel below, and the volume of the completely vaporized liquid is observed to be 63.0 ccm. The barometric pressure at the time is 752 mm. At 100 the density of mercury is 13.35, and its vapor- pressure is negligible. Calculate the density and the molecular weight of the vapor. PRESSURE-VOLUME RELATIONS OF GASES 11 Since gases at atmospheric pressure conform to the perfect-gas laws only approximately, and since gaseous densities are not commonly determined with so great accuracy as the composition of substances, the exact value of the molecular weight of a compound is usually de- rived from the analytical data, the density being employed only to determine what multiple or submultiple of the value so derived is in accordance with Avogadro's principle. Prol. 12. a. A certain oxide contains exactly 72.73% of oxygen. What does this show in regard to its molecular weight? 6. At and 1 atm., one liter of this (gaseous) oxide is found to weigh exactly 1.977 g. What is the molecular weight of the oxide corresponding to this datum? c. What is the exact molecular weight derived by considering the data relating both to the composition and density? d. What do these two values of the molecular weight show as to the percentage deviation of the density from that required by the perfect-gas equation? 17. Derivation of the Atomic Weights of Elements and of the Molecular Composition of Compounds. The exact values of atomic weights are based on analytical determinations of the combining weights, as described in Art. 4. The multiple of the combining weight adopted as the atomic weight of any element is derived by finding the smallest weight of the element contained in one molecular weight of any of its gaseous compounds. This is the true atomic weight only in case some one of the compounds studied contains in its molecule a single atom of the element ; and the adopted atomic weight is therefore strictly only a maximum value, of which the true atomic weight may be a submultiple. The probability that the true atomic weight has been found evidently increases with the number of the gaseous compounds whose molecular weights have been determined. The multiples of the combining weights adopted as the atomic- weights have, however, not been derived solely from molecular-weight determinations. From the laws relating to certain other properties, such as the heat-capacities of gaseous and solid substances (considered in Arts. 118 and 120), independent values of the atomic weights have been obtained, which confirm and extend those derived from molecular weights. The molecular composition of a gaseous substance (that is, the number and nature of the atoms in its molecule) can evidently be derived from its molecular weight, its composition by weight, and the atomic weights of the elements contained in it. The chemical formulas 12 HOLAL PROPERTIES OF GASES of substances whose molecular weights in the gaseous state are known are ordinarily so written as to express this molecular composition. Such formulas are called molecular formulas. Prob. 13. Certain oxygen compounds have molecular weights M (referred to that of oxygen as 32) and percentages of oxygen x as fol- lows : Sulphur trioxide, M = 80.07, x 59.95 ; water, M = 18.02, x = 88.80 ; carbon dioxide, M 44.00, x = 72.73. a. Find the weight of oxygen contained in one molecular weight of each of these oxides. &. State how these values show that the atomic weight of the element oxygen is one-half of the assumed molecular weight of oxygen gas, and therefore that the molecule of oxygen gas consists of two atoms. Prob. Hi. a. Calculate the combining weight of the element contained in the oxide whose composition was given in Prob. 12a. &. What conclu- sion as to its atomic weight can be drawn from this combining weight and the density given in Prob. 12 &? c. What conclusion can be drawn as to the molecular formula of the oxide? Pro 6. 15. a. The chloride of a certain element is found by analysis to contain 52.50% Cl, whose atomic weight is 35.46; and it is found to have at 150 a vapor-density 4.71 as great as that of air. What conclusion can be drawn from these facts as to the exact atomic weight of the ele- ment? &. The hydride of the same element is a gas which contains 5.91% of hydrogen ; and it is found to be produced without change in volume when hydrogen (H 2 ), whose atomic weight is 1.008, is passed over the solid elementary substance. What conclusion can now be drawn as to the atomic weight of the element? c. What are the simplest molecular formulas of the hydride and of the chloride consistent with these con- clusions? Pro6. 16. A certain hydrocarbon is composed of 92.25% of carbon and 7.75% of hydrogen, whose atomic weights are 12.00 and 1.008. respectively. Its density in the form of vapor at 100 and 1 atmosphere is 2.47 times as great as that of oxygen under the same conditions. Calculate its exact molecular weight, and derive its molecular formula. The molecules of elementary substances may consist of single atoms or of two or more atoms. Thus the molecular formulas of some of those whose density in the gaseous state has been determined are: H 2 , N 2 , 2 , F 2 , G1 2 , Br 2 , I 2 , P 4 , As 4 , He, A, Hg, Cd, Zn. At very high temperatures some of these have been shown to dissociate into simpler molecules; thus above 1500 the molecule of iodine consists of a single atom. A knowledge of the molecular formulas of substances is of impor- tance principally because the chemical relationships of different sub- stances are far more clearly brought out by such formulas than by the simpler ones expressing merely composition by weight. The PRESSURE-VOLUME RELATIONS OF GASES 13 structure theory, which underlies the science of organic chemistry, is based upon a knowledge of the molecular weights of substances. 18. Deviations from the Perfect-Gas Laws at Moderate Pressures. Prob. 17. Calculate accurately the percentage deviations at and 1 atin. of the volumes of one mol of oxygen and of one mol of nitrous oxide from that of a perfect gasj At and 1 atin. the density of oxygen is 0.001429, and that of nitrous oxide (N 2 O) is 0.001978.x The deviations from the perfect-gas law, so long as they do not ex- ceed a few percent, are accurately expressed by the equation p v = N E ' T (1 -f- cxp), in which ex is an empirical coefficient which can be determined for each gas at each temperature from a measurement of p v at some one pressure. The following table shows the values of 100 <*, representing the percentage deviation of p v from N R T at one atmos- phere, for various gases at 0, together with their condensation-tem- peratures. Formula of gas. He H 2 N 2 NO CO 2 NH 8 SO, 100 _j_0.06 +0.05 0.04 0.12 0.68 1.52 2.38 Condensation-temp. 269 253 196 151 78 34 10 3 19. Pressure- Volume Relations of Gases at High Pressures. In Figure 1 are plotted the values of p v/T (in arbitrary units) as ordi- 16 V V 40 80 I2O l6o 2OO 240 28O 33O Pressure PIGDBB i 14 MOLAL PROPERTIES OF GASES nates against the values of the pressure (in meters of mercury) as abscissas for one mol of hydrogen at 60, of nitrogen at 60, of carbon dioxide at 60 and at 40, and of a perfect gas (marked P. G. in the figure). At temperatures between and 60, helium (He) and neon (Ne) have curves similar to that of hydrogen (Hj) ; oxygen (O 2 ), carbon monoxide (CO), and nitric oxide (NO) have curves similar to that of nitrogen; and nitrous oxide (N 2 O), ammonia (NH 8 ), and ethylene (C 2 H 4 ) have curves similar to those of carbon dioxide (CO a ). ProT). 18. Summarize the general conclusions in regard to the pressure-volume relations of gases that can be drawn from the curves of Figure 1, the statements in the preceding text, and the condensation- temperatures given in Art. 18. Pro6. 19. Estimate with the aid of the figure the ratio of the volume of one mol of each of the gases at 40 m. (or 53 atm.) to that which one mol of a perfect gas would have at the same temperature and pressure. CHAPTER III PROPERTIES OF SOLUTIONS RELATED TO MOLECULAR COMPOSITION VAPORY-PRESSURE AND BOILING-POINT IN GENERAL 20. Vapor-Pressure. A liquid in contact with a vacuous space vaporizes until the pressure of its vapor in that space attains a perfectly definite value which is determined by the nature of the liquid and by the temperature. If, on the other hand, vapor having a pressure greater than this definite value is brought into contact with the liquid, condensation occurs until the pressure of the vapor falls to that value. In other words, for a given liquid at a given temperature there is only one pressure which its vapor can have and exist in equilibrium with that liquid. This pressure is called the vapor-pressure of the liquid. This is to be distinguished from the pressure of the vapor, which when not in contact with the liquid may have any value from zero up to one somewhat exceeding the vapor-pressure. Solids likewise have definite vapor-pressures, which with certain substances (like iodine) are appreciable even at room temperature. The vapor-pressure of a liquid or solid substance increases rapidly with increasing temperature, as illustrated by the data of Prob. 3 below. When a liquid is in contact with a space containing a gas (for example, when water is in contact with an air space), approximately the same quantity of the liquid vaporizes as if the gas were not present, provided the gas is only slightly soluble in the liquid, and provided its pressure is not much greater than one atmosphere. When the gas is readily soluble in the liquid, or when its pressure is large, considerable deviations from this principle may result. Pro6. 1. At 28 and 1 atm. 25 ccm. of dry air are collected over water, whose vapor-pressure at 28 is 28 mm. a. What is the pressure if the volume is still 25 ccm.? ft. What is the volume if the pressure is still 1 atm.? Pro6. 2. Mr-Bubbling Method of Determining Vapor-Pressure. 2000 ccm. of dry air at 15 and 760 mm. are bubbled through bulbs con- taining a known weight of carbon bisulphide (CS 2 ) at 15, and the mix- 15 16 MOLAL PROPERTIES OF SOLUTIONS ture of air and bisulphide vapor is allowed to escape into the air at a pressure of 760 mm. By reweighing the bulbs, 3.011 g. of the bisulphide are found to have vaporized. What is the vapor-pressure of carbon bisul- phide at 15 ? Steam-Distillation of Liquids Insoluble in Water. Prob. 3. Steam is bubbled through chlorbenzene (C 6 H 8 C1) in a distill- ing flask ; and the vapors, which escape under a barometric pressure of 1 atm., are condensed as a distillate. The steam partially condenses in the distilling flask, and brings the mixture of water and chlorbenzene (which are not appreciably soluble in one another) to that temperature where equilibrium prevails between each liquid and its vapor. Determine this temperature and the molal composition of the distillate with the aid of a plot of the following data, which represent the vapor-pressures of the pure substances at various temperatures : 70 80 90 100 Water .... 234 355 526 760 mm. Chlorbenzene ... 98 145 208 292 mm. Prob. 4- A current of steam is passed at atmospheric pressure through a mixture of water and nitrobenzene (C 6 H 6 NO 2 ). Calculate the tempera- ture of the distilling mixture and the percentage by weight of nitroben- zene in the distillate from the following data: the vapor-pressure of water at 100 is 760 mm. and changes by 3.58% per degree ; that of nitro- benzene at 100 is 20.9 mm. and changes by 5.0% per degree. 21. Relation of Boiling-Point to Vapor-Pressure. The 'boiling- point of a liquid is the temperature at which it is in equilibrium with its vapor when both are subjected to any definite external pressure. In other words, it is the temperature at which the vapor-pressure, which increases as the temperature rises, becomes equal to the external pres- sure. When this temperature is exceeded by an infinitesimal amount, assuming that there is no superheating, the vapor forms throughout the mass of the liquid (not merely at its free surface), giving rise to the familiar phenomenon of boiling. Prob. 5. The vapor-pressure of water at 100 increases 27.2 mm. per degree. What variation of its boiling-point corresponds to a variation of the barometric pressure from 730 to 790 mm. ? 22. Change of Vapor-Pressure with Temperature. The Clapeyron Equation. From the laws of thermodynamics it can be shown that the rate at which the vapor-pressure p of a liquid or solid at the absolute temperature T increases with the temperature is expressed exactly by the Clapeyron equation : dp AH dT = = (v-v )T VAPOR-PRESSURE AND BOILING-POINT 17 In this equation v represents the volume of one mol of the (.saturated) vapor at the pressure p and temperature T, v is the volume of one mol of the liquid or solid substance under the same conditions, and A H is the increase in the heat-content of the substance, equal to the heat withdrawn from the surroundings, when one mol of it vaporizes at the temperature T. The quantity Ajff is called the molal heat of vaporization. Since at the boiling-point of a liquid its vapor-pressure is equal to the external pressure upon the liquid and vapor, the Clapeyron equation also expresses (more clearly in the inverted form) the change of boiling-point with the external pressure. In numerical applications of this equation, the energy quantities &H and (v -- v^)dp must be expressed in corresponding units. The latter quantity will be in ergs when the volumes are in cubic centi- meters and the pressure is in dynes per square centimeter. Three units of energy are commonly used in scientific work the erg, the joule, and the calorie. The erg is the work done when a force of one dyne is displaced through one centimeter. The joule is a decimal multiple of the erg; namely, one joule equals 10 7 ergs. The mean calorie is one one-hundredth part of the heat required to raise one gram of water from to 100. This is identical within 0.02% with the ordinary calorie, which is the heat required to raise one gram of water from 15 to 16. One calorie (1 cal.) is equal to 4.182 (approx- imately 4.18) X 1 7 er gs> this value being the so-called mechanical equivalent of heat. Pro 6. 6. a. Calculate w'ith the aid of the Clapeyron equation the volume of one mol of saturated water-vapor at 100 from the following data: At 100 the vapor-pressure of water increases 27.2 mm. per de- gree, the heat of vaporization of one gram of it is 537 cal., and the specific volume (i.e., the volume of one gram) of liquid water is 1.043. &. Calculate by th$ perfect-gas equation the volume of one mol of satu- rated water-vapor at 100. c. By comparing the two values of this quantity obtained in a and &, determine the percentage error made In assuming that the saturated vapor conforms to the perfect-gas law. Note. The molal volume of the vapor of alcohol saturated at 78.3 (where the pressure is 760 mm.) is 3.6% smaller, and that of the vapor of ether saturated at 12.9 (where the pressure is 330 mm.) is 3.5% smaller, than the molal volume calculated by the perfect-gas equation. Pro 6. 7. a. Derive from the Clapeyron equation the simpler, but less exact, expression - -=-2- by assuming that the volume of 18 MOLAL PROPERTIES OF SOLUTIONS the liquid is negligible in comparison with that of the vapor and that the vapor conforms to the perfect-gas laws. I). What percentage error in dp /d,T would result from each of these assumptions if it be calculated for water at 100 by this approximate equation with the aid of the data of Prob. Gal c. In numerical applications of this equation R and &H must be expressed in corresponding units. Show from the data of Prob. 4, Art. 14, that the value of R is 8.316 (approximately 8.32) in joules per degree, and 1.9885 (approximately 1.99) in calories per degree. Pro&. 8. a. Integrate the approximate Clapeyron equation so as to obtain a relation between the vapor-pressures p t and p 2 at two different temperatures T, and T v Assume that the heat of vaporization does not vary between the two temperatures. &. Calculate by the equation so obtained the boiling-point of water at 92 mm., and its vapor-pressure at 75. Compare these calculated values with the actual ones given above, c. State what inexact assumptions are involved in the equation which would account for the divergence. 23. Solutions. A solution is a physically homogeneous mixture of two or more chemical substances; that is, one which has no larger aggregates than the molecules themselves. Solutions thus defined may be gaseous, liquid, or solid ; but only liquid solutions will be here con- sidered. When one substance is present in large proportion it is called the solvent, and any substance present in small proportion is called a solute. The composition of solutions is often expressed in terms of the mol-fractions of the substances, defined as in Art. 15. In considering the equilibrium of solutions with the vapor or with the solid solvent, the term phase is conveniently employed. The phases of a system are its physically homogeneous parts, separated from one another by physical boundaries. Thus any gaseous mixture or any solution or any solid substance f orms a single phase. A system may consist of any number of such phases. Thus a solution in contact with its vapor, or with the solid solvent, or with the solid solute, is an example of a two-phase system. A .solution in contact both with the vapor and the solid solvent is a three-phase system. With reference to the proportions in which the substances are present, two groups of solutions may be distinguished : dilute solutions, those in which the mol-fraction of the solute is small (not greater than 0.01 or 0.02) ; and concentrated solutions, those in which each sub- stance is present in considerable proportion. There is, of course, no sharp line of demarcation between these two groups of solutions. VAPOR-PRESSURE AXD BOILING-POIXT 11) Some types of concentrated solutions and all dilute solutions con- form approximately more closely as the mol-fraction of the solute approaches zero to certain laws, which, in analogy with the laws of perfect gases, may be called the laws of perfect solutions. The funda- mental laws of perfect solutions are the vapor-pressure laws of Raoult and Henry, and the corresponding laws of distribution between phases and of the osmotic pressure of solutions. To consideration of these laws the rest of this chapter is mainly devoted. VAPOR-PRESSURE AND BOILING-POINT OF DILUTE SOLUTIONS 24. Raoult 's Law of Vapor-Pressure Lowering. The addition of a solute to a solvent causes at any temperature a fractional lower- ing of the vapor-pressure of the solvent equal to the mol-fraction (#) of the solute. That is : Po p _ N Po r No + N " where p is the vapor-pressure of the pure solvenvand p is its vapor- pressure above a solution consisting of N mols of solute and N mols of solvent.* In this expression N is equal to the weight m of the solvent divided by its molecular weight M^ in the vapor, and N is equal to the weight m of the solute divided by its molecular weight M in the solution. The value of M is ordinarily that corresponding to the molecular formula of the solvent; for, as stated in Art. 9, the molecular formula is commonly so written as to represent the molecular weight of the substance in the state of a perfect gas. In the case of very dilute solutions the mol-fraction N/(N + N) may evidently be replaced by the mol-ratio N /N without causing appreciable error. When the solute is so volatile as to have an appreciable vapor pressure, the quantity p in the Raoult equation denotes, of course, the partial vapor-pressure of the solvent, not the total vapor-pressure of the solution. Although Raoult's law is exact only in the case of very dilute solutions, it holds true approximately up to moderate concentrations in all cases, and up 'to very high concentrations in some cases, as will be described in Art. 30. *Throughout this chapter quantities referring to the solvent are represented by letters with the subscript zero, those referring to the solution or solute by letters without subscripts. 20 MOLAL PROPERTIES OF SOLUTIONS Determination of Molecular Weights and Molecular Composition. Pro&. 9. At 30 the vapor-pressure of ethyl alcohol (C 2 H 5 OH) is 78.0 mm., and that of an alcohol solution containing 5% of a non- volatile substance is 75.0 mm. What is the molecular weight of the substance? Pro ~b. 10. The experiment described in Prob. 2, Art. 20, was repeated, using in place of pure carbon bisulphide an 8.00% solution of sulphur in carbon bisulphide. 2.902 g. of carbon bisulphide were found to have vaporized. Calculate the molecular weight of the sulphur, and find its molecular formula. Raoult's law may also be stated in the following simple form, which indicates more clearly its real significance: the vapor-pressure (p) of the solvent in a perfect solution is proportional to its mol-fraction O ) ; that is, representing by p the vapor-pressure of the pure solvent, P = Po a? . ProT). 11. a. Show that the two statements of Raoult's law are mathe- matically equivalent. 6. Show that the second statement of the law requires that the proportionality-factor be the vapor-pressure of the pure solvent, as is assumed in the mathematical expression of it. Raoult's law relates fundamentally to the distribution between the liquid phase and vapor phase of the chemical substance (Art. 5) whose partial pressure in the vapor is under consideration. In other words, from the molecular standpoint, it relates to the distribution of the kind of molecules which give rise to this partial pressure. It shows that the number of these molecules which are present in unit-volume of the Vapor when equilibrium has been reached is proportional to the ratio in the liquid of the number of this kind of molecule to the total number of molecules of all kinds. Moreover, since this molecule-ratio is unity for a solvent which consists solely of the kind of molecule under con- sideration, the proportionality-factor in the expression of the law is the vapor-pressure of/ the pure solvent. Raoult's law in the forms considered above presupposes that the vapor conforms to the perfect-gas law. It can be shown that the devia- tion of the vapor from this law, when expressed (as in Art. 18) by the equation p v = R T (1 + ap), can be substantially corrected for, so long as neither the mol-fraction of the solute nor the correction term OA#A, PB = POB^B, . The method commonly employed for determining the partial vapor- pressures of the components of solutions at any definite temperature is to distil off a small fraction from a large volume of the solution, adjusting the pressure on the liquid so that it boils at this temperature. The composition of the distillate is then determined by chemical analysis or by the measurement of some physical property, such as density; and from this composition and the pressure under which the distillation took place the partial vapor-pressures are calculated, as illustrated in the following problem. \j Prob. 30. A solution of two substances A and B containing A T A mols of A and A 7 B mols of B boils at the temperature T when a pressure of p is exerted upon it. The first portion of distillate consists of tf A ' mols of A and r\ _ N RT , c is the molal concentration, and R is the gas-constant. The osmotic pressure of concentrated solutions to which Raoult's law does not apply cannot be calculated from the composition of the solution. It may, however, be determined not only by direct measure- ment, but also from vapor-pressure measurements, as shown in Art. 37. REVIEW PROBLEMS 4lB Prod. 51. Derive the two equations given in the preceding text. Prob. 52. The lower end of a vertical tube is closed with a semi- permeable wall and is dipped just beneath the surface of a pure solvent. A 0.1 molal solution of cane-sugar (C^H^O^) of density 1.014 is poured into the tube until the hydrostatic pressure at the semipermeable wall is sufficient to prevent water from entering the solution. The temper- ature is 4. What is the height of the column in meters? REVIEW OF THE MOLAL PROPERTIES OF SOLUTIONS 39. Review Problems. Prod. 53. a. Summarize the equations expressing for dilute solutions the approximate relations between molal composition and (1) vapor- pressure, (2) boiling-point, (3) freezing-point, (4) osmotic pressure. State explicitly what each symbol signifies. Pro'b. 54- Summarize the corresponding equations which hold true for perfect solutions of any concentration. Prod. 55. a. Calculate by the laws of dilute solutions, using the data given below, the fractional vapor-pressure-lowering, the boiling- point-raising, the freezing-point-lowering, and the osmotic pressure at 5.5 of a solution A containing 0.1 mol of a nonvolatile solute S in 1000 g. benzene. d. Calculate by the laws of perfect concentrated solutions the values of the same quantities for a solution B containing 2 mols of the solute S in 1000 g. benzene. Calculate the ratio of each of these values to the corresponding one for solution A. (Note that the value of this ratio would be 20.0, if the equations for dilute solutions were applicable to solution B.) c. If the solute in solution B were volatile and at the boiling-point of the solution had a partial pressure of 20 mm., what would be the boiling-point of the solution? Data: The heat of fusion of one gram of benzene at its freezing-point, 5.5, is 30.2 cal. The heat of vaporization of one gram of benzene at its boiling-point, 80, is 93.0 cal. The density at the freezing-point is 0.895. Prod. 56. a. Carbon bisulphide boils at 46 at 1 atm. Its molal heat of vaporization at this temperature is 6430 cal. What is its boiling-point constant? d. A solution of 15.5 g. phosphorus (at. wt.. 31.0) in 1000 g. CS 2 boils 0.300 higher than pure carbon bisulphide. What is the molecular weight and what is the molecular formula of phosphorus in this solvent? Prod. 57. If 127 g. iodine (at. wt., 127) were added to the solution of Prob. 566, how much higher than the boiling-point of pure carbon bisulphide would that of the solution be : a, in case the iodine remained uncombined in the form of I 2 ; 6, in case it all combined with the phos- phorus forming P 4 I 8 ; c, forming P 2 I 4 ; d, forming PI 2 ? 41c MOLAL PROPERTIES OF SOLUTIONS Prol). 58. Human blood freezes at 0.56. What is its osmotic pressure at 37? Prol). 59. At 25 the distribution-ratio of Br 2 between carbon tetra- chloride and water is 38 ; and the pressure of bromine above a 0.05 molal solution of Br 2 in water is 50 mm. If one liter of this solution be shaken with 50 ccm. carbon tetrachloride, what will be the pressure of the bromine over the carbon-tetra chloride phase? (Bromine exists in all three phases only as Br 2 .) Pro 6. 60. When ethyl acetate and water are shaken together at 20, two liquid phases result, one phase containing 1.75 mol-percent. and the other 86.8 mol-percent ethyl acetate, a. Show that, even though Raoult's law may hold for one component in one phase, it cannot hold for the same component in the other phase. o. Calculate the partial vapor-pressures of ethyl acetate and water over the mixture, assuming, since the solutions are moderately dilute, that Raoult's law applies to the solvent in each phase. At 20 the vapor-pressure of pure ethyl acetate is 73 mm., and that of water is 17.4 mm. c. Calculate the partial vapor-pressures of the two components in a solution containing 1.00 mol-percent of ethyl acetate. Prol). 61. Aniline (C 6 H 3 NH 2 ) is distilled with steam at 1 atm. The vapor-pressure of aniline is 46 mm. at 100 and 40 mm. at 97. Water and aniline have limited solubilities in each other; one phase containing at 100 1.5 mol-percent of aniline, and the other 68 mol-percent of aniline. Find the boiling-point of the mixture, and the number of grams of aniline distilling over with each gram of water. Assume that the solu- bilities do not change appreciably within the small temperature interval, and that the vapor-pressure of the aniline in the 68 mol-percent solution is lowered only half as much as Raoult's law requires. (This last assump- tion is a rough estimate based on the considerations that the partial vapor-pressure of the aniline in the 68 mol-percent solution is less than that of pure aniline and that, in . accordance with statement (4) on page 34, it is greater than that required by Raoult's law. In the absence of more definite knowledge it is obviously best to assume that the actual vapor-pressure lies midway between these two limiting values.) Prol). 62. At 1 atm. pure nitric acid has a boiling-point of 86. A solution of nitric acid and water of the composition HNO 8 -f 1.6 H 2 O distils at 1 atm. at a constant temperature of 121. Make a diagram showing the character of the boiling-point-composition curve for this pair of substances. Draw in on the diagram a curve representing in a general way the composition of the vapor in equilibrium with any solu- tion at its boiling-point. State what products would finally result as distillate and residue from the fractionation of the three solutions, HN0 8 + H 2 0, HNO, -f 1.6 H 2 O, HNO 8 + 3 H 2 O. Prob. 63. Upon partial distillation at constant pressure a solution of two components A and B containing 20% A, having a boiling-point of REVIEW PROBLEMS 41o 80, yields a residue containing 15% A; and a mixture containing 75% A, having a boiling-point of 60, yields a distillate containing 70% A. Draw a diagram which will show in a general way the character of the liquid-composition and of the vapor-composition curve. Predict what products would finally result as distillate and residue from the fraction- ation of the 20% and of the 75% solution. CHAPTER IV PROPERTIES OF SOLUTIONS RELATED TO IONIC COMPOSITION EFFECT OF IONIZATION ON THE MOLAL PROPERTIES OF SOLUTIONS 40. Abnormal Effects of Salts on the Vapor-Pressure and Related Properties of Water, and their Explanation by the Ionic Theory. In dilute solution the effect of salts of the uniunivalent type, such as sodium chloride or silver nitrate, on the vapor-pressure of water and on the other related properties is nearly twice as great, and the effect of salts of the unibivalent type, such as potassium sulphate and barium chloride, is nearly three times as great, as it would be if each formula- weight yielded a single mol in the solution. Strong acids and bases show a similar behavior. These and other facts have led to the con- clusion that these substances are largely dissociated in aqueous solu- tion. For example, NaCl dissociates into Na + and Cl~; HNO 3 into IT .and NO 3 -; K 2 SO 4 into K + , K + , and SO 4 "; and Ba(OH) 2 into Ba + \ OH~, and OH~. Tri-ionic substances also dissociate partially into intermediate ions; thus I^SO, into KS0 4 ~ (and K + ), and Ba(OH) a into BaOH + (and OH"). The electrical behavior of the solutions indicates that these dissociation-products differ from ordinary substances in that their molecules are electrically charged. These charged substances are called ions; and to their formulas + or signs are attached, as in the above examples, to indicate the nature and magnitude of the charge. The fraction of the salt dissociated is called its ionization 7. This fraction always decreases with increasing concentration. Prol). 1. A solution of 0.65 formula- weight KC1 in 1000 g. water has at 100 a vapor-pressure of 744.8 mm. Calculate the number of mols of solute per formula-weight of salt and the ionization of the salt. Pro 6. 2. According to an estimate based upon certain of its proper- ties, sulphuric acid in a solution containing 0.05 formula-weights H 2 SO 4 per 1000 g. water at is 58% ionized into H* and HSO 4 ~ and 36% into H+, H + , and SO 4 ". Calculate the total number of mols of solute per formula-weight of the acid and the freezing-point of the solution. (The observed freezing-point is 0.215.) 42 FARADAY'S LAW 43 CONDUCTION OP ELECTRICITY IN SOLUTIONS: FARADAY'S LAW 41. Electrolytic Conduction. Conductors are divided into two classes with reference to the changes that are produced in them by the passage of electric currents. Those which undergo no changes except such as are produced by a rise in temperature are called metallic conductors. Those in which the passage of a current is attended by a chemical change are called electrolytes. Aqueous solu- tions of salts, bases, and acids, and melted salts at high temperatures, are the most important classes of the well-conducting electrolytes. The most obvious chemical changes attending the passage of a current through an electrolyte are those that take place at the surfaces of the metallic conductors where the current enters and leaves the elec- trolyte. The production of such chemical changes by a current from an external source is called electrolysis. The occurrence of such changes, when they themselves give rise to an electric current, is called voltaic action. Those portions of the metallic conductors that are in contact with the electrolyte are called the electrodes; the one at which the current leaves the electrolyte, or the one towards which the positive electricity flows through the electrolyte, is designated the cathode; the other, the anode. 42. Chemical Changes at the Electrodes. The chemical change produced at the cathode is always a reduction; that at the anode, an oxidation. The products resulting when aqueous solutions of certain typical salts, bases, and acids are electrolyzed between electrodes which are unattacked are as follows : Solute Cathode products Anode products Cu(N0 8 ) 2 or AgNO, Cu or Ag O 2 and HNO 8 KNO, H 2 and KOH O 2 and HNO 3 Na 2 SO 4 H 2 and NaOH O 2 and H 2 SO 4 KOH or Ba(OH) 2 H, O a H 2 SO 4 or H 8 PO 4 H a O 2 Dilute HC1 or HNO, H, O 2 Concentrated HC1 H a Cl a Concentrated HNO, NO 2 and NH, O a When the anode is a metal which can react with the anion of the solute, the change at the anode consists only in the dissolving of the metal; thus when a nitrate or sulphate is electrolyzed with a copper anode, copper passes into solution forming copper nitrate or sulphate. In the case of voltaic actions, the chemical changes are of a 44 IONIC PROPERTIES OF SOLUTIONS similar character, most commonly consisting in the solution of the metal composing the anode and in the deposition of another metal or of hydrogen on the cathode (the separation of free hydrogen being, however, often prevented by a secondary reaction between it and the electrolyte or the electrode). Thus, in the Daniell cell, which con- sists of a copper electrode in a copper sulphate solution and of a zinc electrode in a zinc sulphate solution, the two solutions being in con- tact and the two electrodes connected by a metallic conductor, the zinc dissolves and the copper precipitates; and, in the Grove cell, con- sisting of a zinc electrode in dilute sulphuric acid and a platinum electrode in strong nitric acid, zinc dissolves at the anode, and the hydrogen primarily produced at the cathode reduces the nitric acid to lower oxides of nitrogen. The chemical changes involved in voltai" actions do not differ, therefore, essentially from those produced by electrolysis. Pro&. 3. State what chemical change takes place at each electrode when electricity passes through : a, a concentrated solution of NaCl between a carbon anode and an iron cathode ; &, a solution of NaCl be- tween a silver anode and a metal cathode coated with AgCl; c, dilute H 2 SO 4 between a lead anode and a cathode of lead coated with PbO 2 ; d, a solution of ZnSO 4 between a zinc-amalgam anode and a cathode of mercury covered with Hg 2 SO 4 . 43. Faraday's Law. The passage of electricity through an elec- trolyte is attended at each electrode by a chemical change involving a number of chemical equivalents N strictly proportional to the quan- tity of electricity Q passed through and dependent on that alone. That is : Q = F N, where F is a constant with respect to all varia- tions of the conditions, such as temperature, concentration, current- strength, current-density, etc. Such variations often influence the character of the chemical change, but not the total number of equiva- lents involved. The law is applicable to concentrated, as well as to dilute solutions, and to fused salts. The constant F evidently represents the quantity of electricity producing a chemical change involving one equivalent. It is called a faraday, and has the value 96500 coulombs. One coulomb is the quantity of electricity flowing per second when the current is one ampere. The term chemical equivalent in the above statement of Faraday's law signifies the oxidation or reduction equivalent of the substance, FARADAY'S LAW 45 in the sense in which it is used in volumetric analysis. That is, one equivalent of any substance is that weight of it which is capable of oxidizing one atomic weight of hydrogen, or which has the same reducing power as one atomic weight of hydrogen. Prob. 4. Through solutions of AgNO 8 , AuCl 3 , Hg 2 SO 4 , and HgCl r placed in series, 9650 coulombs are passed. How many grains of metal will be deposited on the cathode from each solution? Prob. 5. A Daniell cell, consisting of zinc in zinc sulphate solution and copper in copper sulphate solution, furnishes a current of 0.1 ampere for 100 minutes. How many grams of copper deposit and of zinc dissolve in the cell? Pro ft. 6. How long must a current of 5 amperes be passed through dilute sulphuric acid in order to produce at 27 and 1 atm., a, one liter of oxygen? 6, one liter of hydrogen? Prob. 7. 1930 coulombs are passed through a solution of copper sulphate. At the cathode 0.018 equivalent of copper are deposited. How many equivalents of hydrogen are set free? 44. Relation of Faraday's Law to the Ionic Theory. Faraday's law evidently shows that electricity is transported from solution to electrode, or in the reverse direction, only by the constituents of the electrolyte; and that one equivalent of any constituent carries the same quantity of electricity, namely, one faraday or 96500 coulombs. The dissociated portions of these constituents* are called ions the positively charged ones cations, and the negatively charged ones anions. It is to be noted that the flow of current to the cathode may be brought about either by the deposition of a cation on the electrode or by the dissolving from the electrode of a substance that forms an anion; for flow of negative electricity in one direction produces a current having the same electrical effects as flow of positive electricity in the other. Pro 6. 8. When one faraday is passed through a potassium sul- phate solution, not only one equivalent of hydrogen, but also one equivalent of potassium hydroxide, is produced at the cathode; and not only one equivalent of oxygen, but also one equivalent of sulphuric acid, is produced at the anode. Show that these facts would not be in accord with Faraday's law if the ions were assumed to be (K 2 O) + + and (SO 3 )-~, but that they are in accord with it if the ions are assumed to be K* and SO 4 -~. *The two ion-forming constituents themselves, considered without reference to the extent to which they may be dissociated from each other in the solution, will be hereafter called the ion-constituents. Thus the Ion-constituents of potar slum nitrate are K and NO 3 , since its ions are K> and NO S -. 46 IONIC PROPERTIES OF SOLUTIONS c M ELECTRICAL TRANSFERENCE 45. Phenomenon of Electrical Transference. When a current is passed through a solution of a salt, base, or acid, in addition to the chemical changes taking place at the electrodes in accordance with Faraday's law, a certain quantity of the cation-constituent is trans- ferred from the neighborhood of the anode to that of the cathode, and a certain quantity of the anion-constituent is transferred in the reverse direction. This phenomenon can best be made clear by the consideration of an actual transference deter- mination. Consider, for example, that a 0.02 normal solution of sodium sulphate is electro- lyzed at 185 in an apparatus like that shown in Figure 5, between a platinum cathode (marked ) and a platinum anode (marked +). To avoid stirring of the solution, the electrodes, from which hydrogen and oxygen gases are evolved, are placed near the surface; the anode, around which the solution becomes denser during the electrolysis, is placed near the bottom of the tube; and the apparatus is immersed in a water-bath kept at constant tern- FIGURE 5 perature. The current is stopped before the electrode-products (the sodium hydroxide and sulphuric acid) have migrated beyond the dotted lines in the figure. The three portions of the solution (called the cathode-portion, middle-portion, and anode-portion, and marked 0, M, and A, respectively) are then separately removed from the apparatus, and submitted to analysis. The quantity of sodium and of sulphate present in each portion is compared with the quantity of it originally associated with the weight of water contained in the portion. It is found, if the experiment has been successful, that the middle- portion has undergone no change in composition, that the cathode-portion has increased its sodium-content and decreased its sulphate-content, and that the anode-portion has increased its sulphate-content and decreased its sodium-content. It is found, per faraday of electricity passed through the solution, that the sodium-content has increased in the cathode-portion by 0.39 equivalent and has decreased in the anode- ELECTRICAL TRANSFERENCE 47 portion by the same amount, and that the sulphate-content has in- creased in the anode-portion by 0.61 equivalent and has decreased in the cathode-portion by the same amount. When either of the constituents whose transference is being deter- mined is deposited on or dissolved off the electrode, as is the case when a silver-nitrate solution is electrolyzed between silver electrodes, the quantity of it so deposited or dissolved must evidently be determined either by direct weighing or by calculation with the aid of Faraday's law, and be subtracted from or added to the change in content of that constituent in the electrode-portion. 46. Law of Transference. The sum of the number of equivalents of the cation and anion constituents (N C and N A ) transferred in the two directions is equal to the number of faradays (N ) passed through the solution. That is, N C + N A = N. This equation is illustrated by the data for sodium sulphate given above. In the case of mixtures containing various ion-constituents (Cj, C 2 , . . . AJ, A 2 . . .) all of these are transferred, and the expression of the law of transference is : N Cj + N Cs . . .- + N Ai -f- N A2 . . . = N. 47. Transference-Numbers. The equivalents N C of cation-con- stituent transferred are in general not equal to the equivalents N A of anion-constituent transferred. The experimentally determined ratios N c/( N c + N A ) and N A /(N C + N A ) are called the transference-numbers, T C and T A , of the cation and anion, respectively. These transference- numbers evidently represent (since x c -}- N A = N) the equivalents of cation or anion-constituent transferred per faraday. Thus in the case of sodium sulphate T C = 0.39 and T A = 0.61. Prob. 9. Through a 0.2 normal solution of potassium sulphate be- tween platinum electrodes 0.0075 faraday is passed at 25. The cathode- portion after the electrolysis was found to contain 0.1450 g. more potas- sium than was originally associated with the weight of water in the portion. What is the transference-number of the sulphate-ion? Prob. 10. A current is passed at 25 through a solution of 16.64 g. Pb(NO 3 ) 2 in 1000 g. water between lead electrodes until 0.1658 g. silver is deposited in a coulometer in series with it. The anode portion weighed 62.50 g. and yielded on analysis 1.123 g. PbCrO 4 . What is the transference-number of the lead-ion? Assume that lead dissolves off the anode in accordance with Faraday's law. Prob. 11. A current of 0.1 ampere is passed for 30 minutes through a 0.2 normal Ba(NO s ), solution between platinum electrodes. The 48 IONIC PROPERTIES OF SOLUTIONS transference-number for the barium-ion is 0.45. What changes, expressed in equivalents of the ion-constituents, result in the anode-portion and in the cathode-portion from the electrolysis? from the transference? What is the net result of these two effects on the quantities of the various compounds present in each portion? 48. Transference in Relation to the Ionic Theory. The Mechan- ism of Conduction in Solutions. Just as Faraday's law shows that electricity is carried from the solution to the electrode only by the ion- constituents, so the law of transference (N C -f- N A = N) shows that through the solution the electricity is likewise carried only by the ion- constituents. The only difference is that as a rule only one kind of ion-constituent carries the electricity to the electrode, while all the ion-constituents present take part in the conduction of it through the solution. That the ion-constituents do move through the solution can be shown by placing the solution of a salt, such as copper sulphate or potassium permanganate, whose cation or anion has a characteristic color, beneath a solution of a colorless salt, such as potassium sulphate, and applying a potential-difference at the electrodes. The movement of the ion-constituents is explained by the ionic theory as follows. A certain fraction of the molecules of a salt exists in the state of positively and negatively charged molecules, called cations and anions. When a solution is placed between electrodes that are at different potentials, the ions in virtue of their charges are subjected to an electric force which drives them through the solution the cations towards the cathode, the anions towards the anode; while the unionized molecules, being electrically neutral, are unaffected. The ions are, however, constantly uniting to form union- ized molecules, and the latter are constantly dissociating into ions For this reason, although at any moment only the ions are moving, the resultant effect is that the ion-constituent as a whole moves con- tinuously towards the electrode. The rate at which the ion-constit- uent moves is equal to the rate at which the ion moves multiplied by the ionization of the salt ; for the statement that a certain fraction of the molecules is ionized is equivalent to the statement that any one molecule exists in the form of its ions during that fraction of the time. ELECTRICAL TRANSFERENCE 49 49. Transference in Relation to the Mobility of the Ions. The electric force / acting on any charged body is equal to its charge Q multiplied by the potential-gradient; that is, / = q(dE/dl), where dE is the change of potential in the distance dl. Moreover, the velocity of any body moving through a medium of great frictional resistance is proportional to the force acting upon it. Therefore, since the resistance to the motion of ions through solutions is very great, the velocity u of any given ion is proportional to the potential - gradient; that is: u = u(dE/dl), where u is the velocity under unit potential-gradient, called the mobility of the ion. The velocity of the ion-constituent is evidently also proportional to the potential-gradient. Prob. 12. A solution containing c equivalents of a salt per ccm. of solution is placed in a cylindrical tube of cross-section q sqcm. be- tween electrodes I cm. apart, at which a potential-difference E is applied. The mobilities of the cation and anion constituents in tbis solution are u c and U A , respectively. Sketch a diagram illustrating these conditions, and derive an expression for the number of equivalents N C and N A of the cation and anion constituents which migrate through any cross-section of the solution in tbe time t. Since the migration considered in Prob. 12 takes place also through the cross-sections which separate the middle portion from the two electrode portions, it is evident that the equivalents of the cation and anion constituents transferred are to each other as their mobilities. That is, N C /N A = u c / U A ; and therefore : u and T = _ . U c + U A U c + U A It is also evident that the transference-number of either ion-constitu- ent is the fraction of the current which is carried by that constituent. Prob. 18. An ordinary transference determination is made at 18 with 0.1 normal AgNO 3 solution in a cylindrical tube 4 cm. in diameter between silver electrodes 30 cm. apart. Analysis of the anode-portion shows tbat 0.00207 equivalent of silver have migrated out of it. a. Cal- culate the distance through which tbe silver migrated during the passage of the current. 6. The potential-difference applied at the elec- trodes was 10 volts, and the resulting current of 0.0395 ampere was passed for 3 hours. Calculate tbe mobility in centimeters per second of each of the ion-constituents. (Assume that tbe concentration-changes at the electrodes do not affect the potential -gradient.) C'A 50 IONIC PROPERTIES OF SOLUTIONS 50. The Moving-Boundary Method of Determining Transference. The relation derived in Prob. 12 forms the principle of another method of determining transference. In this method the relative rates are measured at which the two boundaries of a solution of a salt CA move when placed between solutions of two other salts* C'A and CA', arranged as in Figure 6, in which c c and a a represent the original c > positions of the boundaries, c' c' and a' of their positions after a certain time. It is evident that the cation-con- stituent C moves the distance c c' while the anion-con- a ' stituent A moves the distance a of ; and that they are moving under the same potential-gradient, since they are in the same solution. Therefore the ratio c c' /a a' is the ratio of the mobilities U C /U A , and hence of the transfer- , ence-numbers T C /T A . FIGURE 6 The boundaries are most readily seen when the "indicator" ions C' and A' are colored; but even when the ions are all colorless, the boundaries are usually visible because of the different refractive power of the adjoining solutions. Prob. 14 In a moving-boundary experiment an apparatus like that represented by Figure 6 is charged at 18 with solutions of silver nitrate at the bottom, of potassium nitrate in the middle, and of potassium acetate at the top. The lower electrode, which is of silver, is made the anode. In 90 minutes the lower boundary moves 3.00 cm. and the upper boundary 2.88 cm. What transference-numbers can be derived from these facts, and what are their values? 51. Change of Transference-Numbers with the Concentration. The transference-numbers of almost all uniunivalent and unibivalent salts (except the halides of bivalent metals) remain sensibly constant as the concentration increases, so long as it does not exceed a moder- ate value, say 0.1 normal. For example, the values at 18 of the sodium transference-number T Na in sodium chloride solutions at vari- ous concentrations c are as follows : T Na ....... 0.396 0.396 0.395 0.393 0.388 0.369 o ......... 0.005 0.020 0.050 0.100 0.300 1.000 normal. *In order that the boundaries remain sharp it is evidently necessary that the ion C' have a smaller velocity than the ion C, and that the ion A' have a smaller velocity than the ion A ; for otherwise the ions C' and A' would enter the solution of the salt CA, producing a mixture at each boundary. ELECTRICAL TRANSFERENCE 51 The constancy of these numbers in the dilute solutions shows that the ratio of the velocities of the two ion-constituents does not vary, which might be expected to be true so long as they are moving through a medium which is substantially the same as pure water. At higher concentrations the transference-number often changes rapidly with increasing concentration. This may be due to a variety of causes. Thus it may arise from a change in the frictional resistance of the medium; from hydration of the ions, which causes water to be trans- ferred and thus affects the transference value, since this is computed under the assumption that the water is stationary; from existence of complex ions, which in concentrated solutions are more likely to be present in considerable quantity. 52. Determination of the Composition of Ions by Transference Experiments. Proo. 15. When one faraday is passed through a solution of potas- sium silver cyanide (KCN.AgCN) the cathode-portion loses 1.40 equiva- lents of silver and 0.80 equivalent of cyanogen, and gains 0.60 equivalent of potassium. Explain what this shows in regard to the composition of the ions and their transference-numbers. Proo. 16. A transference experiment is made by passing 0.01 fara- day through a solution 0.2 formal in AgNO 3 and 0.6 formal in NH 8 between silver electrodes at 18. The anode-portion is found to gain 0.0053 equivalent of silver and to lose 0.0094 formula-weight NH 3 . Ex- plain what this shows in regard to the composition of the ions. (Silver dissolves at the anode in accordance with Faraday's law.) Proo. 17. Determination of the Hydration of Ions. In a transfer- ence experiment 0.0525 faraday was passed at 25 through a solution placed between silver electrodes and containing 1.12 formula-weights of NaCl and 0.073 formula-weights of raffinose (C^H^O^) in 1000 g. of water. The anode-portion was found by analysis to contain 0.72 g. less water and 1.115 g. less NaCl than was originally associated with the raffinose present in that portion, a. Calculate the number of mols of water and the number of equivalents of sodium transferred per faraday from the anode to the cathode, assuming that the raffinose does not migrate, o. Assuming no hydration of the chloride ion, calculate the number of molecules of water associated with the atom of sodium in the sodium ion. c. Assuming the chloride ion is hydrated with x mole- cules of water, calculate the hydration of the sodium ion. Note. Similar experiments made with other halides have given the following values for the number of molecules of water contained in other ions, assuming the number in the chloride ion to be x molecules : H+ : 0.28 -f 0.185#. K + : 1.3 -f- 1.02#. Cs+ : 0.67 -f 1.03a?. U + : 4.7 -f- 2.29#. 52 IONIC PROPERTIES OF SOLUTIONS ELECTRICAL CONDUCTANCE 53. Conductance, Specific Conductance, and Equivalent Conduct- ance. According to Ohm's law, the current I flowing between two points of a conductor is proportional to the potential-difference E at those points. The ratio of the current to the potential-difference is called the conductance L; and the inverse ratio, the resistance R. That is, I/E = L, and E/I = R. When the current is expressed in ani^eres and the potential difference in volts, the resistance is in ohms and the conductance in reciprocal ohms. An ohm is the resistance, and a reciprocal ohm the conductance, of a column of mercury at one square millimeter in cross-section and 106.3 centimeters long. The conductance of a homogeneous body of uniform cross-section is proportional to its cross-section q and inversely proportional to its length Z. That is, L = L q/l. The proportionality-factor L", which is the conductance when the cross-section is one square centimeter and the length one centimeter, is called the specific conductance. Its reciprocal is called the specific resistance. In the case of a salt in solution, the term equivalent conductance A is employed to denote the conductance of that volume of solution which contains one equivalent of salt, when placed between parallel electrodes one centimeter apart. Thus the equivalent conductance Ap.i of a salt in 0.1 normal solution represents the conductance of 10,000 cubic centimeters of that solution when placed between paral- lel electrodes one centimeter apart. The equivalent conductance of dissolved salts increases with decreasing concentration at first rapidly, then more slowly, and approaches a maximum value as the concentration approaches zero. This is illustrated by the following data at 18 : Fqniv. per liter.... 1 0.1 0.01 0.001 0.0001 0.0 A for NaCl 74.4 92.0 102.0 106.5 108.1 109.0 A for K 2 S0 4 71.6 94.9 115.8 126.9 130.8 133.0 The equivalent conductance of largely ionized solutes increases with rising temperature. At 18 the increase in dilute solution is from 2.1 to 2.5% per degree in the case of salts, and about 1.6% per degree in the case of strong acids. . ELECTRICAL CONDUCTANCE 53 Prol. 18. What is the specific conductance in reciprocal ohms of mercury at ? Prob. 19. a. Derive the algebraic relation expressing equivalent con- ductance in terms of specific conductance and concentration in equiva- lents per cubic centimeter. 6. Calculate the specific conductance of the sodium chloride solutions whose equivalent conductances are given above. Profc. 20. A 0.1 normal solution of silver nitrate at 18 is placed in a tube 4 cm. in diameter between silver electrodes 12 cm. apart. A potential difference of 20 volts at the electrodes produces a current of 0.1976 ampere. Calculate the conductance, the specific conductance, and the equivalent conductance of the solution. Prob. 21. The equivalent conductance of a 0.01 normal CuSO 4 solu- tion at 18 is 71.7 reciprocal ohms. What is the resistance of a column of it 20 cm. long and 5 sqcm. in cross-section? 54. Relation of Conductance to the Mobility of the Ion-Constit- uents. Pro 6. 22. In 0.2 normal NaNO 8 solution the mobility at 18 of the sodium is 0.000505 and that of the nitrate 0.000356 cm. per second. Calculate the conductance of a column of the solution 10 cm. long and 2 sqcm. in cross-section. Note that the conductance of any conductor is the quantity of electricity in coulombs which passes per second when the potential-difference at the ends is 1 volt; and that the conduct- ance of a salt solution is the sum of the conductances of its two ion- constituents. Pro 6. 23. Show that the specific conductance 17 and equivalent con- ductance A are expressed by the equations : L = L C + L A = (U C + U A ) P C, and A = A c + A A = (u c + U A ) F, where c is the concentration in equivalents per cubic centimeter of a solution in which the mobilities of the ion-constituents are u c and U A . 55. Relation of Equivalent Conductance to lonization. The equation A = (u c -f- U A ) F shows that the observed increase of equivalent conductance with decreasing concentration is due to an increase in the mobilities of the ion-constituents. In the case of dilute solutions of di-ionic substances this increase in mobility doubtless arises, in the way described in Art. 48, from a proportionate increase in the ionization of the substance; for up to moderate concentrations the resistance to the movement of the ions is presumably the same as in pure water. The maximum value attained at zero concentration evidently corresponds to complete ionization. The ratio of the equiva- lent conductance A at any, not too high, concentration to the equivalent 54 IONIC PROPERTIES OF SOLUTIONS conductance A O at zero concentration is therefore equal to the ioniza- tion 7 in the case of a di-ionic substance. That is, A / A O 7. The A - value is ordinarily obtained by extrapolation. When the substance is completely ionized the mobilities u c and U A and equivalent con- ductances A C and A A of the two ion-constituents become identical with those, u c +, U A -, A c +, A A -, of the corresponding ions. Therefore A = A C + -f- A A - == (u c + -j- U A -) F - These ion-mobilities and ion- conductances have at each temperature a certain value for each ion, whatever be the other ion in the solution. Proo. 24. In 0.1 normal solution at 18, HC1 and HC 2 H 8 O 2 are 92% and 1.35% ionized, respectively. With what velocity in centimeters per hour does the hydrogen migrate through each solution when the potential-gradient is 10 volts per centimeter? The mobility of the hydrogen-ion at 18 is 0.00326 cm. per second. Pro 6. 25. At 18 the equivalent conductance of 0.1 normal NH 4 OH is 3.1 reciprocal ohms. The equivalent conductance at zero concentra- tion for NH 4 C1, KC1, and KOH is 130.2, 130.0, and 239 reciprocal ohms, respectively. What is the ionization of ammonium hydroxide in 0.1 normal solution? The determination of the ionization of polyionic substances is complicated, even in dilute solutions, by the fact that these substances dissociate in two ways; thus sulphuric acid dissociates partially into H + and HSO 4 ~, and partially into H + , H + , and SO 4 ". The conductance- ratio for such substances has therefore no simple significance. In even moderately concentrated solutions the frictional resistance to the motion of the ions must be appreciably different from that in pure water ; and the conductance-ratio A / A cannot therefore be an exact measure of the ionization. An approximate estimate of the variation of this frictional resistance is given by the ratio of the vis- cosity 77 of the solution to the viscosity 77 of pure water at the same temperature. This viscosity-ratio 77/770 is determined by measuring the relative times required for equal volumes of the solution and of pure water to flow through the same capillary tube, when subjected to the same pressure. A film of the liquid adheres firmly to the walls of the tube ; and the phenomenon of the flow consists essentially in the slipping of the successive cylindrical shells of liquid past one another. Viscosity is therefore a property which depends on the frictional re- sistance to the motion of the molecules of the liquid past one another ; and it may be expected to be roughly proportional to the frictional ELECTRICAL CONDUCTANCE 55 resistance to the motion of ions through the same liquid. By taking the viscosity into account an ionization-value can be derived which is more accurate than that given by the conductance-ratio alone. Pro&. 26. a. At 18 the viscosities of 0.1-normal and 1.0-normal NaCl solutions are respectively 1.009 and 1.086 times as great as that of pure water. With the aid of these data and those given in Art. 53 calculate the ionization*of NaCl at 18 at these two concentrations. 6. Formulate an algebraic expression for the ionization in terms of the conductances and viscosities. Note. The^ viscosity-ratios ?}/TI O for some other salts at 18 at 1-normal are as follows : Salt KC1 LiCl MgCl 2 K 2 SO 4 MgSO 4 ?A 0.982 1.150 1.213 1.101 1.381 The deviation of the ratio from unity at 0.1-normal is approximately one-tenth of that at 1-normal. 56. Mobility and Conductance of the Separate Ions. Although only the sum of two ion-conductances is given by the A - value, the conductances of the separate ions may be obtained by combining it with the transference-number. Pro 6. 27. a. Calculate from the conductance and transference data given in Arts. 51 and 53 for sodium chloride the equivalent conduct- ances of sodium-ion and chloride-ion. 6. At 18 the value of A for KC1 is 130.0, for KNO 3 is 126.3, and for HNO 3 is 377 reciprocal ohms. From these data and the result obtained in a, calculate a value for the equivalent conductance of each of the ions of these salts, c. Cal- culate the cation-transference-number at 18 for nitric acid in dilute solution. (The experimentally found value is 0.839 at 0.005 normal.) The following table contains the values of the equivalent conduct- ance A of some important ions at 18, and the values of its fractional increase 64.5 217 Br- 67.7 215 NH 4 + 64.7 222 I- 66.6 213 Ag+ 54.0 229 NO 3 - 61.8 205 Mg++ 45.9 256 C1O 3 - 55.1 215 Ca++ 51.9 247 BrO 3 - 47.6 216 Ba++ 55.4 238 TO,- 34.0 234 Pb++ 60.8 243 C 2 H 3 O 2 - 35. 238 Cu++ 45.9 254 SO 4 " 68.5 227 Zn + + 47.0 254 C 2 ? / A 8 ? . The average values of this quantity at 18 for the three simplest types of salts are as follows : Type Example 0.001 0.01 0.02 0.05 0.1 normal Uniunivalent KNO 3 0.98 0.93 0.91 0.87 0.84 Unibivalent KO* a95 ' 87 * 84 ' 78 * 73 Bibivalent MgSO 4 0.86 0.64 0.56 0.47 0.41 Few salts of the uniunivalent or unibivalent type have ratios at 0.1 normal differing by more than five per cent, from the average values given above. There are, however, certain salts of the unibiva- lent type which form marked exceptions to the rule ; thus at 0.1 normal the ratio for cadmium chloride is 45%, and the ratios for the three mercuric halides are all less than 0.1%. The value of the ratio for all the types of salts decreases with rising temperature, but only to a slight extent. It has already been stated in Art. 55 that the ratio A^/A ^ is ap- proximately equal to the ionization up to fairly large concentrations in the case of uniunivalent substances. In the case of unibivalent substances this is not true, owing to the fact that they ionize partially in two ways with formation of the intermediate ion as well as with formation of the ultimate ions to which the A -value corresponds. In the case of salts of this unibivalent type the proportion of the inter- mediate ion present is not accurately known; up to concentrations of 0.1-0.2 normal it is, however, so small that only a small error results from neglecting it in calculating from the ratio A? / A O T? O the concen- tration of the ultimate univalent ion; but a much larger error results in thus calculating the concentration of the bivalent ion. Prob. 82. At 18 the specific conductance of a 0.05 formal K 2 SO 4 solution at 18 is 0.00949. a. Formulate an expression for this specific conductance in terms of the equivalent concentrations and conductances of the three kinds of ions that may be present. 6. Calculate with the aid of this expression the equivalent concentrations of K + and of SO 4 ", assuming that these are the only ions present, c. Calculate these equiva- lent concentrations, assuming that 20% of the salt is ionized into KSO 4 - 58 IONIC PROPERTIES OF SOLUTIONS and K + , and that the equivalent conductance of KSO 4 - is 35. d. Tabulate for cases & and c the fraction of the total sulphate which exists as SO 4 ~-, KSO 4 -, and K 2 SO 4 , respectively. Acids and bases, unlike salts, exhibit at any moderate concentra- tion, such as 0.1 normal, every possible degree of ionization between a small fraction of one per cent, and 90 to 95%. There is, to be sure, a fairly large group of monobasic acids and monacidic bases, includ- ing HC1, HBr, HI, HNO 3 , HC1O 3 , KOII, NaOH, LiOH, which have ionization-values comparable with those of the uniunivaler.it salts. with which therefore they may be classed. But outside of thia group all possible values are met with, as illustrated by the following values of the percentage ionization (1007) at 25 and 0.1 formal: H 2 SO 3 , 34% (into H + . and HSCV) J H 3 PO 4 , 28% (into H + and H 2 P(V) ; HN0 2 , 7%; HC 2 H 3 2 , 1.3%; H 2 CO 3 , H 2 S, HC1O, HCN, HB0 2 , all less than 0.2%. Polybasic acids are known to ionize in stages, giving rise to the intermediate ion; and the first hydrogen is almost always much more dissociated than the second, and the second much more than the third. Thus H,SO 3 at 0.1 normal at 25 is about 34% dissociated into IT and HSO 3 ", and less than 0.1% dissociated into H + and SO 8 ". Meth- ods by which the dissociation of the successive hydrogens can be determined will be referred to in Art. 84. 59. Comparison of Ionization- Values Derived from the Conduct- ance-Ratio and from the Freezing-Point-Lowering. It has been shown that the ionization of substances can be determined with the aid of two entirely distinct principles. One of these, based on the increase in the number of mols produced by ionization, involves the assump- tion that the ions and unionized substance have the normal effect on the vapor-pressure and related properties of the solvent. The other principle, according to which the ionization is equal to the conduct- ance-ratio, assumes that the decrease in the mobility of the ion- constituents with increasing concentration is due solely to the decrease in ionization of the substance. A comparison shows that the ioniza- tion-values derived in these two ways agree with each other within 2 or 3% in the case of most uniunivalent substances up to 0.1 normal. The comparison of the ionization-values obtained by the two methods in the case of unibivalent salts is of complicated significance because of the presence of the intermediate-ion. REVIEW PROBLEMS 59 Pro 6. S3 a. Calculate the number of mols per formula-weight of K 2 SO 4 in a 0.05 formal solution of it at 18 from the results obtained in Prob. 32 (under the two assumptions that the salt ionizes without formation of the intermediate ion and that it ionizes with formation of 20% of that ion). b. State how the number of mols per formula-weight of salt actually present in this K 2 SO 4 solution could be experimentally determined. REVIEW OP THE IONIC PROPERTIES OF SOLUTIONS 60. Review Problems. Prob. 34' Give an expression for the transference-number in terms (a) of the number of equivalents transferred; (o) of the mobilities of the ion-constituents; (c) of the conductances of the ions. Prob. 35. Express algebraically the relations (a) between resistance, conductance, specific conductance, and equivalent conductance; (6) be- tween specific conductance and ion-mobilities. Prob. 36. Calculate the number of equivalents of sodium that would be transferred per faraday from the anode to the cathode portion if the solution named in Prob. 30 were electrolyzed. Pro 6. 57. What changes in the hydrogen, chlorine, and platinum content, expressed in terms of the number of atomic weights of each element, take place in the cathode portion when 4825 coulombs are passed through a solution of chlorplatinic acid (H a PtCl 6 ) at 25? Assume that this acid dissociates only into H+ and PtCl 6 ~- ions, whose equivalent conductances at 25 are 350 and 68, respectively. The cathode process consists only in the deposition of platinum. Prob. 38. Sulphuric acid in 0.05 formal solution at 25 consists of 6% H 2 SO 4 , 67% HSO 4 -, 27% SO 4 =, and the corresponding amount of H*. The equivalent conductances at 25 of H*, HSO 4 -, and SO 4 " are 350. 35, and 80, respectively, a. Calculate the specific conductance of the solution at 25. b. Calculate the change in the equivalents of hydrogen in the cathode portion when one faraday is passed. (Note that hydro- gen is transferred both in the form of H + and of HSO 4 ~ ions.) CHAPTER V THE RATE OF CHEMICAL REACTIONS THE EFFECT OF CONCENTRATION 61. Definition of Reaction-Kate. The rate of a chemical reaction between gaseous or dissolved substances may be defined to be the number of equivalents of each of the reaction -products produced per liter in an infinitesimal time divided by that time ; that is, the reaction- rate is dc/dt f where dc signifies the increase of the equivalent concen- tration of each of the reaction-products in the time dt. As to the significance of the terms equivalent and equivalent concentration see Arts. 4 and 27. It is in some cases simpler, especially in dealing with reactions between gases or with reactions that take place in stages, to employ molal concentration (c) in place of equivalent concentration (c) ; but when this is done it will be so specified. Profc. 1. Sodium hydroxide and methyl acetate in dilute solution react with each other (forming sodium acetate and methyl alcohol) at a rate which at any moment is proportional to the concentration of the sodium hydroxide and to the concentration of the methyl acetate at that moment. At 25 in a solution 0.01 normal in each of these sub- stances the rate at which sodium acetate is produced is at the start 0.00118 equivalents per liter per minute. What will be the concentra- tion of the sodium acetate after 10 minutes? 62. The Law of Concentration-Effect. The rate of any chemical reaction which takes place between perfect gases or perfect solute? at a constant temperature completely in one direction is propor- tional to the concentration of each of the reacting substances at the moment in question. This law is exact only for perfect gases or perfect solutions; but, like the other limiting laws, it holds true approximately up to moder- ate concentrations. The statement of the law here given is a pro- visional one which will be made more complete after the mechanism of reactions has been considered. EFFECT OF CONCENTRATION 61 Pro b. 2. Express this law in the form of a differential equation for the general case that any number of substances A, B, C, . . . react with one another. Represent by C OA , C OB > C 0c . ' the concentrations of the reacting substances at time zero, and by c the concentration of the reaction-products which has resulted at any time t. The proportionality-constant which occurs in the equation ex- pressing the law of concentration-effect is called the specific reaction- rate (&). It evidently represents the rate which the reaction would have, under the assumption of proportionality, if the concentrations of all the reacting substances were unity (one equivalent per liter). 63. Eeactions of the First Order. The expression for the reaction- rate is simplest in the case of reactions in which only one substance undergoes a change in concentration. Such reactions, whose rate is expressed by a differential equation of the first degree, are called reactions of the first order. Pro b. 3. Cane-sugar in a dilute aqueous solution containing hydro- chloric acid undergoes hydrolysis according to the reaction: C.A.OU + H 2 = C^O, (glucose) + C 6 H U O 6 ( fructose). Formulate the differential equation expressing the rate at which the hydrolysis takes place. Note. In this case the concentration of the water does not change appreciably, since it is present in such large excess ; and the concentra- tion of the acid does not change at all, since it acts catalytically, accel- erating the reaction without being consumed by it. It is not usual in formulating the reaction-rate equation or in evaluating the specific reaction-rate to take into account the concentrations of substances which, like these just mentioned, do not change with the progress of the reaction. Prob. 4. Integrate the equation obtained in Prob. 3 between the time limits t = and t = t and the corresponding concentration limits. Prob. b. In a solution at 48 containing 0.3 mol of cane-sugar in a liter of 0.1 normal HC1, it is found (by measuring with a polarimeter the change in the optical rotatory power) that 32% of the sugar is hydrolyzed in 20 minutes, a. Calculate the specific reaction-rate and the actual rates at the beginning and at the expiration of 20 minutes. b. Calculate the percentage of sugar that will be hydrolyzed in 40 min- utes, c. Calculate the percentage of sugar that would be hydrolyzed in 20 minutes, if 0.3 mol were dissolved in 10 liters (instead of 1 liter) of 0.1 normal HC1. 64. Reactions of the Second and Third Orders. Prob. 6. a. Formulate the differential equation expressing the rate in dilute solution of the reaction between two suhstances A and B. 62 RATE OF CHEMICAL REACTIONS b. Integrate this equation between the time limits t = and t = t for the case that the initial concentrations of the two substances have the same value c Integrate the equation (with the aid of a table of integrals, if preferred) also for the cas that the initial concentrations C OA and CO B are different from each other. A reaction whose rate is expressed by the differential equation formulated in Prob. 6 is called a reaction of the second order. And in general, the order of a reaction is said to be equal to the degree of the differential equation expressing its rate. The expressions for a reaction of a third order (between three substances A, B, and C) can be similarly derived. For the case that the three substances have the same initial concentration C the inte- grated expression is : - --- 2 k t. ( n r<\2 r< 2 (C C) C Pro&. 7. When 0.01 mol of methyl acetate is dissolved at 25 in 1 1. of 0.01 normal NaOH, 11.8% of the ester is decomposed per minute at the start according to the reaction : CH 3 Ac -f- NaOH NaAc + CH,OH. a. How long will it take for one half of the ester in this mixture to be saponified? 6. How long will it take when 0.01 mol is dissolved in 1 liter of 0.02 normal NaOH? c. Suggest an analytical method by which the rate of this reaction could be followed experimentally. 65. Expressions of the Law of Concentration-Effect in Terms of the Fraction of the Reacting Substances Transformed. Profc. 9. From the integrated expressions of the first, second, and third orders already obtained derive the expressions: in which x represents the fraction of the reacting substances trans- formed, v the volume of the mixture, and N O the equivalents of each of the reacting substances present at the start. Pro&. 10. Show from the expressions derived in Prob. 9 how the specific reaction-rate is related to the time required for the transforma- tion of any definite fraction of the reacting substances in the case of different reactions of the same order, or of the same reaction at different temperatures. Prob. 11. At 25 the specific rate of the reaction between sodium hydroxide and methyl acetate is 1.8 times as great as that of the re- action between sodium hydroxide and ethyl acetate. What is the ratio of the times required for decomposing 90% of the two esters when equivalent quantities are present at the start? EFFECT OF CONCENTRATION 63 Prob. 12. What is the ratio of the times required for decomposing: a. 90% of 1 mol of cane-sugar when it is dissolved in 1 1. and in 10 1. of 0.1 normal HC1 solution? 6. 90% of 1 mol of methyl acetate when it is dissolved in 1 1. of normal NaOH solution and in 10 1. of 0.1 normal XaOH solution? 66. The Mechanism of Reactions. Influence of the Number of Molecules Involved. The rate of certain reactions, such as: 2FeCl3 -f SnCl 2 = 2FeCl 2 + Sn01 4 , and 2AgAc + NaCHO 2 = 2Ag + C0 2 + HAc + NaAc, in which two molecules of one of the substances are involved, has been found to be proportional to the square (instead of to the first power) of the concentration of that substance. In other words, it is found that these reactions are of the third order, instead of the second order. The view that the number of reacting molecules determines the law of the rate is further substantiated by the fact that it harmonizes the law of reaction-rate with the well-established law of chemical equilibrium (see Prob. 26). These considerations justify the conclu- sion that the provisional statement of the law of concentration-effect should be modified so as to state that the rate is proportional to the concentration of each of the reacting substances raised to a power equal to the number of its molecules which interact with the mole- cules of the other substance in the ultimate molecular process on which the occurrence of the reaction depends. Thus, in the case of a reaction aA -f- frB -f- cC = . . ., whose occurrence requires the inter- action of a molecules of A, b molecules of B, c molecules of C, the general law of concentration-effect is expressed by the equation: dc C) a (COB ~ C) 6 (C c ~ G) C . . . Pro 6. 13. Equal volumes of 0.2 normal solutions of silver acetate and sodium formate were mixed at 100 ; and after definite intervals of time samples were removed and the undecomposed silver acetate was titrated with potassium thiocyanate. Its concentration was found to be 0.067 normal after 2 minutes, 0.047 normal after 6 minutes, and 0.032 normal after 14 minutes. Show from these data that this re- action conforms more closely to the expression of the third order than to that of the second order. It is found, however, that many reactions which apparently involve three or more molecules conform to the expression of the second order. This is probably to be explained by the assumption that 64 RATE OF CHEMICAL REACTIONS the reaction expressed by the usual chemical equation takes place in stages, and that the stage which requires appreciable time is a reaction between two molecules. Thus the second-order reaction H 2 O 2 + 2HI = 2H 2 O -f- I 2 may be considered to take place in the two stages, H 2 O 2 + HI = H 2 + HIO, and HIO + HI = H 2 O + I 2 ; the first requiring a measurable time, and the second taking place almost instantaneously as soon as any HIO is formed by the first reaction. It is therefore necessary in the case of complex reactions to know their mechanism (the molecular process by which they take place), in order to predict the law of their rate; and conversely, the law of the reaction- rate throws light on the mechanism of the reaction. Pro 6. 14. a. Suggest an explanation of the fact that the rate of the reaction H+BrO 8 - -f 6H+I- = H + Bi- -j- 3I 2 -f 3H 2 O is propor- tional to the first power of the concentration of the BrO 8 - and of that of the I-. &. Explain also the fact that the decomposition of arsine gas, 4AsH, = As 4 -|- 6H 2 , is a reaction of the first order. A knowledge of the mechanism of reactions is often important \n other ways. Thus, the initial rates* at which an ester is decomposed by different bases of the same concentration are found to be propor- tional to the degrees of ionization of the bases, showing that it is the hydroxide-ion which is directly involved in the reaction; and IJiis knowledge makes it possible to calculate the rate of decomposition of an ester by any solution whose hydroxide-ion concentration is known, or to make the converse calculation. Pro&. 15. At 25 the initial rate of decomposition of ethyl acetate by 0.01 normal NaOH is 9.0 times as great as that by 0.1 normal KGN. The sodium hydroxide is 96% ionized. What is the concentration of hydroxide-ion in the potassium cyanide solution? *The initial rates have to be considered for the reason that the neutral salt which is produced as the reaction progresses has a great influence on the ionization of slightly ionized bases, as will be explained in the next chapter. EFFECT OF CATALYZERS 66 THE EFFECT OF CATALYZERS 67. Catalysis and Catalyzers. A reaction is often greatly accel- erated by the presence of a substance which is not itself consumed by the reaction. This phenomenon is called catalysis, and the sub- stance producing it is called the catalyzer. Although few general principles relating to catalysis have been established, its great practical importance makes it desirable to con- sider the more common types of catalyzers and the ways in which they act. This is done in Arts. 68-72. 68. Carriers. Carriers constitute one of the most common and best understood types of catalyzers. The mechanism of their action is as follows: The catalyzer produces with one of the substances an intermediate compound which reacts with the second substance in such a way as to regenerate the catalyzer; the reaction of the second sub- stance with the intermediate compound taking place more rapidly than that with the first substance. In this way a reaction which does not take place directly at an appreciable rate may be made to take place in stages at a rapid rate. The chamber process of making sul- phuric acid and the technical method of making ether are familiar examples of this type, the fundamental reactions being: a. O 2 + 2NO = 2NO 2 , and SO 2 + NO 2 + H 2 O = H 2 SO 4 + NO. b. C 2 H 5 OH -f H 2 SO 4 = C 2 H 5 HSO 4 + H 2 O, and C 2 H 5 OH -f C 2 H 5 HS0 4 = (C 2 H 5 ) 2 + H 2 SO 4 . 69. Contact Agents. Reactions between gases or solutes are often greatly accelerated by placing the reacting mixture in contact with a suitable solid substance which offers a large surface. The heavier metals are especially likely to be effective; but many other substances have specific effects on definite reactions. The platinum contact- process of making sulphur trioxide from sulphur dioxide and oxygen and the Deacon process of making chlorine by passing hydrogen chloride and oxygen over a porous mass impregnated with copper chloride are examples of contact catalysis. Gas reactions are often catalyzed by solid surfaces to such an extent that the chemical change takes place appreciably only in the gas that is in immediate contact with the walls of the containing vessel or with solid material with which it may be charged. The mechanism of contact actions is little understood. In most 66 RATE Of CHEMICAL REACTIONS cases, the contact action is probably due to an adsorption of the re- acting substances (that is, to a concentration of them on the surface of the solid) and to the fact that in the surface-layer the reaction - rate is greatly increased. Thus, finely divided platinum placed in contact with illuminating gas and air adsorbs a large quantity of the gases, and these then react so rapidly as to cause the gas to take fire. To reactions brought about by a contact agent the law of concen- tration-effect is, in general, not applicable; for the rate of such reactions must evidently be influenced by the rates at which the re- acting substances are adsorbed by the solid surface. 70. Hydrogen-Ion and Hydroxide-Ion as Catalyzers. In aqueous solutions many reactions are accelerated by hydrogen-ion. This is probably true of all reactions in which water is directly involved, such as the hydrolysis of cane-sugar or of esters. It is also true of certain reactions of oxidation and reduction. The rate of such hydrolytic reactions in very dilute solutions is found to be proportional to the concentration of the hydrogen-ion; for example, the specific reaction-rate of the cane-sugar hydrolysis at 48 has been found to be 9.95 times as great in 0.01 normal HC1 as it is in 0.001 normal HC1, while the ratio of the hydrogen-ion concentrations in the two solutions is 9.8. At higher hydrogen-ion concentrations or in the presence of neutral salts considerable devia- tions from proportionality exist; thus, the rate of the cane-sugar hydrolysis is 10.5 times as great in 0.1 normal HC1 as in 0.01 normal HC1, while the hydrogen-ion concentration is only 9.5 times as great. This principle can be employed (as shown by Prob. 18) for de- termining the hydrogen-ion concentration in solutions; for no other ion (except hydroxide-ion in certain cases) exerts a catalytic effect on hydrolytic reactions. Reactions in which water takes part are often accelerated also by hydroxide-ion; thus, milk-sugar, CJI a O u , dissolved in water becomes hydrated (with formation of C 12 H 22 O U .H 2 O) at a rate which is greatly increased by hydroxide-ion. Prob. 16. Name two or three other reactions, besides those here referred to, which are accelerated by hydrogen-ion. Refer, if necessary, to chemical text-books. Prob. n. Show how the catalytic effect of hydrogen-ion on hydro- lytic reactions may be interpreted as a carrier action, assuming that the hydrogen-ion is hydrated. EFFECT OF TEMPERATURE 67 Prob. 18. Diazoacetic ester decomposes in aqueous solution accord- ing to the equation CHN 2 .CO 2 C 2 H 5 4. H 2 O = CH 2 OH.CO 2 C 2 H 5 -f N 2 , and the reaction is catalyzed by hydrogen-ion. At 25 in a solution 0.1 normal in acetic acid, whose ionization is 1.34%, 37.5% of the ester is decomposed in 10 minutes. Assuming that it takes 67 minutes to decom- pose the same percentage of the ester in a solution 0.1 formal in sodium hydrogen tartrate, what is the hydrogen-ion concentration in that solution? 71. Enzymes. Certain complex organic substances called enzymes, which are produced by animal and plant organisms, have an extraor- dinary power of catalyzing certain organic reactions. The effect is highly specific, a particular enzyme being required for a particular reaction. Thus invertase, an enzyme produced by yeast, causes the conversion of cane-sugar into glucose and fructose; and xymase, another yeast enzyme, causes the conversion of glucose, but not of the analogous compound fructose, into ethyl alcohol and carbon dioxide. 72. Water as a Catalyzer. The presence of water in at least minute quantity is essential to the occurrence of almost all reactions. This is shown by experiments upon some of the most energetic chem- ical changes, such as the combination of sodium with chlorine, the union of ammonia and hydrogen chloride gases, the combustion of carbon monoxide with oxygen, and the union of lime and sulphur tri oxide, which are found not to take place when the separate sub- stances are very thoroughly dried before they are brought together. THE EFFECT OF TEMPERATURE 73. The Law of Temperature-Effect. Equal small increments of temperature cause an approximately equal multiplication of the specific rate of any definite reaction. Thus, if the specific rate of a reaction is increased 2.5-fold by raising the temperature from to 10, it will again be increased approximately 2.5-fold by raising the tem- perature from 10 to 20. The deviations from this law are greater, the greater the interval of temperature to which it is applied. The general magnitude of them is illustrated by the ratios of the specific reaction-rates at 10 intervals for the third-order reaction between ferrous chloride, potas- sium chlorate, and hydrochloric acid, which have been found to be 68 RATE OF CHEMICAL REACTIONS as follows: 2.8 between and 10, 2.7 between 10 and 20, 2.4 between 20 and 30, 2.5 between 30 and 40, and 2.2 between 40 and 50. Prob. 19. In a solution 0.01 molal in sodium hydroxide and 0.01 molal in ethyl acetate, 39% of the ethyl acetate is decomposed in 10 minutes at 25, and 55% at 35. How long would it take to decom- pose 55% at 15? Prob. 20. A more general expression of the law of temperature- effect Is that given by the equation dlogfc (A/T*)dT, in which A is a quantity which has nearly the same value at temperatures not far apart. Integrate this equation, and show that the result for a small range of temperature is in accordance with the above stated law of temperature-effect. Prob. 21. Calculate by this more general expression from the data of Prob. 19 how long it would take to decompose 50% of the ethyl acetate at 20. With respect to the effect of temperature, it is further to be noted that in the case of different reactions equal small increments of tem- perature cause not far from the same multiplication of their specific rates. Thus, a 10 rise of temperature multiplies the specific rates of the reactions between the following substances by the following factors : NaOH -f- C,H 5 C 2 H 3 2 (at 27) : 1.9 FeSO 4 -f KC10 3 -f H 2 S0 4 (at 21) : 2.4 NaOH -f- CH 2 ClCO 2 Na (at 100) : 2.5 C^H^O,, -f H 2 (+ HC1) (at 40) : 3.6 Prob. 22. At 100 a certain reaction takes place to an extent of 25% In one hour. Estimate roughly how long it would take for the reaction to proceed to the same extent at 20. THE EFFECT OF SURFACE IN THE CASE OF SOLID SUBSTANCES 74. Solid Substances Involved in Chemical Reactions. When a solid substance is reacting with a solute, the quantity of it dissolved per unit of time is proportional to the surface of the solid, as well as to the concentration of the solute that reacts with it. Thus, when a dilute solution of acetic acid in contact with a compact mass of magnesium oxide is uniformly stirred, the quantity of the oxide dissolved is proportional to the surface of the mass and to the con- centration of the acetic acid. SIMULTANEOUS REACTIONS 69 In reactions with solid substances, it is to be borne in mind that, owing to corrosion, the effective surface is constantly changing, and that the concentration of the solution in contact with the solid is the same as that of the whole solution only when there is adequate stirring. 75. Solid Substances Dissolving in Their Own Solutionr. When a solid substance is dissolving at a definite temperature in its own (partially saturated) solution, the quantity dissolved per unit of time is proportional to the surface of the solid and to the difference between the concentration of the saturated solution and that of the solution in contact with the solid. J'rob. 23. A quantity of a solid substance having a surface-area a and a solubility is stirred uniformly at a definite temperature with a definite volume <; of water, a. Formulate a differential expression for the rate at which the concentration c of the solution will increase. I). Integrate this expression, assuming the surface to remain constant (which it does approximately when the quantity of the solid substance is very large in relation to the quantity of it that dissolves). J'rob. g.'i. a. What are the relative rates at which *a substance dis- solves in its own solution when the solution is 50% and 95% saturated, assuming the surface of the solid and the volume of the solution to be the same in the two cases? It. Tf it takes 2 minutes for the solution to become 50% saturated, how long will it take for the degree of satura- tion to increase from 95 to 98% ? SIMULTANEOUS REACTIONS 76. Law of Independent Reaction-Rates. When two reactions are taking place simultaneously, the rate of each is determined by its own specific reaction-rate and by the concentrations of the substances involved in it, just as if the other reaction were not taking place. Pi-ob. 25. The reaction between methyl oxalate and sodium hydrox- ide takes place in the two stages : (CH 8 ) 2 C 2 4 + NaOH = CH 8 NaC 2 O 4 + CH.OH CH 8 NaC 2 4 + NaOH = Na 2 C 2 O 4 + CH.OH. Formulate the differential equation for the rate (expressed in molal concentration) of each of these reactions for the case that the methyl oxalate and sodium hydroxide are present at the start at the molal concentrations C A and C B , respectively, and that at the time t the sodium methyl oxalate and the sodium oxalate produced have attained the molal concentrations c^ and c 2 , respectively. 70 RATE OF CHEMICAL REACTIONS 77. Reactions Taking Place in Opposite Directions. The law of independent reaction-rates finds an especially important application in the case of reactions which do not go to completion, but take place in one direction when the substances on the one side of the chemical equation are brought together, and in the other direction when those on the other side are brought together. The resultant rate j)f such a reaction must evidently be the difference of the rates of the two opposing reactions. Prob. 26. The gaseous reaction H 2 -f- I 2 = 2HI is an incomplete reaction of the kind just described, which at 400-500 takes place at a measurable rate. For the case that hydrogen, iodine, and hydrogen iodide are present at the start at the molal concentrations C OH , c OI) and c OH i formulate the differential expression for the rate at which the molal concentration of the hydrogen iodide is increasing at the time *, when c mols per liter of hydrogen iodide have been produced. CHAPTEK VI THE EQUILIBKIUM AT CONSTANT TEMPEKATUEE OF CHEMICAL KEACTIONS INVOLVING GASEOUS OK DIS- SOLVED SUBSTANCES AT SMALL CONCENTEATIONS THE LAW OP MASS-ACTION 78. The Equilibrium of Chemical Reactions. When two or more chemical substances capable of reacting with one another are brought together, it is always true after a sufficiently long time (which may vary from a fraction of a second to thousands of years) that the chemical reaction which has been taking place between them prac- tically ceases in other words, that a condition is reached where no further change takes place. The reaction is then said to be in equilibrium. Many reactions (for example, the reaction NH 4 OH -\- HBO 2 = NH 4 BO 2 -f- H 2 O in aqueous solution) take place so incompletely .that at equilibrium the substances on both sides of the equation are present in measurable proportions. But with many other reactions the equilibrium-conditions are such that the change seems to take place completely when the substances on one side of the chemical equation expressing it are brought together, and not to take place at all when the substances on the other side are brought together; thus, this is true of the reaction NaOH + HC1 = NaCl + H 2 O in aqueous solution. In all such cases, however, the gaseous or dis- solved substances on both sides of the equation are really present at some concentration, though this may be so small in the case of some of the substances that it cannot be directly measured. Keactions (like that in a mixture of hydrogen and oxygen at 25) which are taking place so slowly that no appreciable change can be detected within a reasonable time are not to be confounded with those which are in a state of equilibrium. Whether equilibrium- conditions have been attained can be. determined by causing the reaction to take place in the two opposite directions and comparing 71 72 EQUILIBRIUM OF CHEMICAL REACTIONS the concentrations of the substances in the two cases after equi- librium seems to have been reached. The equilibrium-conditions of chemical reactions vary with the temperature. In this chapter and the following one the principles will be discussed which determine equilibrium when the temperature has any constant value. 79. The Mass-Action Law. The effect of concentration in deter- mining chemical equilibrium is expressed by one of the most funda- mental laws of chemistry. This law, which is commonly known as the mass-action law, states that, whatever be the initial concentra- tions of the gaseous or dissolved substances A, B, . . E, F, . . involved in any definite chemical reaction, such as may be represented in general by the equation, flA -f- bB . . = eE -j- /F . . , the reaction always takes place in such a direction and to such an extent that, when equilibrium is reached at any definite temperature, the condi- tions are satisfied which are expressed by the equation : - = K, a constant. c A a c* b . . In this expression C E , C F , . . C A , C B , . . denote the concentrations of the substances E, F, . . A, B, . . in the equilibrium mixture, and e, f, . . a, fc, . . denote the number of mols of them that are involved in the reaction expressed by the chemical equation. The quantity K, which is a constant characteristic of the reaction, is called its equilibrium-constant. Its value is, of course, constant with respect to variations of the initial and equilibrium concentra- tions, but it varies with the temperature. In evaluating it, it is customary to express all concentrations in mols per liter, and to place in the numerator the concentrations of the substances occurring on the right-hand side of the chemical equation written in some speci- fied way. In which direction the chemical equation is written is, of course, arbitrary in the case of reactions at equilibrium. With many types of reactions, however, it is more natural to write the equation in one of the two directions; and in such cases the usage in evalu- ating the equilibrium-constant has become fairly definite. This usage will be illustrated by the examples given in the following articles. The mass-action law has been verified by direct experiments upon a large number of different reactions. It can be derivpd, as will bf> BETWEEN GASEOUS SUBSTANCES 73 shown in a later chapter, with the aid of the laws of thermodynamics, from the perfect-gas equation (p = c R T), or from the osmotic- pressure equation for perfect solutes (P = cRT). It can be also derived from the laws of the rate of reactions, as is illustrated by Prob. 1 below. The mass-action law is exact only in the case of reactions between perfect gases or perfect solutes; but it holds true approximately when applied to gases at moderate pressures or to unionized or slightly ionized solutes at moderate concentrations. The deviations from it in such cases may be expected to be of the same order of magnitude as those from the pressure-concentration law of perfect gases or perfect solutes. As will be seen later, large deviations are met with in the case of largely ionized substances. J'rob. 1. a. Formulate the mass-action expression for the equilib- rium of the reaction 2HI H 2 -f I 2 , and derive it from the differential equation obtained in Prob. 26 of the previous chapter, explaining the substitutions that are made in the derivation. 6. Show what relation exists between the equilibrium-constant and the specific reaction-rates. Note. Although the law of chemical equilibrium can be derived, as here illustrated, from the law of reaction-rate, an important difference between the two laws is to be noted. Namely, it can be shown that, although the expression for the rate of a reaction depends on its mechanism, the same expression is obtained for its equilibrium, what- ever be the process by which it is considered to be attained. In using the equilibrium expression it is only necessary to know the concentra- tions of the substances actually occurring in that expression. APPLICATIONS OF THE MASS-ACTION LAW TO GASEOUS SUBSTANCES 80. Expression of the Law in Terms of Partial Pressures. In applications of the mass-action law to gases, it is usual to substitute for the concentrations of the substances in the equilibrium-mixture their partial pressures. This is admissible since at any definite temperature the two quantities are proportional to one another, in virtue of the relation p = c R T. The general mass-action expression in terms of partial pressures evidently is where K p is a constant, called the equilibrium-constant in terms of 74 EQUILIBRIUM OF CHEMICAL REACTIONS pressure, which has in general a different numerical value from the constant K occurring in the corresponding concentration-expression. In evaluating it the partial pressures are commonly expressed in Prob. 2. Derive the general mass-action expression in terms of pressure from that in terms of concentration, and show the relation between the two equilibrium-constants. 81. Gaseous Dissociation. A chemical change which consists in the splitting of a substance with complex molecules into one or more substances with simpler molecules is called dissociation. Thus, the reactions I 2 = 21, NH 4 C1 = NH 3 + HC1, and 2CO 2 = SCO + O 2 , are examples of dissociation. The fractional extent to which the dissociating substance has been decomposed is called its degree of dissociation, or simply its dissociation (7). The equilibrium-constant of such a reaction is commonly called the dissociation-constant of the dissociating substance. It is characteristic of such reactions that the number of mole- cules increases when the dissociation increases. Since the pressure- volume product of gases increases correspondingly, the degree of dissociation can always be determined by measuring the volume or density of the gas at a known temperature and pressure, and comparing it with that calculated for the undissociated or completely dissociated substance, as is illustrated by Prob. 14, Art. 17, and by Prob. 4 below. Another common method of determining the composition of the equilibrium-mixture is to cool it suddenly to a lower temperature at which the reaction-rate is so small that the original equilibrium is not displaced, and then to analyze the mixture. This method pre- supposes that there is no change in the composition during the short period of cooling. This condition may be practically realized in cases where the rate at which the equilibrium is established is compara- tively slow even at the higher temperature, or in cases where the reaction takes place only in contact with a catalyzer. lu the latter case the equilibrium-mixture can be separated from the catalyzer, and subsequently cooled without the danger of any change taking place. Prob. 8. a. Formulate the mass-action expression for the dissocia- tion of sulphur trioxide into sulphur dioxide and oxygen. 6. At 630 BETWEEN GASEOUS SUBSTANCES 75 and 1 atni. the sulphur trioxide is just one third dissociated. Calculate the dissociation-constant of sulphur trioxide. Prob. 4. Show how the dissociation (7) of sulphur trioxide could be calculated from measurements of the density (d) of the equilibrium mixture at the temperature (T) and pressure (p) in question. Pro 6. 5. A mixture consisting of 1 mol of SO 2 and 1 mol of O 2 is passed at 630 and 1 atm. through a tube containing finely divided platinum so slowly that equilibrium is attained, and the issuing gas is cooled and analyzed by absorbing the sulphur dioxide and trioxide by potassium hydroxide and measuring the residual oxygen gas. At and 1 atm. the volume of this residual gas was found to be 13,780 cc., corresponding to 0.615 mol. a. Calculate the dissociation-constant of sulphur trioxide. b. Calculate the ratio of the mols of sulphur trioxide to the mols of sulphur dioxide in an equilibrium mixture at 630 in which the partial pressure of oxygen is 0.25 atm. /Yob. 6. At a certain temperature a definite quantity of phosphorus pentachloride gas has at 1 atm. a volume of 1 liter, and under these conditions it is about 50% dissociated into PC1 8 and C1 2 . Show by reference to the mass-action expression whether the dissociation will be increased or decreased: a, when the pressure on the gas is reduced till the volume becomes 21.; &, when nitrogen is mixed with the gas till the volume becomes 2 1., the pressure being still 1 atm.; c, when nitrogen is mixed with the gas till the pressure becomes 2 atm., the volume being still 1 1. ; d, when chlorine is mixed with the gas till the pressure becomes 2 atm., the volume being still 1 1. In answering these questions consider whether the first effect of the change in con- ditions (assuming that no reaction takes place) is to increase or de- crease the value of the ratio Pc\ 2 Ppc\ s /Ppc\ 5 > and in which direction the reaction must take^ place in order to restore the equilibrium- value of this ratio. Prob. 7. a. Derive for the dissociation of water- vapor into hydro- gen and oxygen a mass-action expression which will show how the dissociation (7) varies with the total pressure (p). b. At 2,000 the dissociation is 2.0% when the total pressure is 1 atm. How much is the dissociation when the total pressure is 0.2 atm. (as it is approxi mately in the gaseous mixture produced by burning hydrogen with the minimum amount of air)? Solve the equation approximately, neglect ing the (small) value of the dissociation where this is justifiable. 82. Metathetical Gas Reactions. Examples of such reaction* whose equilibrium has been investigated are: 2HI = H, + I 2 *; CO 2 + H 2 = CO + H 2 O; and 4HC! + O 2 = 2C1 2 + 2H 2 O. *This reaction is often called a dissociation, since one substance is con- verted into two others. From a molecular standpoint, however, it consists in an interchange of atoms between molecules, rather than in a splitting of a molocule into simpler ones ; and from a mass-action standpoint it differs from true dissociation in that there is no increase in the number of molecules when the reaction takes place. 76 EQUILIBRIUM OF CHEMICAL REACTIONS An important principle in regard to them, illustrated by Prob. 9 below, is that the equilibrium-constant of any metathetical gas re- action can be calculated from the dissociation-constants of the com- pounds involved in it. Another important principle relates to the effect of pressure. It states that increase of pressure causes the equilibrium of any gaseous reaction to be displaced in that direction in which the number of molecules, and therefore the volume of the gas, decreases. This prin- ciple has already been illustrated by the fact that dissociation is decreased by increase of pressure. It is demonstrated by Prob. 10. Prob. 8. The equilibrium-constant of the reaction CO 2 -f H 2 = CO -f H 2 O at 1,120 is 2.0. What is the ratio of the number of rnols of CO 2 and of H 2 O that will be formed when a "water-gas" consist- ing of 1 mol CO and 1 mol H 2 is burnt at 1,120 with $ mol O 2 ? (Assume that equilibrium is attained and that the quantity of the oxygen which remains uncombined is negligible.) Prob. 9. The equilibrium -constant of the reaction 2CO 2 = 2CO -j- O 2 at 1,120 is 1.4 x 10~ 12 . a. Calculate the partial pressure of the oxygen in the equilibrium mixture of Prob. 8. b. Calculate the dissociation- constant of water-vapor at 1,120. c. Show what relation exists be- tween the dissociation-constants K w and K C02 of water-vapor and of carbon dioxide and the equilibrium-constant K of the reaction C0 2 -f H 2 = CO + H 2 O. Prof). 10. Prove that the equilibrium of any chemical reaction, oA -(- 6B = eE -f- /F, must be displaced in the direction in which the number of molecules decreases when the total pressure p of the equi- librium mixture (in which the substances are present at mol-fractions X A , X B , X E , and # F ) is increased. APPLICATIONS OP THE MASS-ACTION LAW TO DISSOLVED SUBSTANCES 83. lonization 01 Slightly Ionized Univalent Acids and Bases. The mass-action law has been found to be applicable to the ionization (dissociation into ions) of the slightly ionized monobasic acids and monacidic bases. This is illustrated by the following values of the percentage ionization at 18 derived from conductance measurements (as described in Art. 55) and of the ionization-constant K of ammo- nium hydroxide (that is, the equilibrium-constant of the reaction NH 4 OH == NH 4 + + OH") : Cone. 0.300 0.100 0.010 0.001 1007 0.74 1.30 4.05 12.3 10 6 Z 16.6 17.1 17.1 17.1 BETWEEN DISSOLVED SUBSTANCES 77 The ionization-constant varies greatly with the composition and structure of the acid or base, as will be seen from the following values of it for certain acids at 25. Acid 1Q*K Acid 10'K Acid Values of WK HCN 0.0007 HBO, 0.0007 HC1O 0.04 HNO, 500. HCO 2 H 210. CH 3 C0 2 H 18. C 2 H 5 CO 2 H 13. n-C 8 H 7 CO 2 H 15. C 6 H 6 C0 2 H C 6 H 4 OHC0 2 H G 6 H 4 C1C0 2 H C 6 H 4 N0 2 C0 2 H GO. 60. 60. Ortho Mcta Para 1020. 87. 29. 1320. 155. 93. 6160. 345. 396. Prob. 11. At 25 acetic acid in 0.1 normal solution is 1.34% ion- ized, a. Calculate the ionization-constant of acetic acid. &. Calculate its ionization in 0.01 normal solution. Prob. 12. Effect of the Presence of a Substance with a Common Ion. a. Show that the hydrogen-ion concentration in an acetic acid solution is decreased by the addition of sodium acetate approximately in the proportion in which the concentration of the acetate-ion is in- creased, b. State the corresponding principle that would apply to the ionization of ammonium hydroxide in the presence of an ammonium salt. c. Calculate the ionization of acetic acid in a solution 0.1 normal both in acetic acid and in sodium acetate, the sodium acetate in the mixture being 85% ionized. In this and other mass-action problems make any simplifications that will not produce in the result an error greater than 1%. 84. Ionization of Dibasic Acids and Their Acid Salts. Polybasic acids ionize in stages; thus, a bibasic acid H 2 A ionizes according to the equations H^ = H+ + HA~ and HA" = H + + A = . The equi- librium-constants K t and K 2 of these reactions are called the ioniza- tion-constants for the first hydrogen and for the second hydrogen, respectively. The values of K 9 are commonly much smaller than those of R r The values at 25 of the two constants for some impor- tant acids are as follows : Acid K^ K z H 2 SO, 2 x 10-' 5 X 10- 6 H 8 P0 4 1 X 10- a 2 x 10- T H 2 C 4 H 4 O 8 (tartaric) 9.7 x 10-* 4 x 10-" H 2 CO, 3 x 10-' 3 x 10-" H 2 S 9 X 10-" 1 X 10-" From the two ionization-constants of a dibasic acid and its formal concentration c the concentrations of the various substances H,A, HA", A = , and H* present in its solution can be calculated. In solving any such mass-action problem, the best plan is to formulate firs* 78 EQUILIBRIUM OF CHEMICAL REACTIONS the equilibrium equations that must be satisfied thus, in this case the equations (IT) (HA~) == ^(H 2 A) and (IT) (A~) = JC(HA-).* The next step is to formulate the so-called condition equations, which sum up the concentrations of the separate forms in which an element or other constituent exists in the solution. Thus, in this case, for the total molal concentration 2 (A) of the constituent A, and for that 2 (H) of the hydrogen, we have : (H 2 A) + (HA-) + (A=) = 2 (A) = c, and (H + ) -f (HA-) -f 2(H 2 A) = 2(H) = 2c. (It will be noted that one of these equations might be replaced by a simpler condition-equation of another kind; namely, by the equation, (H*) = (HA~) -f- 2(A = ), which expresses the fact that positive and negative ions must be present in equivalent quantities.) We now have four independent equations containing four unknown concen- trations. An exact algebraic solution of these equations is therefore possible. But, as such a solution is often very complicated, it is advisable to try first to simplify the condition-equations, which can often be done by neglecting in them the condensation of some sub- stance which is small in comparison with the concentrations of the other substances. Thus, in this case the concentration (A = ) is small compared with the concentration (HA'), for the reasons that K 2 is small in comparison with K^ and that the hydrogen-ion produced by the ionization of the H,A into H* and HA~ further decreases by the common-ion effect the ionization of the HA~. It is always well to test the correctness of such simplifying assumptions by subsequent calculation of the quantity neglected. Prob. 18. From the ionization-constants given in the table above, calculate by the method just described the molal concentration of each of the substances present in a 0.01 formal solution of tartaric acid at 25. Prob. 14. Determination of the lonization-Constant for the Second Hydrogen by Reaction-Rate Measurements. By measuring the conduct- ance of tartaric acid in 0.06-0.01 formal solution, where the ioiiiza- tion of the second hydrogen is negligible, the ionization-constaut for the first hydrogen of the acid has been found to be 0.00097 at 25. The hydrogen-ion concentration in a 0.1 formal solution of sodium hydro gen tartrate, NaHA, was found in Prob. 18, Art. 70, to be 0.0002 molal. From analogy with other salts of the same type (see Art. 58) the *As is done here, the molal concentrations of substances are often repre- sented by writing their formulas within parentheses. BETWEEN DISSOLVED SUBSTANCES 79 concentration (NaHA) of unionized sodium hydrogen tartrate In this solution may be estimated to be 0.016 molal. In regard to the concen- trations (Na 2 A) and '(NaA~) nothing definite is known, but they may be assumed in this problem to be negligibly small. From these data calculate the concentrations (A=), (HA-), and (H 2 A), and the ioiriza- tion-constant for the second hydrogen of tartaric acid. (In this case no simplification of the condition-equations is admissible, other than the neglecting of the concentrations (Na 2 A) and (NaA~) just men- tioned.) Tabulate the molal concentrations of all the substances present in the solution of the sodium hydrogen tartrate. Prob. 15. Determination of the lonization-Constant for the Second Hydrogen by Distribution Experiments. A 0.1 formal solution of sodium hydrogen succinate NaHA in water is found by experiment to be in equilibrium at 25 with a 0.00187 formal solution of succinic acid in ether. The distribution-ratio of succinic acid between water and ether at 25 is 7.5. a. Find the H 2 A concentration in the aqueous solution of the acid salt. b. Calculate the ionization-constant for the second hydrogen. That for the first hydrogen is 6.6 X 10~ B - Make the same assumptions as in Prob. 14; neglect also the value of (H+) in the condition-equation, and show afterwards by calculating it that it was justifiable to neglect it. 85. The lonization- of Water. Prob. 16. Show that the product (H+) x (OH-) has substantially the same value in any dilute aqueous solution (up to 0.1-0.2 normal) as in pure water. (This constant value of the (H + ) x (OH~) product is called the ionization-constant # w of water.) Note. The application, involved in this problem, of the mass- action law to a constituent so concentrated as the solvent would seem to be inadmissible, and is so in a strict sense. It can be shown, how- ever, with the aid of thermodynamics, that the activity of the solvent, that is, its mass-action effect in influencing equilibrium, is exactly proportional in the case of different solutions to the vapor-pressure of the solvent above those solutions. And it follows from Raoult's law of vapor-pressure that the vapor-pressure of even a 0.5 molal aqueous solution is only about \% less than that of pure water. Prob. 17. At 25 the concentration of hydroxide-ion in pure water is one tenmillionth molal. What is its concentration in 0.1 formal HC1 solution, in which the acid is 92% ionized? 86. lonization of Largely Ionized Substances. In the case of salts and of largely ionized acids and bases the ionization values derived from conductance or from freezing-point measurements do not change with the concentration even approximately in accordance with the mass-action law. Thus, the average ionization values (7 obs.) for uni- univalent salts given in Art. 58, the " ionization-constants " (K) 80 EQUILIBRIUM OF CHEMICAL REACTIONS calculated from them, and the ionization values (7 calc.) calculated conversely from the value of this constant at 0.1 normal are as follows : Cone. 0.001 0.01 0.02 0.05 0.10 7Obs. 0.96 0.92 0.90 0.87 0.84 7 calc. 0.998 0.98 0.96 0.91 0.84 K 0.023 0.11 0.16 0.29 0.44 K/sJc (0.73) 1.10 1.13 1.30 1.39 It has been found empirically that the " ionization-constant " K varies with the concentration c of the salt, and is approximately proportional to the square-root of that concentration. This is illus- trated by the comparatively small variations of the values of JSYy/TT given in the last row of the above table. This principle applies also to mixtures of salts, the "ionization-constant" being in this case dependent on the sum of the concentrations (Sc) of all the salts in the mixture. Thus, for a salt A + B~ the principle is expressed by the equation : (A + ) X (B-) =K const. X V2c. \AB) ProT). 18. a. Show that this principle requires that in a mixture of two uniunivalent salts with a common-ion (such as K+C1- and Na+Cl~) the ionizations of the two salts are equal, whatever be the relative proportions of the two salts in the mixture. 6. Show that the principle further requires that the ionization y 1 or y 2 of either salt in the mixture is equal to the ionization 7 which it has when it alone is present at a concentration c equal to the sum of the concentrations (i + c ) of the two salts. Prob. 19. What are the concentrations of unionized potassium chloride and of unionized sodium chloride in a mixture 0.02 normal in KC1 and 0.08 normal in NaCl? The reason why largely ionized substances do not conform to the mass-action law is not known. One possible explanation of this anomaly is that the mass-action law really holds true in the case of largely ionized substances, but that the conductance-ratio is not an exact measure of their ionization. This would imply that the mobility of the ions does not remain constant even up to moderate concentrations, as was assumed in Art. 55. Another explanation is that the activity either of the ions or of the unionized substance or of both is abnormal in other words, that the mass-action effect is not proportional to the concentration. The simplest way of accounting fairly satisfactorily for the BETWEEN DISSOLVED SUBSTANCES 81 behavior of largely ionized substances in all their relations seems at the present time to be the adoption of the following hypotheses : (1) the conductance-ratio is a substantially correct measure of their ionization; (2) the ions have (approximately) normal activities; and (3) the unionized substance has an activity which is far from being proportional to its concentration. These assumptions will therefore be employed in this book. And in accordance with them, in dealing with mass-action problems, the equilibrium-equations will always be so written as to involve the ions rather than the unionized part of largely ionized substances; and in condition-equations the concen- trations of the unionized parts of such substances will be calculated, not from their ionization-constants, but from the ionization-values which are obtained froin the conductance-ratio or from the average values for salts of the same valence type. Prob. 20. Formulate the equilibrium equation for each of the fol- lowing reactions (taking place in dilute aqueous solution/ : C1 2 + H 2 = HC1 + HC10; and Cl a -f H 2 O = H+ -f 01- -f HC1O. Show that both of these equilibrium equations cannot hold true, since the mass-action law does not apply to the simple ionization of largely ionized acids. Pro 6. 21. When a 0.06 molal solution of Cl s is allowed to stand at till equilibrium is reached, 31% of the chlorine is converted into hypochlorous acid and hydrochloric acid. a. Calculate the equilibrium- constant of the reaction, assuming that the ionization of hydrochloric acid is 95%. 6. Calculate what the initial concentration of the chlorine must be in order that 50% of it may be converted into hypochlorous and hydrochloric acids in the equilibrium mixture. 87. The Hydrolysis of Salts. When either the acid or base of a salt has a very small ionization-constant, the salt in aqueous solution reacts with the water to an appreciable extent with formation of the acid and base. Thus, potassium cyanide in 0.01 normal solution at 25 is decomposed to an extent of 3.5% according to the reaction K + CN- + H 2 O = K+OH- + HON.* This phenomenon is called hydrolysis; and the fraction of. the salt hydrolyzed is called the degree of hydrolysis, or simply the hydrolysis (ft)- The equilibrium- constant of a hydrolytic reaction is called the hydrolysis-constant of the salt. *As is done in this case, it is convenient to indicate largely Ionized sub- stances, whose ionization does not conform to the mass-action law. by attacbine -^ and signs to their ions, and .to omit such signs in the case of slightly ionized substances, whose ionization conforms to that law. 82 EQUILIBRIUM OF CHEMICAL REACTIONS Prob. 22. a. Derive a mass-action expression showing how the hydrolysis of a salt like potassium cyanide varies with the concentra- tion of the salt. b. From the data given in the preceding text calculate the degree of hydrolysis and the concentration of free base in a 0.1 normal KCN solution at 25. Assume that the ionizations of the salt and free base are equal. Prob. 28. Derive a mass-action expression showing how the hydrol- ysis of a salt like ammonium cyanide, of which the acid and base are both slightly ionized, varies with the concentration and ionization of the salt Pro 6. 24> In 0.1 formal solution at 25 ammonium cyanide is 40% hydrolyzed. What is its hydrolysis in 0.01 formal solution? For the ionization values refer to Art. 86. Prob. 25. Show how the hydrolysis-constants of salts like potas- sium cyanide and ammonium cyanide are related to the ionization- constants of water (K w ), of the acid (J A ), and of the base (^ B ). Prob. 26. Calculate the hydrolysis of ammonium acetate in 0.1 formal solution at 25 from the ionization-constants involved. (The ionization-constants of acetic acid and of ammonium hydroxide have the same value (1.8 X 10- 6 ) at 25.) Prob. 27. At 100 the ionization-constant of water is 50 x 10-"; that of acetic acid is 1.1 x 10~ 5 ; and that of ammonium hydroxide is 1.4 X 10~*. Calculate the hydrolysis of ammonium acetate in 0.1 formal solution at 100, and compare it with that at 25. Assume the ioniza- tion of the salt to be the same as at 25. Prob. 28. Determination of Hydrolysis by Distribution Experi- ments. A 0.05 formal solution of Na 2 NH 4 PO 4 in water is found to be in equilibrium with a 0.00173 formal solution of NH 3 in chloroform at 18. What is the hydrolysis of the salt? The distribution-ratio for NH 8 between water and chloroform at 18 is 27.5. Prob. 29. Determination of Hydrolysis by Reaction-Rate Experi- ments. The specific reaction-rate at 100 of the sugar hydrolysis has been found to be 0.0386 in a solution 0.001 normal in HC1, and to be 0.0946 in a solution 0.01 formal in A1C1 8 . What fraction of the salt is hydrolyzed (into Al(OH), and 3HC1)? Prob. 80. Determination of Hydrolysis by Conductance Measure- ments. The specific conductance at 100 of a 0.025 formal solution of NH 4 Ac (ammonium acetate) is 0.00685 reciprocal ohms; and that of a solution 0.025 formal in NH 4 Ac and 0.025 formal in NH 4 OH is 0.00717 reciprocal ohms. a. Calculate the hydrolysis of the ammonium acetate in the first solution, assuming that in the second solution the hydrol- ysis of the salt and the conductance of the base are both negligible. b. Calculate the hydrolysis of the salt in the second solution, c. Calcu- late the specific conductance of the free base in the second solution. The A -value for NH 4 OH at 100 is 647. BETWEEN DISSOLVED SUBSTANCES 83 88. Displacement of One Acid or Base from its Salt by Another. One of the most important types of equilibrium in aqueous solution is the partial displacement of one acid or base from its salt by an- other ; for example, that of acetic acid from sodium acetate by formic acid, or that of ammonium hydroxide from ammonium chloride by sodium hydroxide. This phenomenon has been studied experimentally by the methods illustrated by Probs. 35-37. Before the mass-action relations involved were fully understood, the extent of the displace- ment was taken as a measure of the relative strengths of different acids or bases; those which are largely displaced from their salts being called weak acids or bases, and those which cause such displace- ment being called strong acids or bases. It is shown by Probs. 31- 33 that the mass-action law and ionic theory give a comparatively simple explanation of this phenomenon in the case of not largely ionized univalent acids or bases; also that the relative strengths of different acids or bases are determined by their ionization, weak ones being those which are slightly ionized and strong ones those which are largely ionized. Proft. 31. To a 0.1 normal solution of KNO 2 is added at 25 an equal volume of a 0.1 normal solution of acetic acid. a. Calculate the fraction of the potassium nitrite that is converted into potassium ace- tate. &. Calculate the fraction that would be so converted if the acetic acid solution were 1.0 normal (instead of 0.1 normal). Pro 6. 82. a. For the general case expressed by the equation B+A- + HA' B+A'- + HA where the solution is originally c-forinal in B+A- and cK-formal in HA', derive an expression for the fraction x of the salt B+A- converted into the salt B+A'-, assuming that the acids are not largely ionized. 6. Derive from this expression the relation be- tween the ionization-constants of the acids and the fractions of the basic constituent that are combined with the two acidic constituents for the case that c = c' ; and state the principle fully in words. Pro&. 33. To a liter of 0.1 normal HC1 is added at 25 a liter of 0.1 normal ammonia (K = 0.000018) and a liter of 0.1 normal methyl- amine (K 0.00050). Calculate the fraction of the acid which com- bines with each base. Prob. 34. A 0.1 formal solution of acetic acid is added to an equal volume of 0.1 formal NaHCO 8 solution at 25, the carbon dioxide pro- duced being kept above the solution at a pressure of 1 atm. Calculate the concentrations of the two salts and two acids in the resulting solu- tion (looking up the data needed). Compare this result with that calculated for the case where no carbon dioxide is allowed to escape from the solution. 84 EQUILIBRIUM OF CHEMICAL REACTIONS Pro&. 35. Determination of Displacement ~by Volume Measure- ments. When 1000 g. of a solution containing 1KOH is mixed with 1000 g. of a solution containing 1CH 8 CO 2 H there is an increase of volume of 9.52 ccm. When the former solution is mixed with 1000 g. of a solution containing 1HCO 2 H there is an increase of volume of 12.39 ccm. When 2000 g. of a solution containing 1HCO 2 K is mixed with 1000 g. of one containing 1CH 3 CO 2 H there is a decrease of volume of 0.74 ccm. Determine what fraction of the formic acid is displaced from its salt by the acetic acid. Compare this result with that calcu- lated from the ionization-constants. Note. The displacement may also be determined experimentally by measuring the quantities of heat evolved, instead of the changes of volume, when solutions are mixed as in the preceding problem. Prol). 36. Determination of Displacement ly Color Measure- ments. Paranitrophenol is a slightly ionized acid whose solution is colorless, while solutions of its salts have a yellow color which in- creases proportionately with the quantity of salt. 25 ccm. of a 0.01 normal solution of paranitrophenol are placed in each of two tubes of the same diameter ; to one tube are added 25 ccm. of a 0.01 normal solution 'of potassium acetate; and to the other tube is added 0.01 normal KOH solution until on looking down through the tubes the colors are seen to be the same, 1.65 ccin. of the KOH solution being required. Calculate the fraction of the paranitrophenol which exists in the first mixture in the form of its salt; and calculate the ionization- constant of the paranitrophenol. Prol. 37. Suggest a method (different from those already men- tioned) by which the displacement of ammonium hydroxide from am- monium chloride by sodium hydroxide might be determined. 89. Neutralization-Indicators. An acid neutralization-indicator (such as litmus, paranitrophenol, or phenol phthalein) is a mixture of two isomeric acids* (Hln' and HIn") in equilibrium with each other, one of which (HIn") is present in much smaller proportion but is so much more ionized than the other (HIn') that when a base BOH is added the salt produced is almost wholly of the form B + In"~. The substances HIn' and B + In'~ are different in color from the substances HIn" and B + In"~, the color being determined only by the molecular structure of the group In. It follows from these conditions that the *Thus the formulas commonly assigned to phenol phthalein and its potas- sium salt are : C 8 H 4 C and r B H 4 C CO O C H * OH COOK X CfiH4 = BETWEEN DISSOLVED SUBSTANCES 85 indicator acid changes color when converted into its salt. It will also be evident from Prob. 38 that the indicator behaves as if it were a single acid Hln whose salt has a different color from the acid itself. Prob. 38. Show by formulating the equilibrium equation that in the case of an indicator such as has been described r= const. (Hln') Note. This constant, which is a product of two equilibrium-con- stants, will be called the indicator-constant K lt Pro 6. 39. Derive a relation between the hydrogen-ion concentration (H + ), the indicator-constant K^ jEnd the fraction a? of the indicator which is converted into the colored form existing in alkaline solution. (This fraction will hereafter be called simply the fraction of the in- dicator transformed.) In indicator problems the ionization of the indi- cator salt, as well as that of any other salt present, may be regarded as complete, since only approximate results are ordinarily desired. The relations in the case of basic indicators (existing almost wholly in the two differently colored forms In'OH and In" + A~) are entirely similar to those in the case of acid indicators. By formulat- ing the mass-action equations, it can, in fact, be shown that the relation derived in Prob. 39 between the hydrogen-ion concentration and the fraction (#) of the indicator transformed holds true also in the case of basic indicators. By adopting as the value of the indicator- constant that calculated by the equation K\ = a basic indi- 1 x cator may therefore be treated as an acid indicator; and it will be so treated in the following problems. Prob. 40. Determination of the Indicator-Constant. To 100 ccrn. of a solution 0.1 normal in K>AI,-. Calculate its solubility in 0.01 normal KI solution at 25. Pro ft. 53. Hypochlorous acid is prepared by the action of chlorine solution on solid mercuric oxide. What is the expression for the equilib- rium of the reaction? 93. Solubility Effects in the Case of Largely Ionized Sub- stances. The mass-action law evidently requires, in all dilute solu- tions saturated with the same solid substance, that the concentration of the unionized substance have the same value, and also that the product of the concentrations of its ions have the same value, what- ever other substances may be present (at small concentrations) in the solution. Owing, however, to the considerable deviations from the mass-action law (referred to in Art. 86) which largely ionized substances exhibit even at moderately small concentrations, it is clear that one or both of these principles must be inexact when ap- plied to such substances. Experiment has shown that, in tbis case as in others, a fairly good agreement with the facts is secured (up to about 0.2 normal) by employing tbe principle involving the ions that expressing tbe constancy of the ion-concentration product. In solutions saturated witb tri-ionic substances, such as silver sulphate or calcium hydroxide, the relations are further complicated, except in very dilute solutions, by the probable presence in consid- erable proportion of intermediate ions, such as AgSO 4 ~ or CaOH*. Owing to lack of knowledge of the proportion in which such ions are present, the principle of the constancy of tbe ion-concentration product (for example, of the products (Ag + ) a X (SO 4 = ) and (Oa i+ ) X (OH") a ), combined with tbe assumption that the conductance-ratio 90 EQUILIBRIUM OF CHEMICAL REACTIONS gives the ionization of the salt into the simple ions, can be used with fairly satisfactory results only when the solubility of the solid sub- stance and the concentration of any other substance present are both very small.* Solubility-Decrease by Substances with a Common-Ion. Pro b. 54' 0" Derive an expression for the solubility s of silver chloride in a dilute sodium chloride solution of concentration c in terms of its solubility s in water, assuming the salts to be completely ionized, b. Calculate the ratio s/s for c = s , fo~ c = 2s , and for c = 106v Prob. 55. The solubility at 25 of thallous chloride (T1C1) is 0.0161 normal. Calculate its solubility in a 0.05 normal solution of potassium chloride. Estimate the ionization-values involved with the aid of the average values for uniunivalent salts given in Art. S6. Compare the calculated solubility with the value (0.00592 normal) found by experiment. Prob. 56. The solubility of magnesium hydroxide in water at 1S is 1.4 x 10-' formal, a. Calculate its solubility in 0.002 formal NaOLJ solution. 6. Calculate its solubility in 0.001 formal MgCl 2 solution. (Assume that the substances are completely ionized.) Prob. 57. Solubility-Increase through Complex-Formation. The solubility of silver chloride in water at 25 is 1.1 x 10~ B formal. Cal- culate its solubility in 0.1 formal NH 3 solution. There is formed a complex ion by the reaction Ag + -|-2NH 3 = Ag(NH 8 ) 2 + , its equilibrium- constant (commonly called the complex-constant) having the value 1.4 x 10 7 . Assume the ionization of the salts to be complete, and make any other justifiable simplification. Prob. 58. Solubility-Increase through Metathesis. The solubil- ity of magnesium hydroxide in water at 18 is 1.4 x 10~ 4 formal. Cal- culate its solubility in 0.002 formal NH 4 C1 solution. Assume that the salts and the magnesium hydroxide are completely ionized. Neglect the concentration of OH- in comparison with that of NH 4 OH. Conversion of One Solid Substance into Another. Prob. 59. The solubility of silver thiocyanate is 1.2 x iO' 6 formal and that of silver bromide is 0.6 x 10-' formal at 25. a. Calculate the equilibrium-constant of the reaction AgSCN -f K+Br~ = AgBr -j- K>SCN- *How small these concentrations must be in order that the solubility in the presence of an added salt may be calculated from that in water with an accuracy of a few percent depends on the valence type of the added substance. When this substance has the same univalent ion as the substance saturating the solution (e. g., when Ag+NO 3 - is added to Ag+ 2 SO 4 =, or Na+OH- to Ca+-MOH-) 2 ), the total concentration may be as high as 0.05 normal; but when the added substance has the same bivalent ion (e. g., when K 2 +SO4= is added to Ag+ 2 SO 4 = or Ca++(NO 3 -) 2 to Ca+ J (OH-) 2 ) the total concentration must not be greater than 0.002 normal. INVOLVING SOLID SUBSTANCES 91 in dilute solution. 6. If 8.3 g. of solid silver thiocyanate are treated with 200 cc. 0.1 formal KBr solution, what proportion of the silver salt is converted into bromide? c. What volume of the 0.1 formal KBr solution would convert the silver thiocyanate completely into bromide? d. With what solutions of potassium thiocyanate and potas- sium bromide could the silver thiocyanate be treated without any change taking place? Pro 6. 60. Determine the ratio of carbonate to hydroxide in the solution obtained by digesting at 25 a 0.1 formal Na,CO, solution with excess of solid Ca(OH) a (as in the technical process of causticizing soda). The solubility of calcium hydroxide is 0.020 formal, and the product (Ca~) x (CO,=) has the value 3 x 10~ 9 in water saturated with calcium carbonate. Assume that the substances are completely ionized. CHAPTER VII EQUILIBRIUM OF CHEMICAL SYSTEMS IN RELATION TO THE CHARACTER OF THE PHASES PRESENT 94. Fundamental Conceptions. In this chapter are considered the principles relating to the number, state of aggregation, and composi- tion of the phases (defined as in Art. 23) which coexist in equilibrium with one another when a given substance is subjected, or when mix- tures in various proportions of two or more given substances are subjected, to different temperatures and pressures. The kinds of phenomena to be considered are illustrated by the following examples. The state in which the substance water exists is determined by the temperature and pressure thus, whether it exists in the form of a single phase as ice, as liquid water, or as vapor; in the form of two phases as ice and liquid water, as ice and vapor, or as liquid water and vapor; or in the form of the three phases, ice, liquid water, and vapor. So also in the case of two substances, such as carbon bisulphide and acetone, there are, as shown by the vapor- pressure-composition and boiling-point-composition diagrams of Arts. 32 and 33, definite conditions of pressure and temperature at which any definite mixture of the two substances forms two phases a liquid phase and a vapor phase; and under these conditions the composition of each phase is also definite. Certain fundamental conceptions may first be presented. The combination of matter under consideration is called the system. A system is therefore defined when the nature and quantity of the substances of which it is composed are specified. Throughout this chap- ter are to be considered the conditions which determine the state of systems in equilibrium composed of the same kinds of substances in various proportions; that is, of a series of systems having the same qualitative, but varying quantitative composition. Any pure substances (Art. 2) which, put together in suitable propor- tions, will produce each and every phase with such varying composition as it may have in the systems under consideration are called the components, the pure substances being so chosen that the systems can be produced out of the smallest number of them. 92 FUNDAMENTAL CONCEPTIONS 93 Any phase may be produced out of the components either directly or as a result of equilibria established between them. For example, a gaseous phase containing hydrogen, oxygen, and water-vapor at high temperatures where chemical equilibrium is established between these substances can be produced by putting together only hydrogen and oxygen; and the phase is therefore said to consist of these two com- ponents. Such a gaseous phase at low temperatures, where no chemical reaction takes place between the hydrogen and oxygen, can be produced only by putting together with these two substances also water ; so that in this case the phase is said to consist of these three components. It will be clear from this example that the number of components is determined not only by the substances present, but also by the equi- libria which are established between them. Another aspect of the matter may be considered. In a gaseous phase where hydrogen, oxygen, and water are in chemical equilibrium, one particular system could be produced by taking only the one substance water; but the phase considerations of this chapter have reference always to a series of systems of the same qualitative, but varying quantitative composition; and such a series containing hydrogen, oxygen, and water-vapor in every possible proportion cannot be produced out of water alone, but can be produced out of hydrogen and oxygen. As another example consider a gaseous phase containing HC1, O 2 , C1 2 , and H,0. By taking HC1 and O 2 as two components, the other two substances can be produced out of them by the reaction O, -h 4HC1 = 2C1 2 + 2H 2 O, but only in equivalent proportions. To produce a system containing the four substances in any proportion whatsoever, it is necessary to make use of a third component, either Cl, or H 2 O. Thus, either by adding Cl, to a mixture containing O 2 and HC1 in any proportions and C1 2 and H 2 O in equivalent propor- tions, or by removing C1 2 from such a mixture, any composition what- ever can evidently be secured. The systems are therefore said to consist of three components. In the above description O 2 , HC1, and C1 2 have been used as the components; but evidently any other three of the four substances might be equally well employed. It is, how- ever, preferable to select as components substances of as simple a composition as possible; thus in this case, O 2 , HC1, and C1 2 rather than O 2 , HC1, and H 2 O. 94 PHASE EQUILIBRIUM IN CHEMICAL SYSTEMS Profc. 1. With the aid of the preceding considerations, specify the components which will produce phases containing the following chemical substances, assuming chemical equilibrium to be established between them. In answering this question, write chemical equations showing how all the chemical substances can be produced from the pure substances used as components, a. Gaseous H 2 , O 2 , CO, CO 2 , H 2 O. 6. Solution contain- ing H 2 O, (H 2 O) 2 , NaC12H 2 O, NaCl, Na + , C1-. c. Solution containing H 2 O, NH 4 CN, NH 4 % CN~, NH 8 , NH 4 OH, HCN. .Proft. 2. Specify the components of the systems that exist in the following groups of phases, assuming chemical equilibrium to be estab- lished between the substances named : a. Solid NH 4 C1, gaseous NH 8 and HC1. 6. Solid CaCO 8 , solid CaO, gaseous CO 2 . c. Solid MgSO 4 .7H 2 O, solid MgSO 4 .6H 2 O, solution of Mg+^SCV in water, water- vapor, d. Solid carbon ; gaseous H 2 O, H 2 , CO, and CO 2 . e. Solid iron, solid FeO, gaseous CO and CO 2 . /. Solid iron, solid FeO, solid carbon, gaseous CO and CO 2 . It is a fundamental law of the equilibrium between phases that the absolute quantity of the different phases does not influence their com- position. Thus the composition of a solution in equilibrium with a solid salt is not dependent on the quantity of the solid in contact with the solution. The composition of the vapor in equilibrium with a definite liquid solution is not dependent on the quantities of liquid and vapor in contact with one another. The proportion of hydrogen and water-vapor in equilibrium with solid iron and solid ferrous oxide is not dependent on the quantities of these solid phases in contact with the gaseous phase. The rate at which equilibrium is established between phases is, however, greatly increased by increasing the extent of the surfaces between them. A system is determined by its composition, that is, by the quanti- ties of its components, as described in the preceding paragraphs. The state of a system is determined when the nature and quantity of each of its different phases are specified and when the specific properties of each phase, such as its density, specific conductance, index of refrac- tion, etc., have definite values. In order to determine fully the specific properties of any one phase of a system, it is necessary to specify, in addition to the proportions of its components, any external factors which affect these properties. The only external factors which com- monly have an appreciable influence are the pressure and temperature ; and, in the following considerations relating to the equilibrium of phases, these factors alone are taken into account, and their values are assumed to be uniform throughout all the phases of a system. ONE-COMPONENT SYSTEMS 95 Briefly stated, the purpose of this chapter is to show how the nature, the quantity, and the composition of the phases of a system may be represented when the composition of the system as a whole and the pressure and temperature are given. Systems are classified according to the number of their components. In this chapter one-component systems are first considered; then a general principle, known as the phase rule, applicable to systems with any number of components, is presented; and finally two-component and three-component systems are discussed. ONE-COMPONENT SYSTEMS 95. Representation of the Equilibrium-Conditions by Diagrams. In the case of one-component systems the conditions under which the different phases exist in equilibrium with each other are con- veniently represented by pressure-temperature diagrams; for the state of any phase of such a system is evidently determined when the pressure and temperature are specified. 0.05 , 0.04 0.03- 0.02 0.01 RHOMS/G H MO/VOCL //V/& 7 _ u _--V c /G 90 95 100 105 110 FIGURE 7 115 120 125 Figure 7 shows a part of the temperature-pressure diagram for the component sulphur, which forms not only liquid and gaseous phases, but also two solid phases, known from their crystalline forms 96 PHASE EQUILIBRIUM IN CHEMICAL SYSTEMS as rhombic and monoclinic sulphur. In the diagram the (vapor- pressure) curve AB represents the pressures at which rhombic sul- phur and sulphur-vapor are in equilibrium at various temperatures; the (vapor-pressure) curve BC represents the pressures at which monoclinic sulphur and sulphur- vapor are in equilibrium at various temperatures; and the (transition-temperature) curve BE represents the temperatures at which monoclinic and rhombic sulphur are in equilibrium at various pressures, Temperatures, like these, at which two solid phases are in equilibrium with each other are called transi- tion-temperatures. The point of intersection B of these three curves shows the only temperature and pressure at which rhombic sulphur, monoclinic sulphur, and sulphur-vapor are in equilibrium with one another. A point, like this, at which three phases coexist is called a triple point. The (vapor-pressure) curve CD represents the pressures at which liquid sulphur and sulphur-vapor are in equilibrium with each other at various temperatures; and the (melting-point) curve OF repre- sents the temperatures at which monoclinic sulphur and liquid sulphur are in equilibrium at various pressures. The point C is evi- dently a second triple point at which monoclinic sulphur, liquid sulphur, and sulphur-vapor co-exist. As indicated on the diagram, the fields between the different lines show the conditions under which the sulphur exists as a single phase. Profc. 3. Describe with the aid of the diagram the changes that take place, a, when sulphur is heated in an evacuated tube in contact with its vapor from 90 to 125; 6, when sulphur is allowed to cool from 125 to 90, the pressure being kept constant at 0.04 mm. The curves BE and OF in Figure 7 are very nearly vertical lines ; for increase of pressure always produces a relatively small change in the transition or melting temperature. Thus the transition-tem- perature of the two solid forms of sulphur increases about 0.04, and the melting-point of monoclinic sulphur increases about 0.03, per atmosphere of pressure. With certain substances the effect of pressure is to decrease the transition or melting temperature; thus the melting-point of ice is lowered by 0.0076 by an increase of pres- sure of one atmosphere. ONE-COMPONENT SYSTEMS 97 Prob. 4. At what temperature and pressure is there a triple point in the system composed of water? The freezing-point at 1 atm. is 0. For the other data needed refer to Art. 34 and the preceding text. 96. Unstable Forms. Prob. 5. a. To what equilibria do the curves BG, GO, and GH and the point G in Figure 7 correspond? b. Considered with reference to these equilibria, to what form of sulphur do the fields EBGH, HGCF, and GBC correspond? c. In what sense are these equilibria unstable? Prob. 6. Draw a sulphur diagram extending to pressures above the triple-point between rhombic, monoclinic, and liquid sulphur, which lies at 151 and 1280 atm. Prob. 7. It will be noted that the unstable form at any temper- ature has the greater vapor-pressure. Prove that this must be so by showing what would happen if the two forms were placed beside each other in an evacuated apparatus. Prob. 8. Prove by a similar consideration that the unstable form must also have the greater solubility in any solvent, such as carbon bisulphide. (Note that a substance is commonly present in the same molecular form in the solutions produced by dissolving its different solid forms. ) How great the tendency is for a substance to remain in the same form after passing through a melting-temperature or transition -tem- perature and thus to exist in an unstable form depends in large measure on the nature of the substance. The following general statements in regard to it can, however, be made. A crystalline solid cannot as a rule be heated appreciably above its melting- point; thus, ice always melts sharply at (under a pressure of 1 atm.). On the other hand, a liquid (like water) can ordinarily be cooled to a temperature considerably below the freezing-point if agitation and intimate contact with solid particles, especially with the stable solid phase, is avoided. Still more pronounced is the tendency of solid substances to remain in the same form upon being heated or cooled through a transition-temperature; thus rhombic sulphur can be heated to its melting-point (110), although this is about 15 higher than the transition-temperature (95.5) at which it should go over into monoclinic sulphur; and monoclinic sulphur can be cooled to room temperature without going over into the rhombic form, provided this be done quickly and without agi- tation. The rate at which an unstable phase goes over into the stable 98 PHASE EQUILIBRIUM IN CHEMICAL SYSTEMS one tends to increase with the distance from, the transition-tempera- tures ; but when the substance is below its transition-temperature this tendency may be more than compensated by the greatly reduced rate of reaction which a considerable lowering of temperature always pro- duces; thus white phosphorus is an unstable form, but the rate at which it goes over into the stable red form at room temperature is so small that it may be preserved unchanged for years; similarly, diamond is an unstable form of carbon at room temperature, but it does not go over into graphite or amorphous carbon. An effective means of causing an unstable form to go over into a stable one is to mix it intimately with the stable form. The transi- tion may be further accelerated by moistening the mixture of the two forms with a solvent in which they are somewhat soluble. These facts are made use of in the determination of transition-tempera- tures. Thus the transition-temperature of sulphur has been deter- mined by charging a bulb with a mixture of rhombic and monoclinic sulphur, filling it with carbon bisulphide and oil of turpentine, keep- ing it for an hour first at 95 and then at 96, and noting whether the liquid rose or fell in the capillary stem attached to the bulb. The volume was found to decrease steadily at 95 (owing to the transition of the monoclinic into the rhombic form), and to increase steadily at 96 (owing to the reverse transition), showing that the transition-temperature lies between 95 and 96. ProT). 9. Suggest an explanation of the catalytic action of the solvent in accelerating the transition of the sulphur. Pro 6. 10. Outline a method by which the transition- temperature of sulphur could be determined by quantitative solubility measurements. THE PHASE RULE 97. Concept of Variance. Profc. 11. If sulphur is kept at a specified pressure of 0.04 mm., at what temperatures is it stable, a, in a single phase as rhombic sul- phur, as monoclinic sulphur, and as liquid sulphur? 6, in two phases, as rhombic and monoclinic sulphur, and as monoclinic and liquid sulphur? c, in the three phases, rhombic sulphur, monoclinic sulphur, and sulphur- vapor? It will be noted that, in order to determine the position on the diagram and therefore the state of the system, the values of two THE PHASE RULE 99 determining factors, namely, the values of both the pressure and the temperature, must be specified when there is only one phase; that the value of only one of these factors, either the temperature or pressure, need be specified when any two phases coexist; and that no condition can be arbitrarily specified when any three phases coexist. The number of determining factors whose values can and must be specified in order to determine the state of a system consisting of definite phases and components is called its variance*; and, corre- sponding to the number of such factors, systems are said to be non- variant, univariant, bivariant, etc. It is evident from the preceding statements that when a one- component system consists of only one phase the system is bivariant, when it consists of two phases it is univariant, and when it consists of three phases it is nonvariant. In other words, the sum of the variance and number of phases is always three for a one-component system. Prob. 12. Discuss with reference to the principle just stated and to the sulphur diagram the possibility of the coexistence, a, of the three phases, rhombic, monoclinic, and liquid sulphur; b, of the four phases, rhombic, monoclinic, liquid, and gaseous sulphur. 98. Inductive Derivation of the Phase Rule. Pro 6. IS. In order that the specific properties of a liquid mixture of ethyl alcohol and water, such as its density, specific heat-capacity, or refractive index, may have definite values, we must evidently state not only that it is at some definite temperature and pressure (say 20 and 1 atm.), but also that it contains some definite proportion of alcohol or water (say 30% of alcohol). What is the variance of a system consist- ing of such a mixture? What are the number of phases and the number of components? Prob. 14. In the case of each of the following systems, state of what factors the values might be specified in order that all its specific properties may be fully determined; and state the number of compo- nents, the number of phases, and the variance of the system: a. Two solutions produced by shaking together water and bromine. 6. The two solutions of water and bromine and their vapors, c. The two solu- tions of water and bromine, their vapors, and ice. d. A solution of water, *Some authors use the expression degrees of freedom, Instead of the term variance. 100 PHASE EQUILIBRIUM IN CHEMICAL SYSTEMS ethyl alcohol, and acetic acid. e. A solution of water, ethyl alcohol, and acetic acid and their vapors. Prob. 15. a. Make a table showing the number of components, the number of phases, and the variance of each of the systems named in Probs. 11, 13, and 14. &. State the effect of increasing the number of phases in a system consisting of a definite number of components. c. State the effect of increasing the number of components in a system with a definite number of phases, d. Add to the table a column showing the sum of the number of phases and of the variance for each of the systems. These problems show that in every case the sum of the number of phases (P) and of the variance (V) is greater by two than the number of the components (0) ; that is, P + V = C -f- 2. This principle, which is called the phase rule, is a general one, applicable to systems consist- ing of any number of components and of any number of phases. The phase rule furnishes a basis for the classification of different types of equilibrium. It also enables the number of phases that can exist under specified conditions to be predicted. The usefulness of the phase rule itself is, however, often exaggerated. Of primary importance in the treatment of the equilibrium conditions of systems in relation to the phases present are the methods of representing those conditions by diagrams, as described in later articles of this chapter. ProT). 16. Sodium carbonate and water form solid hydrates of the composition Na 2 CO 3 .H 2 O, Na 2 CO 3 .7H 2 O, and Na 2 CO 3 .10H 2 O. a. How many of these hydrates could exist in equilibrium with the solution and ice under a pressure of 1 atm.? 6. How many of these hydrates could exist in equilibrium with water- vapor at 30 ? 99. Derivation of the Phase Rule from the Perpetual-Motion Principle.* In order to determine fully the state of one-phase systems consisting of any number C of components (as defined in Art. 94), evidently the composition of the phase and in addition any external factors that determine its properties must be specified. The compo- sition of the phase is fully determined by specifying the mol-fractions of all but one of the components, that is, by specifying 1 quantities. The only external factors which commonly affect the properties of a phase of specified composition are temperature and pressure; but in special cases certain other factors, some of which are mentioned below, have an appreciable influence- Representing the number of * This Article may be omitted in briefer courses. THE PHASE RULE 101 such external determining factors by n, the variance or total num- ber of quantities (independent variables) that must be specified is C 1 -f- n. In this case, therefore, where the number of phases P is one, the sum of the number of phases and the variance V is equal to -j" n > that is, P -f- V = C + n; or, for the common case where pressure and temperature are the only external determining factors, P -f- V = C + 2. For example, the state of one-phase systems con- sisting of three components (1, 2, 3) is fully determined by specifying four quantities, namely, the mol-fractions (x v and x 2 ) of any two of the components and the pressure (p) and temperature (T) ; hence any specific property whatever of the system, such as its density d, can be expressed as some function of these four variables, such as d = f (ajj, x z , p, T). In the general case, where the number of compo- nents is C and the number of external factors is n, the function becomes d i(x lt x 2 , . . x c _ v p, T, . .). The derivation of the phase-rule now consists in showing that the equation P -f- V = C + n, which for one-phase systems is a result of the definitions of components and of external factors, still holds true whatever be the number of phases. In order that this may be so, it is evidently necessary only that each new phase introduced into the system shall diminish the variance by one; for then P + V will still have its former value C + n. That this is the case can be shown as follows. Consider that a system of C components (1, 2, 3, . . C) exists as a gaseous phase, and that a new, liquid or solid, phase is developed in it (for example, by varying the pressure or temperature). Now the partial pressures (p lf p 2 , . . p c ) of the separate components in the gaseous phase are, like any other property of the phase, functions only of the mol-fractions (x lf x 2 , . . x c _J and of the external factors (p, T, . .). This conclusion may be expressed mathematically as follows :* P! = fi(#!, # 2 > X C 1> Pt T, . .) p 2 = f,(x lt x 2 , . . x c -i, P, T, . .). p c = f c (x v x 2 , . . x c . l9 p, T, . .). * In the case of perfect gases, functions of the simpler form Pi = a?i p hold true ; but the other variables a? 2 , T, . . affect pi when the gases do not conform to the perfect-gas law. 102 ' PtiASE E'QUlLlB'RWM IN CHEMICAL SYSTEMS When the new, liquid or solid, phase is present in equilibrium with the gaseous phase, the partial pressure of each component in the gaseous phase is determined by the mol-fractions (#/, #/, . . x' c _j) of the liquid or solid phase, by the temperature, and by any other external determining factor which has an appreciable influence ; for, if a liquid or solid phase of such composition as to be in equilibrium with the gase- ous phase could also be in equilibrium at the same temperature with some other gaseous phase with different partial pressures, perpetual motion of the kind described in Art. 28 could be realized. This conclu- sion from the perpetual-motion principle that the partial pressures must be fully determined by the mol-fractions in the liquid phase and by the external factors may be expressed mathematically as follows : P C = <*, ,.. a c . lt p, ,... By equating these two sets of expressions for p lt p 2 , ..p c , the following functional relations between the mol-fractions in the gaseous phase and those in the liquid phase are obtained : i,(x l9 x v . . x c . l9 p, T, . .) = f/, <, . . *Vi, P> T > -)- f 2 . Give the value of A A 7 in this expression for the change of state involved in each of the three preceding problems. Pro 6. 8. Calculate the work in calories produced when 1 mol of oxygen at pressure p and temperature T is heated at constant pressure through one degree. 110. Heat-Effects attending Changes in Temperature. A compara- tively simple kind of change of state is that which a system undergoes when its temperature is increased. The ratio of the quantity of heat dQ absorbed when its temperature rises from T to T -\- dT to the rise of temperature dT is called its heat-capacity ((7) at T ; that is, C = dQ/ 'dT. The heat-capacity is substantially equal to the quantity of heat absorbed when the temperature rises one degree. When the heat- ing takes place without change of pressure, the ratio dQ/dT is called the heat-capacity at constant pressure C p . When the heating takes place without change of volume, the ratio is called the heat-capacity at constant volume C v . The heat-capacity of any system is the sum of the heat-capacities of its homogeneous parts; and the heat-capacity of any such part is the product of its weight by the heat-capacity of one gram of it, which is called its specific heat-capacity C. The heat- capacity of one atomic weight, one mol, or one formula-weight of a pure substance is called its atomic, molal, or formal heat-capacity. FUNDAMENTAL PRINCIPLES 123 Profc. 9. When 100 g. of silver at 100 are immersed in 1000 g. of water at 15.000 the temperature rises to 15.475. Calculate the heat- capacity of this weight of silver and the specific and atomic heat-capacities of silver, assuming these quantities to be constant between 15 and 100. Prob. 10. The molal heat-capacity of oxygen at constant pressure at the temperature T is given by the expression M "C p = 6.50 -f 0.0010T. Calculate the heat absorbed in heating 2.24 liters of oxygen at and 1 atm. to 100, the pressure remaining constant. 111. Heat-Effects attending Changes in State at Constant Temper- ature. The heat withdrawn from or imparted to the surroundings when a change takes place in the state of a system which is kept at a constant temperature is experimentally determined by calorimetric measurements, which involve the principles illustrated by the follow- ing problem. Pro 6. 11. Into a calorimeter containing 50H 2 O at 20.00 1KC1 at 20.00 is introduced, and the temperature falls to 15.11. a. What change in state takes place in this process, considering the calorimeter to be a part of the system ; and what is the change in the energy-content of the system, neglecting the small quantity of work produced and any loss of heat by radiation? b. What change in state takes place when 1KC1 is dissolved in 50H 2 O at 20 ? c. In order to calculate the heat-effect attend- ing this change in state, what other change in state must be combined with that occurring in the calorimeter? d. State what additional data would be needed to calculate this heat-effect, and formulate an expres- sion by which the calculation could be made. e. Calculate the value of this heat-effect with the aid of such of the following data as may be needed: the heat-capacity of the calorimeter is 19 cal. per degree; the specific heat-capacity of solid potassium chloride is 0.166, of water is 1.00, and that of the solution of 1KC1 in 50H 2 O is 0.904 cal. per degree. Changes in state at constant temperature may take place either without change of volume or without change of pressure. The different heat-effects attending these two different changes in state are known as the heat-effect at constant volume Q and the heat-effect at constant pressure Q p . The heat-effect at constant pressure is the one which is usually called the heat of reaciion f and the one which is commonly recorded in tables of constants. There is difference of usage regarding the algebraic sign of the heat of reaction: when heat is actually evolved by the reaction, the heat of reaction is usually taken positive in thermo- chemical considerations, but negative in thermodynamic considera- 124 THERMOCHEMISTRY tions. In this book a uniform convention, corresponding to the thermo- dynamic one, is employed throughout; heat-effects being always con- sidered positive when heat is absorbed (as in the vaporization of water) and negative when it is evolved (as in the combustion of hydrogen and oxygen). 112. Changes in Energy- Content and in Heat-Content attending Changes in State. Having discussed the determination of the work and heat produced in the surroundings by processes involving a change in the state of a system, the corresponding change taking place in the energy-content of the system itself may be now considered. When the change in state takes place at constant volume, it is evident that no work is involved, and that therefore the heat absorbed from the V surroundings is equal to the increase U 2 U 1 in the energy-content of the system. When, on the other hand, the change takes place at constant pressure, there is not only a quantity of heat Q p withdrawn from the surroundings, but also a quantity of work equal to p (v 2 vj produced in them. The increase U 2 U^ in the energy-content of the system is then equal to Q p p (v 2 vj. The change in energy- content may thus be determined for changes in state which can be made to take place either at constant volume or at constant pressure. Instead of determining and recording the values of U 2 U l for various changes of state, it is generally more convenient to employ the values of the quantity (U 2 + p 2 v 2 ) (U t + p^ vj. The quan- tity U + p v, like the quantity U, is a property of the system which always has a definite value when the system is in a definite state, and which always changes in value by a definite amount when the system changes from one state to another, whatever be the process by which the change in state is brought about ; for it is evident that the pressure p and the volume v, as well as the energy-content U, have values which are determined by the state of the system. In other words, a law of initial and final states applies to the change in the quantity U + p v, just as it does to the change in the quantity U. For brevity, this quantity U -\- p v will be represented by a single letter H, and will be called the heat-content of the system, it being understood that this term is a purely conventional one which does not imply that the energy quantity denoted by it is a heat quantity, any more than the term energy-content implies it. FUNDAMENTAL PRINCIPLES 125 For any change in* state taking place by any process whatever, the change in heat-content may be found by subtracting from the heat- effect attending the process the work produced by it diminished by the increase in the pressure-volume product; that is, in general: = Q [W AO)]. For a change in state for which the initial and final pressures p, and p 2 have the same value p, the increase AH in the heat-content is equal to the heat-effect Q p when the change takes place at the constant pressure p; that is, it is equal to the heat of reaction commonly employed. Pro&. 12. Derive the two conclusions stated in the last two para- graphs of the preceding text. Pro&. 13. a. When the reaction 2 CO -f- O 2 = 2 CO 2 takes place without change of temperature or pressure in a system consisting of 2 mols of CO and 1 mol of O, at 20 and 1 atm., the heat-effect is 136,000 cal. What is the increase in the heat-content, and what is the increase in the energy-content, of the system? 6. When the reaction takes place in the same mixture without change of temperature or volume, the heat-effect is 135,420 cal. What is the increase in the heat-content and what is the increase in the energy-content in this case? c. In what re- spect does the final state of the system in a differ from that in 6, and what conclusion can be drawn from a comparison of the results obtained in a and & as to the change in energy-content and in heat-content that would attend the change from the final state in a to the final state in &? (It will be seen later that such a conclusion is justifiable only when the system consists of a perfect gas. ) 113. Expression of Heat-Effects by Thermochemical Equations. In order to express the changes in heat-content that attend changes in state, especially those involving chemical reactions, at any constant temperature and pressure, equations are conveniently employed in which the heat-contents of the various substances involved are repre- sented by their chemical formulas, and in which the change in heat- content is shown by placing a numerical term on the right-hand side of the equation. For example, the expression. Fe 2 3 + 3 CO = 2 Fe -f 3 C0 2 + 9000 cal. (at 20) signifies that at 20 and 1 atm. (this pressure being understood unless some other pressure is stated) the heat-content of one formula-weight of ferric oxide plus that of three formula-weights of carbon monoxide 126 THERMOCHEMISTRY is equal to the heat-content of two formula-weights of iron plus that of three formula-weights of carbon dioxide plus 9000 cal. ; 9000 cal. being the decrease ( A5T) in the heat-content of the system, which is equivalent to the heat evolved by the system when the reaction takes place at a constant temperature and pressure. Such expressions are called thermochemical equations, or specifically, heat-content equations. As indicated in the preceding equation, the fact that a substance is in the solid state is shown by black-face type, and the fact that it is in the gaseous state by italics. The fact that a substance is liquid is denoted by ordinary type, and the fact that a substance is dissolved in x formula-weights of water is shown by attaching the symbol #Aq to the formula of the substance. Thus the equation KOI + 100 Aq == KCllOOAq 4400 cal. (at 20) signifies that when at 20 one formula- weight of solid potassium chloride is dissolved in 100 formula-weights of water there is an in- crease of 4400 cal. in the heat-content of the system, corresponding to an absorption of 4400 cal. from the surroundings. When the sub- stance is dissolved in so large a quantity of water that the addition of more water produces no appreciable heat-effect, the symbol c Aq may be attached to the formula of the substance. These thermochemical equations can evidently be treated strictly as algebraic equations, and can be combined with one another by addi- tion or subtraction; for every quantity in them has a definite value (namely, that of the heat-content of the substance represented by the formula), irrespective of the other quantities that occur with it in the equations. Profc. 14. The union at 20 and 1 atm. of 1 g. of aluminum with oxygen is attended by a heat-evolution of 7010 cal. ; and the union of 1 g. of carbon with oxygen to form carbon monoxide is attended by a heat- evolution of 2420 cal. Express these data in the form of thermochemical equations ; and calculate from them the heat of the reaction A1 2 8 -{- 3 C = 2 Al -f 3 CO at 20. Prob. 15. a. Express the following data in the form of thermo- chemical equations, employing the conventions described in the preceding text : The heat of formation of 1 mol gaseous HC1 from the elementary substances is 22,000 cal. Its heat of solution in 100 formula-weights of water is 17,200. The heat of the reaction between 1 mol gaseous chlorine and a solution of 2 formula-weights HI in 200 formula-weights FUNDAMENTAL PRINCIPLES 127 of water, forming solid iodine and a dilute HC1 solution is 52,400 cal. The heat of solution of 1 mol gaseous HI in 100 formula-weights of water is 19,200 cal. &. By combining these equations calculate the heat of formation of 1 mol gaseous HI from gaseous hydrogen and solid iodine. Although it is true that only changes in heat-content can be deter- mined, yet it is convenient to employ an arbitrary scale of heat-content which has as its zero-points the heat-contents of the various elementary substances at the temperature under consideration, at a pressure of one atmosphere, and in the form which is most stable at this temperature and pressure. Under this convention the heat-content (T) of any com- pound substance is evidently equal to the increase in heat-content (AH) which attends its formation out of the elementary substances; and in any thermochemical equation the formula of a substance may evidently be replaced by the numerical value of its heat of formation. For example, the heat-content at 20 of one formula- weight of gaseous hydrogen bromide, or the numerical value of the formula ~\.HBr, is 8500 cal. ; for 8500 cal. is the heat absorbed when it is formed out of gaseous hydrogen and liquid bromine at 20 and 1 atmosphere. Similarly the heat-content of a potassium chloride solution represented by the formula KCllOOAq is 101,200 cal.; for this is the sum of the heat-effect ( 105,600 cal.) attending the formation of one formula- weight of solid KC1 out of solid potassium and gaseous chlorine at 20 and of the heat-effect (4,400 cal.) attending its solution in 25 formula- weights of water at 20. This example of the heat of formation of a solution, which may be expressed by the equation K -\- %C1 2 + 100 Aq = KCllOOAq + 101,200 cal., shows that in thermochemical equations the symbol #Aq (not attached to another formula) has, like the formu- las of elementary substances, the value zero; for the water represented by it has not been formed out of its elements. (Water which has been so formed is represented by the formula #H 2 O). It is evident that the employment of heats of formation greatly simplifies the task of determining and systematizing thermochemical data; for, instead of measuring and recording the change in heat- content attending every chemical reaction, it suffices to do this for the formation of every compound out of the corresponding elementary substances. The numerical values of the heats of formation so deter- mined may then be substituted in any thermochemical equation, and 128 THERMOCHEMISTRY the change in heat-content attending the reaction expressed by it may thus be calculated. Prol). 16. Calculate the heat-effect that attends at 20 the reaction PbS + 2PbO 3Pb 4- &O 2 , from the following heats of formation at 20 : PbO, 50,300 cal. ; PbS, 19,300 cal. ; 8O y 70,200 cal. Prob. 17. At 20 the heat of combustion of one mol of acetylene (C,H,) is 313,000 cal. Calculate its heat of formation. The heat of formation of 1H 2 O is 68,400 cal., and that of 1CO 2 from charcoal and oxygen is 96,600 cal. Prol). 18. a. State just what heats of formation are denoted by the second and third terms in the following equation : Zn -|- 2HClooAq = ZnCl,coAq 4. H 3 -f- 34,200 cal. (at 20). 6. Write the complete thermochemical equations which express these heats of formation, taking into account the facts that the heat of forma- tion of 1 mol of gaseous HC1 is 22,000 cal. and that its heat of solution in a large quantity of water is 17,300 cal. 114. Indirect Determination of the Heat-Effects of Chemical Changes. On account of radiation-errors the heat-effect can be directly determined by calorimetric measurements only for those chemical changes which take place completely within a few minutes, and for such changes only when the temperature is not greatly different from the room-temperature. It is, however, possible to calculate the heat-effects of many other changes from those which have been directly measured, by applying the law of initial and final states. This is illus- trated for changes at the room-temperature by the problems of the preceding articles and by the following problems, which are solved by combining the proper thermochemical equations in such a way as to eliminate the heat-content of all the substances except those involved in the reaction under consideration. The method commonly employed for determining heat-effects at temperatures much higher or lower than the room-temperature is described in the next article. Prob. 19. Calculate the heat of formation of one formula-weight of solid zinc hydroxide from the following equation and from the other necessary data, which have been given in preceding problems : Zn(OH), 4- 2HC1 Aq = ZnC^ooAq 4. 2H 2 O -f 19,900 cal. Prob. 20. Calculate the heat of formation at 20 of one formula- weight of H 2 SO 4 from the following data at 20 and those given in pre- ceding problems: the heat of solution in a large quantity of water of 1 mol gaseous SO, is 8000 cal., and that of 1 formula-weight H 2 SO, is 18,000 cal. One mol gaseous Cl, acting OD a dilute solution of 1 mol FUNDAMENTAL PRINCIPLES 129 SO, with formation of a dilute solution of HC1 and H a SO 4 produces a heat-evolution of 73,900 cal. (Note that when x formula- weights of H 2 O are involved in the chemical reaction the symbol a?H 2 O must appear In the thermochemical equation, even though o>Aq may also occur in it.) Profc. 21. A direct determination of the heat-effect of the reaction CH 8 CH,OH oo Aq -{- O 2 = CH 8 CO 2 H co Aq -f H 2 O is not practicable. State what measurements could be made which would enable this heat-effect to be calculated ; and show how it would' be calculated from the results of such measurements. Pro 6. 22. Suggest a series of chemical reactions from whose heat- effects, which must be readily determinable in a calorimeter, the heat of formation of Na 2 CO 3 .10H 2 O at 20 could be calculated. Write the thermochemical equations, and indicate how they would be combined to yield the desired result. Prob. 23. Calculate the heat-effects of the following reactions at room-temperature, referring to Landolt-Bornstein Tabellen for the heats of formation, of combustion, and of solution needed : a. H 2 8 -f 2Ag = Ag 2 S -f H 1). CgH^Oc (glucose) = 2C 2 H 6 OH -f 20O 2 . c. Zn -f- CuSO 4 400Aq = Cu -f- ZnSO 4 400Aq. 115. Influence of Temperature on the Heat-Effects attending Chemical Changes. The heat-effect at constant pressure of a chemical change taking place at any temperature can be derived from the heat- effect at constant pressure at any other temperature by the following consideration of two different processes resulting in the same change in state. In one process cause the chemical change (for example, the union of 1 mol of CO with mol O 2 ) to take place at a pressure p and temperature 2\, and heat the products (the carbon dioxide) under the pressure p to the other temperature T 2 ; and, in the second process, heat the reacting substances (the carbon monoxide and oxygen) under the pressure p from T l to T v and cause them to combine (forming carbon dioxide) at the pressure p and the temperature T v Since in each of these processes the system changes from the same initial state (1 mol of CO and \ mol O 2 at p and T,) to the same final state (1 mol C0 2 at p and T 2 ), the total change in heat-content must be the same in the two processes ; and therefore the heat-effect at constant pressure p and temperature T 2 must differ from the heat-effect at constant pres- sure p and temperature T^ by the same amount as the heat absorbed in heating the reaction-products differs from that absorbed in heating the reacting substances from T t to T, at the constant pressure p. 180 THERMOCHEMISTRY Pro&. 24> Demonstrate the truth of the principle stated in the pre- ceding paragraph by considering the increase in heat-content attending each step of the processes there described. Pro 6. 25. Calculate the heat of formation of one formula- weight of PbO at 200 from its heat of formation ( 50,300 cal.) at 20 and from the mean specific heat-capacities at constant pressure of lead (0.032), of oxygen (0.212), and of lead oxide (0.052). Profe. 26. Calculate the heat of formation of 1 mol gaseous water at 1000 from the following data. The heat of formation of 1 mol liquid water at 20 is 68,400 cal. Its heat of vaporization at 100 is 9670 cal. The molal heat-capacity at constant pressure at T is 6.50 + 0.0010 T for hydrogen or oxygen, and 8.81 0.0019T -f 0.000,002,221* for water. Pro&. 27. Calculate the heat of the reaction K (liquid) -f $Cl t KC1 at 160, referring to Landolt-Bornstein Tabellen for the necessary data. Proft. 28. a. State what quantity is equivalent to the change per degree of the heat-effect of a reaction at constant pressure. 6. Formu- late an integral which is a general expression for the difference In the heat-effects Q 2 and Q a of any reaction at two temperatures !F 2 and 2\, when no change takes place in the state of aggregation between those temperatures. RESULTS WITH PERFECT OASES 131 THE RESULTS OF THERMOCHEMISTRY 116. Constancy of the Heat-Content of Perfect Gases at Constant Temperature. Experiments have shown that when a perfect gas ex panels at a constant temperature without producing any work (for example, when it expands within a calorimeter from one vessel into another vessel previously evacuated), there is no heat-effect in the surroundings (no change of temperature in the calorimeter). Such experiments have established the important law that the energy- content U of a definite quantity of a perfect gas at any definite tem- perature has the same value, whatever be its volume and pressure; in other words, that AZ7 when a perfect gas changes its volume and pressure at a constant temperature. The sama,, principle is evidently true of the heat-content H of a perfect gas ; for this is by definition equal to U -j- p v, and p v does not change in value when a perfect gas changes its volume and pres- sure. Therefore, Aff = when a perfect gas changes its volume and pressure at a constant temperature. The law that the energy-content of a perfect gas at a constant tem- perature is independent of the volume leads to the conclusion that, when the expansion of a perfect gas at a constant temperature is attended by the production of work, there must be a quantity of heat absorbed by it equal to the work produced; for it follows from the equation AC7 = Q - - W that Q = W when AZ7 = 0. In the case of actual gases there are small deviations from these principles at moderate pressures, and large deviations at high pressures, in the direction corresponding to an increase in energy-content with increase of volume. This increase is most accurately determined by experiments, like that described in the following problem, in which the gas is caused to expand without producing work (except that equiva- lent to the change in its pressure- volume product) and without taking up heat from the surroundings. It undergoes thereby a decrease in temperature, which is often called from its discoverers the Joule- Thomson Effect. From this decrease in temperature and from heat- capacity data the quantity of heat which must be imparted to the gas to heat it in its expanded state to its original temperature is then calculated. 132 THERMOCHEMISTRY ProJ). 29. Determination of the Change in Heat-Content ~by Porous- Plug Experiments. Carbon dioxide at pressure p l (e.g., 2 atin.) and temperature T^ (e.g., 20.00) is caused to flow continuously through a well-insulated hardwood tube containing a porous plug of cotton. On passing through the plug its pressure falls to p 2 (e.g., 1 atm.), and it emerges from the tube at this pressure. After the gas has flowed so long that every part of the apparatus has assumed the temperature of the gas in contact with it, the expansion of the gas takes place without ex- change of heat with the surroundings. Its temperature after passing through the plug is found to be T 2 (e. g., 18.86). a. What other process must be combined with this one in order that the net result of the two processes may be the expansion of one mol of the gas from volume v t to volume v t at a constant temperature T x ? b. Formulate expressions for the work produced W, the heat absorbed Q, the change in energy-content A [7, and the change in heat-content A.H for each of these two processes. (Note that in the first process a volume v t of the gas disappears on one side of the plug under a constant pressure p lt and that a certain volume 17,' of the gas is produced on the other side of the plug under a constant pressure p 2 .) G - Combine these results so as to give an expression for the change in energy-content and the change in heat-content that attends a change in volume from t^ to t> 2 of one mol of the gas at a constant tem- perature T v d. Calculate in calories the change in energy-content and the change in heat-content attending the expansion of 1 mol of carbon dioxide from a pressure of 2 atm. to a pressure of 1 atm. at 20. Use the following data in addition to those given above: the molal volume of carbon dioxide at 20 and 2 atm. is 11,890 ccm., and at 20 and 1 atm. is 23,920 ccm. ; its molal heat-capacity at 20 and at a constant pressure of 1 atm. is 8.92 cal. per degree. Note. The decreases in temperature that have been observed in similar experiments with carbon dioxide at a series of temperatures are : 1.35 at 0, 1.14 at 20, 0.83 at 60, and 0.62 at 100 ; and that observed with air, a more nearly perfect gas, is 0.25 at 20. It will be noted that this last value signifies that, when a quantity of air expands at 20 from a pressure of 2 atm. to one of 1 atm., the increase in its heat-content is equal to the increase in its heat-content which takes place when it is heated at constant pressure through 0.25. This gives a general idea of the magnitude of the effect under consideration. 117. The Heat-Capacity of Perfect Gases in Relation to Pressure and Volume. From the law that the heat-content of a perfect gas at a definite temperature does not vary with changes in its pressure and volume, the following principles can be derived : the heat-capacity of a perfect gas both at constant pressure and at constant volume at a definite temperature has the same value whatever be the pressure and RESULTS WITH PERFECT OASES 133 volume; and the molal heat-capacity of a perfect gas at constant pres- sure is greater than that at constant volume by an amount equal in value to the gas-constant R, whatever be the gas and whatever be the temperature. Prob. SO. Derive the first principle stated in the preceding text by considering that a perfect gas changes by two different processes from a volume v^ at pressure PJ and temperature 2\ to a volume v a at pressure p 2 and temperature T r Pro 6. 31. Derive the second principle stated in the preceding text by a consideration similar to that employed in the last problem. 118. The Heat-Capacity of Perfect Gases in Relation to Compo- sition and Temperature. The molal heat-capacity of gases at constant volume depends primarily on the complexity of the molecules of the gas. It has the smallest value for gases with monatomic molecules, such as mercury and helium; and it has the same value, namely fR or 2.98 calories per degree, for all such gases at all temperatures. It has a considerably larger value for gases with diatomic molecules a value which is approximately the same for nearly all such gases, namely, for H,, O 2 , N 2 , NO, CO, HC1, HBr, and HI, and one which varies appreciably, but not very greatly, with the temperature. Its value for these gases at any temperature T is given by the expres- sion MV V = 4.50 -f- O.OOIOT". The corresponding expressions for the molal heat-capacity of perfect gases at constant pressure, which has been seen to be always_greater than that at constant volume by the gas-constant R, are MC p = 4.97 for monatomic gases, and MG p = 6.50 -j- 0.0010 T for most of the diatomic gases. A few diatomic gases, namely, C1 2 , Br 2 , I a , and IC1, have at room-temperature larger values of the heat-capacity than do the other diatomic gases, and the values increase more rapidly with the temperature ; thus, though these heat- capacities have not been satisfactorily determined, the incomplete data that exist may be expressed roughly by the equation MC p = 6.5 + 0.004 T. The only general statement that can be made in regard to the heat-capacities of triatomic and other polyatomic gases is that the values are much larger than those for the diatomic gases and that they increase with the complexity of the molecule; thus the value of MC p at 100 is 6.9 for 17, and O 2 , 8.3 for H 2 O and H a S, 9.4-9.9 for CO,, S0 2 , and N 2 0, 9.0 for NH., 21 for alcohol (C a H 6 O), and 34 for ether 134 THERMOCHEMISTRY (C 4 H 10 O). In the case of these polyatomic gases the increase of the heat-capacity with the temperature has to be expressed by quadratic or cubic functions; thus in the case of carbon dioxide and sulphur dioxide, MC P = 7.0 -f 0.0071T 0.00,000,186r 2 . The heat-capacity of water-vapor can be expressed by a similar function (see Prob. 26). Prob. 32. Determination of the Complexity of the Molecules of a Gas from the Heat-Capacity-Ratio. From the experimentally determined velocity of sound in a gas the heat-capacity-ratio C P /C V can be calcu- lated. This ratio has been thus found to be 1.67 in the case of argon. a. Compare this value with the values of the ratio calculated for a monatomic gas and for a diatomic gas at 20 with the aid of the state- ments made in the preceding text. &. Show how the atomic weight of argon can be obtained by combining this result with the value of another experimentally determined property of the gas. 119. Heat-Effects at Constant Pressure and at Constant Volume attending Reactions involving Perfect Gases. The law that the heat- content of a perfect gas at a definite temperature is independent of its pressure leads to the conclusion that in the case of reactions involving gases at small pressures the heat-effect Q p at constant pressure is greater than the heat-effect Q v at constant volume by an amount approximately equal to the work W produced when the reaction takes place at constant pressure. This work has already been shown to be approximately equal to A.ZV R T, where A7V denotes the increase in the number of mols of the gaseous substances present. The derivation and application of this principle are illustrated by the following problems. Pro 6. 33. Show that R T is the difference between the quantities of heat absorbed when at T a mixture of 2 mols of CO and 1 mol of O, at 1 atm. unites to form CO 2 in one case at constant pressure (for example, in an open calorimeter) ; and in another case at constant volume (for example, in a bomb calorimeter). Note that the change of state which results when the reaction takes place at a constant pressure of 1 atm. could also be brought about by a process consisting of two steps, namely, by causing the mixture of CO and O 2 at 1 atm. to change at constant volume to CO 2 (whereby the pressure would become atm.), and then compressing it till its pressure becomes 1 atm. Prob. 34. When 1 mol of naphthalene (C 10 H 8 ) is burnt with oxygen at 20 in a bomb-calorimeter 123,460 cal. are evolved. Calculate its heat of combustion at constant pressure. HEAT-CAPACITY OF SOLID SUBSTANCES 135 120. The Heat-Capacity of Solid Substances. Experiments have shown that the atomic heat-capacity of all solid elementary substances is relatively small at temperatures below 100, that with rising tem- perature it increases at different rates in the case of different sub- stances, and that, after room-temperature is reached, it increases as a rule only slowly with further increase of temperature. At room- temperature the atomic heat-capacity has approximately the same value in the case of all elements with atomic weights above 35, and in the case of the metallic elements of still lower atomic weight. The average value at room-temperature is 6.2 calories per degree. Deviations of db 0.4 unit are not uncommon ; and deviations of +0.7 to -J-0.9 unit are exhibited by some elements (for example, by sodium, potassium, and iodine) which at room-temperature are not much below their melting- points. In the case of the non-metallic elements with smaller atomic weights than 35 the atomic heat-capacity has a value much smaller than 6.2 at room-temperature; thus the value for boron is 2.6, for graphite 1.9, for silicon 4.8, for phosphorus 5.6, and for sulphur 5.5. This princip^ in regard to the approximate constancy of the atomic heat-capacities of solid elementary substances is known as the law of Dulona and Petit. The following table, which contains experimentally determined values of the atomic heat-capacity, illustrates the preceding statements : Element At Wt. 2^0 100 50 50 10CP 20CP 300 Aluminum 27.1 2.0 4.6 5.3 5.6 5.8 6.0 6.3 6.7 Antimony 120.2 4.2 5.4 5.7 5.9 6.0 6.1 6.2 6.3 Cadmium 112.4 4.3 RjR 6.1 6.2 6.2 6.3 6.8 9.4 Chromium 52.0 2.0 4.1 4.9 5.4 5.7 5.8 6.1 6.4 Copper F3.6 3.3 5.0 5.5 5.8 5.9 6.0 6.1 6.2 Iron 55.8 3.0 4.6 5.2 5.8 6.2 6.6 7.0 7.4 Lead 207.1 5.6 5.0 6.0 6.2 6.3 6.5 6.7 7.0 Magnesium 24.3 3.5 4.9 5.4 5.8 6.0 6.2 6.9 7.7 Silver 107.9 4.0 5.6 6.0 6.3 6.5 6.6 6.9 7.1 Sodium 23.0 5.1 5.0 6.3 6.7 7.1 7.5 Boron 11.0 . 1.6 2.0 2.4 2.8 3.2 4.0 .. Diamond 12.0 0.1 0.5 0.8 1.1 1.6 2.3 3.2 3.9 Sulphur 32.1 2.5 4.3 4.9 5.3 5.6 5.8 , . . . The law of Dulong and Petit, even though it is only an approxi- mate principle, may evidently be employed for determining what multiple of the combining weight of an element is its atomic weight; 136 THERMOCHEMISTRY and the application of this law was in fact one of the most important methods by which the present system of atomic- weight values was originally established. A simple principle has also been discovered in regard to the formal heat-capacity of solid compound substances at room-temperature. It has been found, namely, that this property is approximately an additive one, that is, one whose value can be approximately calculated by add- ing together certain values representing the heat-capacity of the ele- ments contained in the compound. This principle is expressed by the following equation, which shows at the same time the values of the constants (the so-called atomic heat-capacities) for all the common elements : MC P = 6.2 TIE + 4.0 no + 2.3 tin + 1.8 n c + 5.4 ns + 2.7 n& + 5.4 n P + 3.8 n S i. In tljis equation MC p represents the formal heat-capacity of the com- pound at constant pressure at room-temperature; no, nu, nc, n& KB* n P> an d n s\ are the number of atomic weights of oxygen, hydro- gen, carbon, sulphur, boron, phosphorus, and silicon present in one formula-weight of the compound; and ?? E is the number of atomic weights of any other element so present. The values given for the constants are average values derived from heat-capacity measurements with solid compounds. The following table illustrates the degree of correspondence which exists between the values of the formal heat-capacity so calculated and those measured experimentally : H 2 (ice) A1 3 0, . Cole. Meas. Substance Cole. Meas. 8.6 9.7 PbN 2 O 6 42.6 38.8 24.4 20.5 CaSiO 8 22.0 21.3 24.4 25.6 K 4 Fe(CN) fl 79.0 78.8 28.6 28.7 CuSO 4 .5H 2 O 70.6 67.2 12.4 12.4 A1K(S0 4 ) 2 .12H 2 158. 165. 18.6 18.5 C 10 H 8 36. 40. 20.0 20.2 H 2 C 2 4 24.2 25.1 Sb,S, KC1 PbCl, CaCO, It will be observed that differences of ten percent between the calcu- lated and measured values are not uncommon. Pro&. 35. Calculate an approximate value at 20 of the specific heat- capacity at constant pressure of a, platinum ; &, silver bromide ; c, potas- sium sulphate ; d, benzophenone, C^H^O. Find the percentage deviations CHANGES IN STATE OF AGGREGATION 137 of these values from the measured values, which are, a, 0.032 ; &, 0.074 ; c, 0.190 ; and d, 0.31. Prob. 36. Determination of Atomic Weights from Heat-Capacity Measurements. Calculate the exact atomic weight of an element whose specific heat-capacity is 0.092, and whose oxide contains 88.82% of the element. 121. Heat-Effects attending Changes in the State of Aggregation of Substances. The heat of vaporization of liquid substances, the heat of fusion of solid substances, and the heat of transition of one solid substance into another (as of rhombic into monoclinic sulphur) are quantities which are important in themselves and which are fre- quently involved in calculations of the heat of chemical reactions. The general statement can be made in regard to them that the conver- sion of the form that is stable at lower temperatures into that stable at higher temperatures (for example, of ice into water, or of rhombic into monoclinic sulphur) is always attended by a positive heat-effect, that is, by an absorption of heat. The following simple principle, known as Trouton's rule, has been discovered in regard to the values of the heat of vaporization: the ratio of the molal heat of vaporization of a liquid at its boiling-point to its boiling-point on the absolute scale has approximately the same value (namely, about 20.5) for all liquids except those whose molecules are associated; that is, M L/T = approx. 20.5. The actual values of these quantities in the case of five very different liquids are shown in the following table : Substance ML T Bromine 67GO 332 20.4 Benzene 7350 353 20.8 Carbon bisulphide 6380 319 20.0 Ethyl ether 6260 308 20.3 Ethyl formate 7J80 327 22.0 Substances containing the hydroxyl group, such as water, alcohols, and acids, whose molecules in the liquid state are for other reasons believed to be associated (see Art. 26, Note at end), form marked exceptions to Trouton's rule. Thus the value of M~L/T is 25.9 for water, 27.0 for ethyl alcohol, and 14.9 for acetic acid. The molal heat of vaporization of a liquid or solid substance can also be calculated by the approximate form of the Clausius' equation (Art. 22) from the change of its vapor-pressure with the temperature 138 THERMOCHEMISTRY The heat of solution of substances is another important quantity, In determining and expressing it the quantity of solvent in which a definite weight of the substance is dissolved must be taken into con- sideration. The two limiting cases are the heats of solution in a very large quantity of solvent and in that quantity of solvent which forms with the substance a saturated solution. These two heat-effects some- times have different signs. They evidently differ by the heat of dilu- tion of the saturated solution with a large quantity of water. The dissolving of gaseous substances in solvents is always, and that of liquid substances is usually, attended by an evolution of heat; and the dissolving of solid substances in solvents is usually attended by an absorption of heat. The heat of dilution of substances in solution is also important; for it enables the heat of formation of a solution of one concentration to be calculated from that of a solution of another concentration, and thus enables heats of reaction to be calculated at different concentra- tions. The heat-effect attending the addition of an equal volume of water to a concentrated aqueous solution is often large ; but it becomes less as the concentration diminishes; and, after a moderately small concentration (such as 0.2 formal) has been attained, there is usually only a very small heat-effect on adding even a very large quantity of water. For example, on adding at 18 to 1 formula- weight of gaseous HC1 or of solid ZnCl 2 successively AA T formula-weights of water, there are evolved the following quantities of heat ( Q} in calories: Atf 5 5 10 30 50 100 200 for HC1... 14960 1200 600 360 120 50 QforZnCl,.. 7740 1850 1300 2170 1490 820 390 Proft. 57. The heat of fusion of 1 g. of ice at is 79.7 cal. What other data would be needed to calculate the heat-effect attending the melting of 1 g. of ice into 1000 g. of a normal solution of sodium chloride at its freezing-point, 3.42, and how would the calculation be made? Prob. 38. The heat of solution at 20 in a large quantity of chloro- form of 1 at. wt. rhombic sulphur is -j-640 cal. and of 1 at. wt. monoclinic sulphur is -(-560 cal. Show by the law of initial and final states what other heat-effect can be derived from these data, and what its value is. Prob. 39. Change of Heat of Reaction with the Concentration. The heat of solution at 18 of IZn in HC1 200Aq is 34,200 cal. Find its heat of solution in HC1 5Aq, a, by applying the law of initial and final states, and &, by formulating the thermochemical equations involved. REACTIONS IN AQUEOUS SOLUTION 139 There is another kind of heat of dilution or solution that is of great importance in the thermodynamic treatment of electromotive force and of chemical equilibrium (considered in Chapters VIII and IX). This is the heat-effect attending the addition of one formula-weight of the solvent or solute to an" infinite quantity of the solution. The value of this quantity, which is known as the partial heat of dilution or solution, can be derived from actual data like those given in the preceding table in the ways illustrated by the following problems. Pro 6. 40. Determine with the aid of a plot the heat-effect attending the addition of 1H 2 O to an infinite quantity of ZnCl 2 100H s O at 18. Prob. 41. The measured heat-effect Q attending the mixing of 1H 2 SO 4 with ;VH 2 O at 18 is expressed by the empirical equation, Q = 17860 N/(N Jf- 1.80) for values of N not exceeding 20. Calculate the heat-effect attending the addition of 1H 2 O to an infinite quantity of H a SO 4 10H 2 O. Prob. 42. Derive a relation between the heat-effect attending the dissolving of !ZnCl 2 in 100H 2 O, that attending the addition of 100H 2 O to an infinite quantity of a solution of the composition !ZnCl 2 100H a O, and that attending the addition of IZnCl, to an infinite quantity of such a solution. Prob. 43. Calculate the heat-effect attending the addition of: a, !ZnCl 2 to an infinite quantity of lZnC! 2 100H 2 O. 6, 1H 2 SO 4 to an infinite quantity of 1H 2 SO 4 10H 2 O. Prob. 44. a. The heat-content of !ZnCl 2 is 97,300 cal. Calculate the heat-content of !ZnCl 2 in !ZnCl 2 100H 2 O (by which is meant the increase in heat-content in starting with IZn, 1C7 2 , and an infinite quan- tity of a solution of the composition ZnCl 2 100H 2 O, and ending with IZnClj in that solution), b. Give the values of the other symbols in the thermochemical equations : ZnCl 2 100H 2 O = ZnCl 2 (in ZnCl 2 100H 2 O ) -f 100H 2 O (in ZnCl 2 100H 3 O ) -f cal. ZnCl 2 100Aq = ZnCl 2 (in ZnCl 2 100H 2 O) -f 100Aq(in ZnCljlOOI^O) -f cal. 122. Heats of Reaction in Aqueous Solution. The investigations made of the heat-effects attending chemical reactions in aqueous solution between substances present at fairly small concentrations (0.1-0.3 normal) have established the following principles : 1. On mixing solutions of two neutral salts which do not form a precipitate by metathesis (for example, solutions of potassium chloride and sodium sulphate) there is scarcely any heat-effect. Exceptions to this principle are met with in the few cases in which an unionized 140 THERMOCHEMISTRY salt is produced by the metathesis. Thus the metathetical reaction 2K+C1- + Hg* + (NO 8 -) 2 = 2K*N0 3 - + HgCl 2 is attended by a heat- effect of 12,400 cal. 2. The heat of neutralization- of a solution of any largely ionized monobasic acid with a solution of any largely ionized monacidic base (for example, of hydrochloric acid, nitric acid, etc., with sodium hydroxide, potassium hydroxide, etc.) has approximately the same value, whatever be the acid or base. At 18 this nearly constant value averages 13,810 cal. per equivalent when the acid and base solutions are 0.12 to 0.25 normal. 3. When the base or acid is only partly ionized (as in the case of ammonium hydroxide or hydrofluoric acid), the heat-effect attending its neutralization with a largely ionized acid or base is often much larger or smaller than that observed when both acid and base are largely ionized; thus the heat of neutralization of one equivalent of ammonium hydroxide with one of hydrochloric acid is 12,300 cal., and that of one equivalent of hydrofluoric acid with one of sodium hydroxide is 16,300 cal. 4. When one formula-weight of a dibasic acid is neutralized in steps by adding first one equivalent of a largely ionized base and then a second equivalent, the heat-effects for the two equivalents of base are usually different; for example, at 18 the two heat-effects are 14,600 and 16,600 cal. in neutralizing 0.28 normal sulphuric acid with 0.28 normal sodium hydroxide, and they are 11,100 and 9,100 cal. in neutralizing carbonic acid with that base. 5. When certain polybasic acids are neutralized, there is sometimes scarcely any heat-effect when the second or third equivalent of base is added. Thus the successive heat-effects when phosphorous acid (H,PO S ) is treated with sodium hydroxide at 18 are: 14,800 cal. with the first equivalent, 13,600 cal. with the second equivalent, and only 500 cal. with the third equivalent ; and with hypophosphorous acid (H,PO a ) there is a heat-effect of 15,200 cal. with the first equiva- lent of sodium hydroxide, and only 110 cal. with the second equivalent. 6. With certain polybasic acids there is a considerable heat-effect when to the solution of the neutral salt another equivalent of base is added; thus, there is a heat-effect of 1200 cal. on mixing a solution REACTIONS IN AQUEOUS SOLUTION 141 containing one equivalent of sodium hydroxide with one containing one formula-weight of sodium phosphate (Na 3 PO 4 ). Frol. 45. a. Explain the first principle stated in the preceding text with the aid of the ionic theory, assuming that the solutions are very dilute. 6. What conclusion as to the heat of ionization of neutral salts can be drawn from the fact that this principle holds true even at fairly high concentrations (e. g., 0.3 normal)? c. Write a thermochemical equation corresponding to the ionic reaction to which the heat-effect is mainly due in the reaction cited as an exception to the principle. Pro&. 46. Show that on mixing solutions of two salts (such as lead nitrate and potassium iodide) which form a precipitate by metathesis there must be a heat-effect which is substantially equal, but opposite in sign, to the heat of solution of the precipitated substance (the lead iodide) . Pro&. 47. a. Write the ionic reaction to which the nearly constant heat of neutralization of largely ionized acids and bases corresponds. &. What other heat-effect is involved in the neutralization of ammonium hydroxide (a slightly ionized base) with a largely ionized acid, and what is its value? c. What is the heat-effect that attends the reaction between 1NH 4 C1 and INaOH in 0.2 normal solution? Prol). 48. Calculate the heat of ionization of one formula-weight of HF from the facts that its heat of neutralization in 0.28 normal solution with 0.28 normal NaOH solution has been found to be 16,300 cal., and its ionization in 0.28 normal solution is estimated to be 5.2%. Pro&. 49. Calculate the heats of ionization at 18 that can be derived from the heats of neutralization of carbonic acid given in the text. Carbonic acid is a very slightly ionized acid ; sodium hydrogen carbonate solution is practically neutral; and the sodium carbonate in the 0.07 formal solution produced by the neutralization is 6.3% hydrolyzed. Pro 6. 50. Conductance, transference, and reaction-rate measurements have shown that sodium hydrogen sulphate in 0.1 formal solution at 18 consists approximately of 8% NaHSO 4 , 12% Na 2 SO 4 , 44% HSO 4 -, 36%-SO 4 =, and of the corresponding amount of Na+ and H + . A calorimetric measurement at 18 has given the result expressed by the equation : NaHS0 4 601Aq + NaOH200Aq = Na 2 SO 4 801Aq -|_ H 2 O + 16,620 cal. Calculate the heat of the reaction HSO 4 - = H + 4. SO 4 =, assuming that the heats of ionization of Na 2 SO 4 and of NaHSO 4 (into Na* and HSO 4 -) are zero. Pro 6. 51. What do the heat-effects observed in the neutralization of phosphorous acid and of hypophosphorous acid show as to the existence of salts of these acids in solution? Pro&. 52. Explain the fact stated in the last paragraph of the pre- ceding text. 142 THERMOCHEMISTRY 123. Applications of Thermochemical Principles. Determination of the Change of the Heat of Reaction with the Temperature. Prob. 53. Derive a numerical expression for the heat of ionization of water as a function of the temperature from the following data. The heat evolved on mixing at 18 a solution of INaOH in 200H 2 O with one of 1HC1 in 200H,O is 13,810 cal. The specific heat-capacity of the sodium hydroxide solution is 0.9827, that of the hydrochloric acid solution is 0.9814, and that of the sodium chloride solution produced is 0.9887. (Tt will be noted that the heat-capacity data must be known with great accuracy in the case of reactions between solutes in dilute solution.) Prob. 54. The heat of formation of 1 mol of gaseous HI at 18 is 4-6200 cal. Estimate the value of its heat of formation at 300 with the aid of the principles relating to the values of heat quantities stated in this chapter, and from the fact that the vapor-pressure of solid iodine is 70 mm. at 109 and 89 mm. at its melting-point, 114. Prob. 55. Heat Evolved by Continuous Processes at High Temper- atures. A mixture of oxygen and hydrogen chloride in the proportion $O, : 1HC1 at 18 is passed continuously into a vessel at 386 containing a suitable catalyzer. The gas is passed so slowly that equilibrium is established, 80% of the hydrogen chloride being converted into chlorine and water. Calculate the heat that will be actually given off from the equilibrium vessel per mol of HC1 passed through. Maximum Temperature Producible by Chemical Changes. Prob. 56. Calculate the maximum temperature that could theoret- ically be attained in the flame produced by burning at 18 a "water-gas" consisting of equimolal quantities of hydrogen and carbon monoxide with twice the quantity of air required for complete combustion. Assume that there is no loss of heat to the surroundings and that the reaction- products are not appreciably dissociated. Prob. 57. a. Calculate the maximum temperature and pressure that could be produced by the explosion within a bomb of a mixture consist- ing of 1 mol H 2 , \ mol O 2 , and 1 mol N 2 , at 18 and 100 mm., assuming that the water produced is not appreciably dissociated, b. The maxi- mum pressure produced in an actual experiment was found to be 840 mm. Show how from this result the degree of dissociation of the water- vapor and the temperature of the mixture at the moment of the explosion can be calculated. (The equations should be formulated, but they need not be solved numerically. ) Prob. 58. Determination of Chemical Equilibria by Thermochemical Methods. Describe a thermochemical method of determining the extent to which acetic acid displaces hydrofluoric acid from sodium fluoride in dilute solution. (The heat of neutralization of acetic acid with sodium hydroxide at 18 is 13,230 cal.) CHAPTEK IX ELECTROCHEMISTRY: THE PRODUCTION OF ELECTRI- CAL ENERGY BY CHEMICAL CHANGES AND OF CHEMICAL CHANGES BY ELECTRICAL ENERGY THE SECOND LAW OF ENERGETICS 124. The Second Law of Energetics. Before the production of electrical energy by chemical changes can be adequately considered, familiarity with certain aspects of another general principle relating to energy, the so-called second law of energetics, is essential. The first law of energetics states that when one form of e.nergy is converted into another the quantity of the form of energy that is pro- duced is equivalent to the quantity of the form that disappears; but it does not indicate that there is any other restriction as to the trans- formability of the different forms of energy, Experience has shown, however, that while the various forms of work can be completely transformed into one another and into heat, the transformation of heat into work is subject to certain limitations. Thus, no system (com- bination of matter) has ever been discovered which, as a result of any process taking place continuously in it, can produce work in unlimited quantity merely by withdrawing heat from the surroundings. An ideal process which is conceived to produce this result may be called perpetual motion of the second kind; and the experience just men- tioned may be expressed by the statement that perpetual motion of the second kind is impossible. This is the perpetual-motion principle that was stated and illustrated in Art. 28. Perpetual motion of the second kind, by which work is conceived to be produced out of heat, is to be distinguished from perpetual motion of the first kind (described in Art. 107), by which work is conceived to be produced without con- suming energy of any kind. Experience with processes taking place at different temperatures has led to the conclusion that this principle is a consequence of a drill more general law, known as the second law of energetics, which may be expressed as follows: a process whose final result is only a transformation of a quantity of heat into work is an impossibility. 143 144 ELECTROCHEMISTRY Prob. 1. a. State what energy-effects occur when a perfect gas ex- pands at a constant temperature against an external pressure. 6. Explain why this is not a contradiction of the second law of energetics. This law does not imply that heat cannot be transformed into work, but only that its transformation must be attended by some other change. This attendant change may consist in a permanent change in the state of the system by which the energy-transformation is brought about; or, when the system undergoes no permanent change in state, it consists in the passage of an additional quantity of heat from a higher to a lower temperature. The latter effect, which can take place only when there is difference of temperature in different parts of the surroundings, will be considered later. In this chapter will be considered only the production of work by changes in state taking place in surroundings of constant temperature. When the process by which any change in state takes place is one that produces a quantity of work equal to (or differing by only an infinitesimal amount from) the quantity of work which must be ex- pended in order to restore the system to its original state, the process is called a reversible process. When the process is one that produces a smaller quantity of work than that which must be expended in restoring the system to its original state, it is called an irreversible process. Similarly, when work has to be expended in changing the state of a system, the process is said to be reversible when the amount of work expended is equal to that which can be produced when the change in state takes place in the opposite direction, and to be irreversible when the amount of work expended is larger than that which can be so produced. It is to be noted that the term reversible is always employed, in the manner just defined, to designate a process of such a character that it is possible to restore the original condition of things both in the system and its surroundings. After an irreversible process has taken place, it is in general possible to restore the system to its original state, but only by withdrawing from the surroundings a larger quantity of work than was produced in them, so that the original condition in the surroundings is not reproduced. When an irreversible change has once taken place, it is not possible by any means whatever to reproduce in their entirety the conditions that previously existed. THE SECOND LAW OF ENERGETICS 145 Prol). 2. A 7 mols of a perfect gas having a pressure of 2 atm. are enclosed within a cylinder placed in a thermostat at a temperature T and provided with a weighted piston. The weight on the piston is suddenly reduced^so that it exerts on the gas a pressure of 1 atm., and the gas expands till its own pressure becomes 1 atm. Explain why this process is irreversible; and state what would have to be true of the pressure exerted by the piston on the gas during its expansion in order that the process might be reversible. Prol). 8. a. Derive an expression in terms of A 7 , R, and T for the work produced by the irreversible process described in Prob. 2. ft. Derive a corresponding expression for the work produced when the same change in state is brought about by a reversible process, c. Find the numerical ratio of these two quantities of work. Prol). 4. Two Daniell cells, each having an electromotive force of 1.10 volts, are connected in series and are used for charging a lead storage-cell having a (counter) electromotive force of 2.10 volts. Ex- plain why the process is not reversible; and state how a number of Daniell cells and a number of storage-cells could be so arranged that the latter might be charged reversibly. (In this case the Daniell cells may be regarded as the system, and the storage cells as a part of the surroundings in which the electrical work is produced and stored.) As illustrated by the preceding problems, in order that the process by which a change in state is brought about may be reversible, the pressure externally applied must be substantially equal to the pressure exerted by the system itself, or the applied electromotive force must be substantially equal to the electromotive force of the cell. For the change in state will take place in one direction when the applied pres- sure or electromotive force is only infinitesimally less than that of the system, and in the other direction when it is only infinitesimally greater; but, if the applied pressure or electromotive force were less by a finite amount than that of the system, the .quantity of mechanical or electrical work produced in the surroundings would evidently not suffice to restore the system to its initial state; and if the applied pressure or electromotive force were greater than that of the system, more work would be withdrawn from the surroundings than the system would be capable of reproducing on reverting to its original state. 125. Application of the Second Law to Isothermal Changes in State. The Concept of Free Energy. Prol). 5: In a voltaic cell consisting of one platinum electrode in contact with a 0.1 normal HC1 solution and with hydrogen gas at 1 atm. and of a second platinum electrode in contact with the same HC1 solution 146 ELECTROCHEMISTRY and with hydrogen gas at 0.1 atm., hydrogen is found to go into solu- tion (as hydrogen-ion) at the first electrode and to be evolved at the second electrode, as a result of the electromotive force which is produced. o. Name the change in state that takes place in such a cell when one faraday of electricity passes through it at 18, specifying all the factors determining the change in state (see Art. 108, first paragraph), but dis- regarding the transference in the solution, which in a cell of this kind will be shown later to be attended by no energy-effect. 6. Describe how the same change in state could be brought about reversibly by a process not involving voltaic action, c. Show by the perpetual-motion principle that the quantity of work attending this process is equal to the work produced when the same change in state takes place reversibly in the cell. The principle, illustrated by the preceding problem, that the quantity of work produced or expended when a definite change in the state of a system takes place at a constant temperature by a reversible process is the same whatever be the nature of the reversible process, is of great importance in chemical considerations. This principle shows that any system in any definite state has a certain power of producing work, and that this power changes by a definite amount when a definite change in state takes place. The power of producing work is therefore, like the energy-content and the heat-content, a quan- tity determined by the state of the system; and, in analogy with the names given to them, it may be called the work-content (A} of the system. The absolute value of this quantity cannot be determined; but the change in its value when any definite change in state takes place at a constant temperature is measured by the work (TF R ) produced when the change in state takes place reversibly. Thus, representing by A l the work-content of the system in the initial state, and by A 2 that in the final state, the decrease in work-content 4 A, = Al = W R . It will be noted that, while the First Law requires that there be a quantity of energy produced in the surroundings equal to the decrease of the energy-content of the system, the Second Law does not require that there be a quantity of work (W) produced equal to the decrease in work-content (as was illustrated by Prob. 3). The Second Law requires only that the quantity of work produced be not greater than the decrease in work-content. That is, W > AA for no process whatever; W = A A for a reversible process; and W < A A for an irreversible one. THE SECOND LAW OF ENERGETICS 14? Just as it is more convenient in chemical considerations to con- sider the heat-content rather than the energy-content of -systems, so there are many advantages in considering in place of the work-content a quantity which differs from it, just as the heat-content differs from the energy-content, by the value of the pressure-volume product. This quantity, which may be called the free-energy content (F), or simply the free-energy of the system, is defined by the equation F = A -\-pv. Its value is determined by the state of the system, since the values of A and of p v are so determined ; and the decrease in its value when any change in the state of the system takes place is evidently equal to the work produced when the change takes place reversibly, dimin- ished by the increase of the pressure-volume product; that is, ^ " P, = W R (p.t;, p,*,), or -AF= W R -A(pv). The difference (F^ F 2 ) between the free-energy-content of a system in its initial state and that in its final state will be called the free-energy-decrease ( AF 7 ) attending the change in state, irre- spective of the sign of its numerical value, which may be either positive or negative. The change in state is always considered to take place at some constant temperature. Prob. 6. a. What is the decrease in joules in the work-content and in the free-energy-content of one formula-weight of water when it changes from liquid water at 100 and 1 atm. to gaseous water at 100 and 1 atm., referring to Prob. 4, Art. 109, for the data needed? 6. What would be the decrease in these two quantities if the liquid water at 100 and 1 atm. changed to gaseous water at 100 and 1 atm. by a process which produces no work, for example, by introducing the liquid water into an evacuated vessel having a volume equal to that of the saturated vapor? In cases where different parts of the system are under different pressures, as in the cell of Prob. 5, the decrease in free energy is defined to be the quantity obtained Vy subtracting from the work pro- duced when the change takes place reversibly the difference between the sum of the p v values for all the parts of the system in its final state and the sum of the p v values for all its parts in its initial state. Therefore, in general : F, F, = W R 2 (p 2 v, Pl tO ; v or AF = W R 2(Apv). The principle that the free-energy-decrease attending any process is determined solely by the change in state of the system is the funda- 148 ELECTROCHEMISTRY mental principle on which is based the treatment presented in this book of the subjects of Electrochemistry and Thermodynamic Chemistry. The first step in any application of the principle is therefore to formu- late exactly the change in state .of the system in terms of its initial and final states; and the next step is to formulate an expression for the free-energy-decrease attending it. FREE-ENERQY CHANGES 149 FREE-ENERGY CHANGES ATTENDING CHANGES IN PRESSURE AND CONCENTRATION 126. Free-Energy Changes attending Changes in Volume and Pressure. When a system which is under an external pressure equal to its own, so that equilibrium prevails, undergoes an infinitesimal change of volume dv and of pressure dp, the work produced dW R is equal to pdv (Art. 109). The free-energy-decrease dF attending such a change in state at a constant temperature is, however, by defini- tion equal to d\V d(pv}. In virtue of the mathematical relation d(pv) = pdv -f- vdp, the free-energy-decrease attending an infinites- imal change in volume or pressure at a constant temperature is there- fore given by the expression : dF = vd. Prob. 7. a. Derive an expression for the free-energy-decrease ( attending the change in state of N mols of a perfect gas when at the temperature T its volume and pressure change from v t and PI to v a and p 2 . b. Formulate an algebraic expression for the decrease of the free energy of N mols of hydrogen when at 18 its pressure changes from 100 atm. to 1 atm. The pressure-volume relations of hydrogen up to high pressures are expressed by the equation p (v Nb) = NRT, in which & is a constant. 127. Free-Energy Change attending the Transfer of a Substance from a Solution of One Concentration to One of Another Concentration. Prob. 8. a. Formulate an exact expression (not involving the perfect- gas law) for the free-energy-decrease attending the introduction at the temperature T of m grams of a pure substance at a pressure equal to its vapor-pressure p into an infinite quantity of a solution in which the substance has a vapor-pressure p. Note that this change in state can be brought about by the following reversible process : vaporize the m grams of the substance at the temperature T under a pressure p ' change the pressure of this vapor to p ; and at this pressure condense the vapor into the solution. I. Derive from this expression an equation which holds true when the vapor conforms to the perfect-gas law. c. Formulate an expression for the free-euergy-decrease attending the transfer of N mols of a substance from an infinite quantity of a solution in which its vapor- pressure is p, into an infinite quantity of a solution in which its vapor- pressure is /> 2 . Prob. 9. Show how the equation formulated in Prob. 8c may be modified so as to contain the mol-fractions x l and x 3 of the substance in the two solutions, , when the vapor-pressures p a and P 2 conform to Raoult's law: b, when these vapor-pressures conform to Henry's law. 150 ELECTROCHEMISTRY c. State in the case of a solution consisting of two substances A and B the conditions of composition under which Raoult's law and under which Henry's law (and therefore under which the corresponding free-energy expressions) hold true approximately, d. Show as in Art. 27 that when the mol-fractions x^ and x z are small, their ratio may be replaced by the ratio of the concentrations c x and c 2 , which denote the number of mols of the solute per unit-volume of the solvent. In the preceding problems the following expressions have been derived for the free-energy-decrease which attends the transfer at the temperature T of that quantity of a substance which is N mols in the state of a perfect gas from an infinite quantity of a solution in which its vapor-pressure is p v its mol-fr action x lt and its concen- tration Cj, into an infinite quantity of another solution in which its vapor-pressure is p 2 , its mol-fraction x v and its concentration c a : ** = # T log ?L(1); -&F = NRT\og^(2); P2 #2 &F = NRT log^i (3). c z Equation (1) holds true whatever be the mol-fractions or concentra- tions; but it involves the assumption that the vapor conforms to the perfect-gas law. Equation (2) holds true when, in conformity either with Raoult's law or with Henry's law, the vapor-pressures are pro- portional to the mol-fractions. Equation (3) holds true when, in conformity with Henry's law, the vapor-pressures are proportional to the mol-fractions and when the latter are small enough to be sub- stantially proportional to the concentrations; it is the expression commonly employed when a solute is transferred from one dilute solution to another. Since Henry's law applies to the concentrations of a definite molecular species, it is evident that equation (3) must be applied separately to the transfer of each molecular species, not to the transfer of the substance as a whole, if it exists in the solution as two or more different molecular species, as is the case with H + C1" or any other partially ionized substance in aqueous solution, or with ac^ic acid in benzene solution in which it exists as C 2 H 4 O a and (C a H 4 O 2 ) 2 . More- over, although equation (3) was derived by the consideration of ft process involving the vaporization of the solute and with the aid of the assumption that the pressure of the vapor was small enough FREE-ENERGY CHANGES 151 to conform to the perfect-gas law, yet that equation relates only to the transfer of the substance from one solution to another; and it would be remarkable if its validity depended on whether the substance were volatile or on how large its vapor-pressure might be. In fact, it can be shown, by deriving equation (3) through a consideration of an osmotic process of transferring the substance from one solution to the other, that the equation is exact, provided only that the substance behaves as a perfect solute, as shown by its conformity to the osmotic- pressure equation P = c R T considered in Art. 38. Prob. 10. a. Calculate the free-energy-decrease attending the transfer at 20 of 1NH, from an infinite quantity of a solution of the composition 1XH 3 8^H 2 O in which its vapor-pressure is 80 mm. into an infinite quan- tity of a solution of the composition 1NH R 21H 2 O in which its vapor- pressure is 27 mm. ft. Calculate the free-energy-decrease attending the dissolving t 20 of 1NH, at 1 atm. in an infinite quantity of a solution of the composition 1NH 3 21H 2 O. Note. Throughout this chapter free-energy values are to be expressed in joules. For the gas-constant R its value 8.32 in jou'es per degree (derived in Prob. 24, Art. 22) must therefore be used. Prob. 11. Calculate the free-energy-decrease attending the transfer at 20 of INHs from an infinite quantity of 0.1 formal NH 4 OH solution into an infinite quantity of 0.001 formal NH 4 OH solution. The ammonium hydroxide is 1.3% ionized in the 0.1 formal and 12.5% ionized in the 0.001 formal solution, and the remaining ammonium hydroxide is m percent dissociated into NH 3 and H 2 O in each solution. Prob. 12. The transfer at the temperature T of one formula-weight of NaCl from an infinite quantity of a dilute solution in which its formal concentration is c^ and its ionization is y t into an infinite quantity of another dilute solution in which its formal concentration is c 2 and its ionization is y^ can be brought about by so transferring either 1 mol Na* and 1 mol Cl- or 1 mol unionized NaCl. a. Formulate an expression for the free-energy-decrease attending each of these processes. b. Show that one of these expressions can be derived from the other with the aid of the mass-action equation for the ionization of the salt. c. Calculate the free-energy-decrease by both expressions for the case that the tempera- ture is 18, the concentrations are 0.0334 and 0.00167 formal, and the ionizations are 0.900 and 0.970. The large divergence between the values of the free-energy-decrease calculated in Prob. 12 by the two expressions corresponds to the devia- tions (considered in Art. 86) from the requirements of the mass-action law exhibited by largely ionized substances. It shows that either the 152 ELECTROCHEMISTRY ions or the unionized substance or both deviate considerably from the laws of perfect solutions (thus from the requirements of the osmotic-pressure equation P = cRT). It will be seen later (in Prob. 23, Art. 131) that the true value of the free-energy-decrease attending the transfer of the sodium chloride can be directly derived from the electromotive force of a cell containing the two sodium chloride solutions. This true free-energy-decrease is larger by 26 per- cent of its value than that calculated by considering the transfer of the unionized substance; and it is smaller by 3.0 percent of its value than that calculated by considering the transfer of the ions. This result is representative of the behavior of largely ionized uniunivalent substances in general. It shows that the unionized substance deviates very greatly from the laws of perfect solutions, and that a great error would be made if it were assumed to conform to those laws in mass- action or in free-energy calculations. The comparison also shows that the ions deviate considerably from the behavior of perfect solutes ; the deviation up to 0.1 formal is, however, not so great as to cause a very serious error in approximate calculations. Therefore, throughout this book, in the case of largely ionized substances, the ions, rather than the unionized substance, are always considered; and the deviation of the activity of the ions of uniunivalent substances from that of perfect solutes up to a concentration of 0.1 formal is disregarded. In the case of unibivalent salts, such as zinc chloride or sodium sulphate, for which the ionization values are uncertain owing to the probable presence of intermediate ions (referred to Art. 58), the free- energy-decrease attending their transfer from one solution to another can be calculated even approximately only when the concentrations are so small (say less than 0.01 formal) that no great error is made by regarding the ionization as complete. VHANGES IN STATE IN CELLS 153 CHANGES IN STATE AND PRODUCTION OF WORK IN VOLTAIC CELLS 128. Changes in State in Voltaic Cells. A change in state can be made to yield electrical energy when it can be brought about by a process of reduction occurring in one place and a process of oxidation occurring in another place; for it is an inherent characteristic of reduction processes that they are attended by a liberation of positive electricity or by an absorption of negative electricity, and of oxidation processes that they are attended by the opposite electrical effects. Thus the reduction of copper-ion to metallic copper may be represented by the equation Cu ++ = Cu + 2 ; and the oxidation of metallic zinc to zinc- ion, by the equation Zn -f- 2 = Zn" 1 " 1 ", where the symbol denotes one faraday of positive electricity. If these two processes occur at the same place, as is the case when metallic zinc is placed in a copper-ion solution, no electrical effect is observed. If, however, the zinc is placed in a zinc-ion solution and the copper in a copper-ion solution, and if the two solutions are placed in contact, the reduction- process tends to occur at one place and to liberate positive electricity there, and the oxidation-process tends to occur at another place and to absorb positive electricity there, thereby producing a difference of potential or an electromotive force between the two places ; and if they are now connected by a metallic conductor, a current of electricity will flow through -it, and this current can be made to produce work, for example, by passing it through an electric motor. Such an arrange- ment as that here described, in which an electric current is produced by causing an oxidation-process and a reduction-process to occur at two different places, is known as a voltaic cell. It has been shown in Art. 125 that the quantity of work that can be produced by any process is determined solely by the change in state of the system. Hence the maximum quantity of work producible by the action of a voltaic cell is fully determined by the initial and final states of the substances of which it consists; and in any case under consideration these states, or the corresponding change in state, must be exactly specified. Thus, there must be stated, in addition to the temperature, not merely the chemical reaction that takes place in the cell, but also the conditions of pressure and concentration under which the substances involved in it are produced and destroyed and 154 ELECTROCHEMISTRY any transfers of substances from solutions of one composition to those of another composition. In order that there may not be a finite change in the concentrations of the solutions, and therefore in the electromotive force of the cell, during the occurrence of the- change in state, it will always be assumed that the solutions are present in infinite quantity, so that when a finite quantity of any substance is introduced into or withdrawn from one of the solutions of the cell there is only an infinitesimal change in its concentration. The character of the cell under consideration will be shown by writing the symbols of the pure substances and solutions in the order in which they are actually in contact with one another, commas being inserted to indicate the junctions at which, as will be explained later, an electromotive force is produced. The conventions described in Art. 113 will be employed to indicate the state of aggregation of sub- stances and the composition of solutions. Thus a cell consisting in series of metallic zinc, of a zinc chloride solution of the composition ZnCl 2 100H 2 O, of solid mercurous chloride, and of metallic mercury, at a pressure of one atmosphere (as is always understood unless some other pressure is specified), will be represented by the expression: Zn, Zn01 2 100H 2 O, Hg 2 Cl 2 -f-Hg. Similarly, a cell whose electrodes are an inert metal M in contact with hydrogen gas at 0.1 atmosphere and the same metal in contact with chlorine gas at 0.05 atmosphere, and whose electrolyte is a 0.1 formal HC1 solution, will be represented by the expression: M + fl,(l atm.), HC1(0.1 1), CZ a (0.05 atm.) + M. So also a lead storage-cell whose electrodes are lead and an inert metal coated with lead dioxide, and whose electrolyte is a sulphuric acid solution, say of the composition H 2 SO 4 10H 2 O, saturated with lead sulphate, will be represented by the expression: Pb 4- PbS0 4 , H 2 SO 4 10H 2 O, PbS0 4 + Pb0 2 + M. The change in state taking place in a cell may he expressed by an equation whose left-hand member represents the initial state, and whose right-hand member represents the final state, of the substances involved in the change. Thus the changes of state occurring when at 25 two faradays pass from left to right (as is always understood CHANGES IN STATE IN CELLS 155 unless the opposite is specified) through the first two cells just formu- lated are shown by the equations : Zn + Hg 2 CL=: 2Hg + ZnCl 2 (in ZnOl.lOOH.O) at 25 H 2 (l atm.) -f C7 2 (0.05 atm.) = 2HCl(at 0.1 f.) at 25. In determining the change in state in the cell it is often well to consider separately the reactions that take place at the two electrodes (see Art. 42) and the ion-transferences that occur in the solutions. The electrode-reactions take place in accordance with Faraday's law (Art. 43), and the transference-effects in accordance with the principles of transference (Arts. 45-49). 1'rob. 13. a. State the changes at each electrode and the ion-transfer- ences that attend the passage of two faradays through the first cell formulated above; and show that the net result of these changes is the change in state expressed by the first of the preceding equations, ft. Do the same for the second cell formulated above. Prol). llf. a. State the electrode changes and ion-transferences that occur when two faradays pass at 20 through the lead storage-cell formu- lated above. I). Express the resultant change of state that takes place in the cell by an equation. When, as in these cells, a gaseous or a solid non-metallic substance is in contact with a metal electrode, the adjoining solution is under- stood to be saturated with that substance. In such cells there are there- fore in reality two different solutions, even though for the sake of brevity only one may be written in formulating the cell; thus these cells ought strictly to be written : Zn, ZnCL100H 2 0, ZnCl 2 100H 2 O -f Hg 2 Cl 2 (s. f.),* Hg 2 Cl,4-Hg. M + # 2 (1 atm.), H01(0.1f.) -f H 2 (0.00076 f.), HC1(0.1 f.) + Cl a (0.0044f.), CZ,(0.05 atm.) + M. Pb + Pt>S0 4 , H 2 SO 4 10H 2 O+PbSO 4 (s.f.), H 2 SO 4 10H 2 O+PbS0 4 .(s.f.) + Pb(SO 4 ) 2 (10- 8 f.), PbS0 4 + Pb0 2 + M. The recognition of the fact that there are in such cells two solutions of slightly different composition is of great importance in considering the mechanism of voltaic action and in evaluating the separate poten- tials discussed in Arts. 133-137. When, as in the three cells above considered, the cell contains two solutions of substantially the same composition (in the respects that *The symbol s. f. represents the solubility (in formula-weights per 1000 g. of water) of Hg 2 Cl 2 in a solution of composition ZnCl a 100H a O. 156 ELECTROCHEMISTRY in both solutions the ions present at considerable concentrations are the same, that the concentrations of these ions are substantially the same, and that the solvent-medium as a whole is substantially the same) , the ion- transferences need not be considered ; for, though there is transference from the solution around one electrode to that around the other electrode, there is not an appreciable free-energy change, since the two solutions have substantially the same composition. Pro 6. 15. Specify the electrode-changes and ion- transferences and the resultant change in state taking place when one faraday passes through each of the following cells: a. # 2 (1 atm.), HC1 (0.01 f.), HC1(0.1 f.), #,(1 atm.). &. CZ 2 (1 atm.), HC1(0.01 f.), HC1(0.1 1), Cl 2 (l atm.). Note that the transference-effects in these cells are similar to those described in Art. 45, the left-hand solution being regarded as the anode- portion and the right-hand solution as the cathode-portion. 129. Production of Work in Voltaic Cells. According to a funda- mental principle of the science of electricity, when a quantity of positive electricity Q flows between two places, such as the electrodes of a voltaic cell, between which there is a potential-difference or electro- motive force E, a quantity of work equal to the product EQ can be produced. In order that the work produced by a voltaic cell may be the maximum which can be obtained from the change in state taking place in it, the electromotive force at the electrodes must be such that, when it is increased by an infinitesimal amount, the change in state under consideration takes place in the opposite direction. It is this value of the electromotive force which is considered throughout the following pages, and which is called the electromotive force (E) of the cell. In an experimental determination of this electromotive force the assurance must be obtained, by making measurements under vari- ous conditions, that the measured electromotive force really corre- sponds to the change of state under consideration, and not to some incidental process taking place at the electrodes. Now, according to Faraday's law, the quantity of electricity flow- ing through a voltaic cell is strictly proportional to the number of equivalents N that are involved in the chemical change at each elec- trode ; that is, Q = N F, where F represents the quantity of electricity (96,500 coulombs) that passes when a reaction involving one equiva- lent of each of the reacting substances takes place at each electrode. PRODUCTION OF WORK IN CELLS 157 The electrical work that can be produced by a change in state taking place in a voltaic cell and involving the passage of N faradays of electricity is therefore equal to E N F. The numerical value of the electromotive force E of the cell will in this book be given a positive sign when the cell tends to produce a current of positive electricity through the cell in the direction in which it is written, and a negative sign when the cell tends to pro- duce such a current in the opposite direction. The symbol N will denote the number of faradays of positive electricity that are con- sidered to pass through the cell from left to right, its numerical value being given a negative sign when positive electricity is considered to pass in the opposite direction. The value of W calculated by the equation will then have, as usual, a positive sign when the cell pro- duces electrical work, and a negative sign when external work is expended upon the cell. The work will be in joules when the electro- motive force is in volts and the quantity of electricity in coulombs. Prol). 16. The electromotive force at 15 of the Daniell cell Zn, ZnS0 4 7H 2 -f ZnSO 4 w'H 2 O, CuSO 4 n"H 2 O + CuS0 4 5H 2 0, Cu is 1.093 volts. What must be the values of NF in coulombs and of W in joules in order to precipitate 1 at. wt. of zinc ; and what does the sign of each of these quantities signify? When a voltaic cell acts reversibly there is ordinarily produced, in addition to the electrical work, a quantity of mechanical work corresponding to the changes in volume of the different parts of the cell taking place under their respective pressures. The total work W B is therefore E N F -f- S(joAv). Since by Art. 125 the free-energy- decrease is equal to W^ (Ap t>), it is equal simply to the elec- trical work that can be produced. That is, for any change in state taking place in a voltaic cell under constant pressures : AF = ENF. Prol). 17. Calculate exact values of the electrical work and the mechanical work that are produced when 1 faraday passes under re- versible conditions through the cell -Hj(l atm.), HC1(0.1 f., 1 atui.), Cl, (0.05 atm.) at 25. The electromotive force of this cell is 1.451 volts. The increase in the volume of an infinite quantity of 0.1 f. HC1 solution caused by introducing 1HC1 into it is 18.7 ccm. What is the correspond- ing free-energy-decrease? 158 ELECTROCHEMISTRY ELECTROMOTIVE FORCE OP VOLTAIC CELLS IN RELATION TO CONCENTRATION 130. Change of the Electromotive Force of Voltaic Cells with the Concentration of the Solutions. The considerations of the preceding articles make it possible to calculate the change that is produced in the electromotive force of a cell by varying the concentration of the solutions contained in it. Thus the difference between the electro- motive force of the cell H 2 (l atm.), HC1(0.01 f.), Cl 2 (l atm.) and that of the cell H 2 (l atm.), HC1(0.1 f.), Cl 2 (l atm.) can be derived by considering that one faraday passes through these two cells arranged in series in opposition to each other, by noting what the resultant change in state is, and by equating the two expressions for the attend- ant free-energy-decrease derived in preceding articles. Pro b. 18. Calculate the difference between the electromotive forces at 18 of the two cells named in the preceding text. The ionizations of the acid in the two solutions are 0.972 and 0.925. Prob. 19. Calculate the difference between the electromotive forces at 25 of the cells J/ 2 (l atm.), HC19H 2 O, Cl 2 (l atm.) and # 2 (1 atm.), HC116.7H 2 O, C1 2 (1 atm.). The vapor-pressures of HC1 in the two solu- tions at 25 are 0.0550 and 0.0044 mm. Prob. 20. State what data are needed in order to calculate the electromotive force E t at 18 of the cell If 2 (1 atm.), H,SO 4 (10f.), O a (latm.) from that E 2 of the cell H 2 (l atm.), H 2 SO 4 (0.01 f.), O,(l atm.) ; and formulate an expression by which the calculation could be made. Prob. 21. The cell Ag-f AgCl, HCl(0.1f.), CZ 2 (latm.) has at 25 an electromotive force of 1.142 volts. How much would its electromotive force be changed by substituting HC1(0.01 f.) for the HC1(0.1 f.)? 131. The Electromotive Force of Concentration-Cells. A cell which consists of two identical electrodes and of two solutions con- taining the same substance at two different concentrations is cal'ed a concentration-cell The cell Zn, ZnCl 2 (0.01 f.), ZnCi(0.001 f.), Zn is an example of a concentration-cell; so also are the cells formulated in Prob. 15. Prob. 22. a. Derive an algebraic expression for the electromotive force of the cell # 2 (1 atm.), HCl(c' f.), HCl(c" f.), # 2 (1 atm.), where the concentrations c' and c" are small, by considering (as in Prob. 15) the change in state that occurs in the cell when N faradays pass through it and by considering the two expressions for the attendant free-energy- decrease derived in preceding articles. b. Calculate the electromotive force of the cell at 18 when the concentrations c 7 and c" are 0.01 and 0.1 formal. For the ion-conductances see Art. 56. CONCENTRATION-CELLS 159 Profi. 23. a, b. Answer the same questions as in Prob. 22 for the cell atm.), HCl(c' f.), HCl(c" f.), Cl z (l atm.). Prob. 2.'f. Formulate a numerical expression by which the electro- motive force of the cell Zn, ZnCl 2 (0.01 f.), ZnCl, (0.001 f.), Zn at 18 can be calculated. Consider the ionization of the salt in both solutions to be complete. (See the last paragraph of Art. 127.) Prob. 25. Calculate the electromotive force at 18 of the cell Ag -{- AgCl, NaCl (0.0334 f.), NaCl (0.00167 f.), AgCl + Ag, using the data of Prob. 12c. (The electromotive force of this cell has been experimentally determined and found to be 0.05614 volt.) Another type of concentration-cell is that in which there is a single aqueous solution in contact with electrodes consisting of some metal dissolved at two different concentrations in mercury. For example, the cell ZnlOOHg, Zn01 2 100H 2 O, Zn200Hg is of this type. Since most of the metals dissolved at small concentrations in mercury have been shown by vapor-pressure and freezing-point measurements to be nearly perfect solutes having monatomic molecules, the electromotive force of cells having such solutions as electrodes can be calculated with the aid of the preceding considerations. Another similar type of concentration-cell is that in which the two electrodes consist of some inert metal surrounded by the same gas at two different pressures, as is the case in the hydrogen cell of Prob. 5 and in the oxygen cell M + 2 (1 atm.), KOH(1 f.), 2 (0.1 atm.) + M. Pro b. 26. Calculate the electromotive force at 18 of the zinc-mercury cell formulated in the preceding text. Prob. 27. Calculate the electromotive force at 25 of the oxygen cell formulated in the preceding text. 160 ELECTROCHEMISTRY . ELECTROMOTIVE FORCE OF CELLS INVOLVING CHEMICAL CHANGES. ELECTRODE-POTENTIALS AND LIQUID-POTENTIALS 132. Chemical Changes in Voltaic Cells. In the cells thus far considered the change of state consists only in a transfer of one or more of the substances from one pressure or concentration to another. In most voltaic cells, however, a chemical change takes place. Thus in the Daniell cell the reaction Zn -f CuS0 4 = Cu -}- ZnS0 4 occurs. Other examples of cells in which similar chemical changes take place are given in Art. 128. A chemical change somewhat different in character from those just considered takes place in cells whose half -cells consist, not of solid or gaseous elementary substances in contact with solutions of their ions, but of an inert metal electrode in contact with two solutes in different stages of oxidation. Thus the electrode-reactions in the cell M, Fe^Cl-.CcJ+Fe^+Ol-^c,), Cr(e.) + Cl a (c 4 ), M, attending the passage of one faraday are Fe ++ + = Fe ++ + and iCl, = Cl-+e, and the whole reaction is Fe ++ Cl- 2 -f iCl 2 = Fe +++ Cl- 3 . Such cells do not, however, require special treatment; for the principles applicable to other cells can be readily extended to them, as will be seen in the following articles. Evidently, if the free-energy-decrease attending the chemical change in any cell could be determined by any independent method, the electromotive force of the cell could be found with the aid of the equation A^ ENF. There is, in fact, such a method by which the free-energy-decrease attending a chemical change can be calcu- lated when the pressures or concentrations involved are small. This method will be considered in the following chapter. In this chapter will be presented only the electromotive-force side of the problem. There remains therefore to be considered only the partial electromotive forces at the different junctions within the cell. The consideration of these is important in two respects : first, it shows more clearly the separate factors which determine electromotive force : and secondly, it enables the electromotive force of a great number of cells to be calculated from a relatively small number of experimentally determined constants. 133. The Electrode-Potentials of Voltaic Cells. The electromotive force produced by a voltaic cell is the sum of the electromotive forces ELECTRODE-POTENTIALS 161 produced at the junctions between the electrodes and the solutions and of those produced at the junctions between the different solutions that may be present in the cell. Thus the electromotive force of the Daniell cell is the algebraic sum of the electromotive force from the zinc to th zinc-sulphate solution, that from the zinc-sulphate solution to the copper-sulphate solution, and that from the copper- sulphate solution to the copper. The electromotive forces at the elec- trodes are commonly called electrode-potentials; and those at the junctions of the solutions, liquid-potentials. Although attempts have been made to determine by certain experi- mental methods the absolute values of electrode-potentials, yet the measurements made by these methods are not nearly so accurate nor reliable as those of the electromotive force of ordinary cells. It is therefore customary to adopt as the value of the electrode-potential of any half -cell (such as Zn, ZnCl,100H 2 O) the electromotive force of the whole cell which consists of the half-cell under consideration combined with a standard half-cell. The electrode-potential of the standard half-cell is thereby arbitrarily assumed to be zero. Three different half -cells are in use as such standards of reference; namely, the half-cell Hg, Hg 2 Cl_, + K01(l n.), the half-cell H 2 (l atm.), H 2 SO 4 (2 n.), and the half -cell 7J 2 (1 atm.), H + (l m.). The first of these cells, which is commonly called the calomel-electrode, has certain experi- mental advantages. The last of these cells, which will be here called the molal hydrogen-electrode, has the simplest theoretical significance. The exact difference between the electrode-potentials of these standard half-cells will be shown later. The electrode-potential of any half-cell is therefore equal to the electromotive force of the whole cell formed by combining it with one of these standard half-cells. Thus the electrode-potential of the half -cell Zn, ZnCl 2 100H 2 O referred to the calomel-electrode is equal to the electromotive force of the whole cell Zn, ZnO^lOOHO 1 1 KC1(1 n.) -f- Hg^Cl,, Hg. Similarly the electrode-potential of the half-cell C1 2 (latm.), HCl(0.1f.) referred to the molal hydrogen-electrode is equal to the electromotive force of the cell C1 2 (1 atm.), HCl(0.1f.) | , H*(lm.), #2 (latm.). It is, however, further understood that in evaluating this electromotive force the liquid-potential has been sub- tracted from the measured electromotive force of the whole, cell a 162 ELECTROCHEMISTRY fact which is indicated in the symbolic representation of the cell by inserting parallel lines, instead of a comma, at the liquid junction. It will be noted that the electrode-potential has a positive sign when positive electricity tends to flow from the electrode to the solution, and a negative sign when it tends to flow in the opposite direction. In the case of liquid-potentials the same convention as to sign is adopted as in the case of whole cells; namely, the liquid- potential is given a positive sign when positive electricity tends to flow from the solution whose symbol is written on the left-hand side to the solution whose symbol is written on the right-hand side; and it is given a negative sign in the reverse case. (In using data from outside sources, it is to be noted that foreign electrochemists employ the opposite convention as to the sign of electrode-potentials.) Prob. 28. Measurements of the electromotive forces at 25 of the cells # 2 (1 atm.), HC1(0.1 n.), KC1(0.1 n.), KC1(1 n.), H&Clj-fHg, C'Z,(1 atm.), HCKO.l n.), KC1(0.1 n.), KC1(1 n.), Hg^ + Hg, have given the values 0.3740 and 1.114 volts, respectively. The liquid- potential of KC1(0.1 n.), HC1(0.1 n.) has been found to be 0.0283 volt; and that of KCHO.l n.), KC1(1 n.) is probably about 0.001 volt. a. Find the values of the electrode-potentials that can be derived from these data. I). Calculate the electromotive force of the cell H 2 (l atm.), HC1(0.1 n.), 134. Change of Electrode-Potentials with the Ion-Concentrations. Concept of Specific Electrode-Potentials. Just as the total electro- motive force of a cell is determined solely by the change in state that takes place in it, so any electrode-potential is determined solely by the change in state that takes p.ace at the electrode. With the aid of this principle and the general expressions for free-energy-decrease derived in Arts. 127 and 129, the change of an electrode-potential with the concentration of the ions or with that of any other gaseous or dis- solved substance involved in the electrode-reaction can readily be calculated, provided the concentrations are so small that the dissolved substances conform to the laws of perfect solutes. Pro&. 29. Derive an algebraic expression for the difference E X E X between the electrode-potentials of the two half-cells involved in each of the following cells, by considering the change in state that occurs when one faraday is passed through the cell and assuming that the dissolved substances behave as perfect solutes up to a concentration of 1-molal : ELECTRODE-POTENTIALS 163 a. O a (patm.), OH-(cm.), OH-(lm.), O 2 (latm.). the two electrode-potentials involved in each of the following cells : c. Hg + Hg a S0 4 , S0 4 =(cm.), SO 4 =(lm.), Hg 2 S0 4 + Hg. (j. Ag-f AyCl, 01- (cm.), Cl-(lm.), AgCl -f- Ag. e. CZ a (patm.), Cl-(cm.), Cl-(lm.), C7 2 Clatm.). Pro&. 5(7. Derive an algebraic expression for the difference between a. Ag, Ag + (cm.), Ag + (l m.), Ag. b. Zn, Zn+ + (c in.), Zir + U m. ), Zn. c. Hg,Hg a ++(cm.),Hg !l ++ (lm.),Hg. d. I,, I-(om.), I-(l m.), I* d. M, Cl,(c l m.)+Cl-(c 1 m.), Cl 2 (lm.) + Cl-(lm.), M. e. M, Fe ++ (Cim.)-hFe +++ (c 2 m.), Fe ++ (l m.)-|- Fe +++ (lm.), M. The electrode-potential calculated as in the preceding problems for the case that the concentrations of the ions or other solutes in- volved in the electrode-reaction are 1 molal and the pressure is one atmosphere is called the specific electrode-potential. It will be here represented by the symbol E followed by a subscript showing the ele- ment involved in the electrode-reaction, or, when necessary for clear- ness, by a parenthesis showing both substances involved in that reaction; for example, by such symbols as E Zn , E H , (1%, 01"), S(Fe + % tV ++ ), E(Ag + AgCl, 01"). Its value is not that actually observed when the concentrations are 1 molal, owing to the large deviation (discussed in Art. 86) of the activity of ions from that of perfect solutes at such high concentrations. It is to be regarded as an em- pirical constant derived from measurements at small concentrations, from which conversely the true electrode-potential at any small con- centration can be calculated. It is evident from the preceding problems that in general the electrode-potential E(A, B, . .) when the concentrations of the solutes A, B, . . involved in the electrode-reaction are c c , . . is related to the specific electrode-potential E(A, B, . .) in the way expressed by the equation : E (A,B;. .) =E(A, B, . .) -X A logc A -N log C B - .. where N^, N^ . . represent the number of mols of the solutes A, B, . . which are produced when one faraday of positive electricity passes from the electrode to the solution, the value being given a negative sign for any solute which is destroyed instead of being produced. When one of the substances involved in the electrode-reaction is a gas its concentration may be replaced by its pressure, the specific electrode- 164 ELECTROCHEMISTRY potential in that case being the electrode-potential when the gas is at a pressure of one atmosphere. For convenience in numerical calcula- tions it may be noted that the constant 2.303 R/F (equal to 2.303 X 8.32/96500) has the value 1.984 X 10" 4 . Prob. 31. Determination of Specific Electrode-Potentials. Calculate the specific electrode-potentials at 25, referred to the calomel-electrode, of a, H z , H* ; b, C1 2 Cl~ ; c, Ag -f AgCl, 01- ; d, Ag, Ag- 1 -. For the electro- motive-force values needed see Probs. 28 and 21. The ionization of HC1 at 0.1 normal is 92%. The solubility of silver chloride in water at 25 is 1.30 X !0~ 6 normal. The specific electrode-potential of H v H* referred to the calomel- electrode has been calculated in the preceding problem. From data for similar cells with more dilute HC1 solutions its value has been more accurately determined, and found to be +0.277 volt at 25. This potential is evidently the electrode-potential of the molal hydrogen- e'ectrode, Z7,(l atm.), H + (l m.), mentioned in Art. 133 as one of the standard half -cells. It is, to be sure, not the actual potential of this half -cell; but it is the potential that is calculated for it by the usual logarithmic formula from the electrode-potential of any half-cell # 2 (latm.), H + (cm.) in which the concentration c is very small. This molal hydrogen-electrode will be used as the standard of reference throughout this book, because of its simple theoretical significance. Its potential at 25 is substantially identical, so far as can be judged from existing data, with that of the half -cell T,(1 atm.), H,SO 4 (2n.), which is often employed as the standard hydrogen- electrode. The value of any electrode-potential at 25 referred to this standard is evidently 0.277 volt smaller than the value of the same electrode-potential referred to the calomel-electrode. 135. Values of the Specific Electrode-Potentials. The following table contains some of the more accurately determined values of the specific electrode-potentials at 25 and one atmosphere, referred to that of the molal hydrogen-electrode taken as zero. Li, Li* 3.027 Cu+, Cu*+ 0.20 Fe++, Fe* + + 0.740 K,K> 2.931 Ag, AgCl -f 01- 0.224 Ag, Ar 0.793 Na, Na+ 2.721 Hg, HgjClj-f Cl~ 0.270 Hg, Hg/+ 0.80 Zn, Zn++ 0.76 Cu, Cu ++ 0.34 Hg/+, Hg** 0.92 Fe, Fe+ + 0.43 O 2 , OH- 0.393 Br 2 ( 1 in. ) , Br- 1.10 Cd, Cd++ 0.40 Cn, Cu+ 0.51 C1 2 , 01- 1.357 Pb. Pb+* 0.13 I 2 , 1- 0.533 C1 2 , 01- 1.388 ELECTRODE-POTENTIALS 165 Specific electrode-potentials which are related to one another can be calculated in the ways illustrated by the following problems. Prol). 32. State what data would be needed to calculate one of the following specific electrode-potentials from the other, and formulate an expression by which the calculation could be made : a, C1 2 , Cl~ from Cl v Cl- ; 6, Hg, Hg 2 Cl 2 + Cl- from Hg, Hg 2 ++ . Relation between the Specific Electrode-Potentials of an Element that exists in More than Two States of Oxidation. Prob. 38. By considering the effect of passing electricity through the cells Cu, Cu + (lm.) Cu + + (lrn.), Cu, and Cu, Cu ++ (l m.) 1 1 Cu + (lm.) -fCu ++ (lm.),M, derive a relation between the specific electrode-poten- tials of Cu, Cu+; Cu+, Cu ++ ; and Cu, Cu ++ ; and calculate the last of these potentials from the first two. Pro 6. 33 A. a. By considering appropriate cells derive an expression for the specific electrode-potential of Hg, Hg*+ in terms of those of Hg, Hg a ++ and of Kg/*, Hg++. b. Calculate the value of the first of these potentials from the values of the other two which are given in the pre- ceding table. 136. Calculation of the Electromotive Force of Cells from the Specific Electrode-Potentials. The electromotive force of cells in which the liquid-potentials are negligible or are to be disregarded can be calculated from the specific electrode-potentials in the way illustrated by the following problems. Prob. 34. Calculate the electromotive force at 25 of the cell Zn, ZuCl 2 ( 0.001 f.), AgCl + Ag. Consider the ionization of the zinc chloride to be complete. Pro b. 35. Calculate the electromotive force at 25 of the cell H, (0.1 atm. ), H 2 S0 4 (0.05 f.), PbS0 4 -f Pb. The saturated solution of PbSO 4 at 25 is 0.00014 formal. In regard to the ionization of sulphuric acid see Prob. 38. Art. 60. Prol). 36. Formulate numerical expressions for calculating the elec- tromotive force at 25 of each of the following cells : a. M-f CuCl, CuCl 2 (0.01f.), C7 2 (0.1atm.). b. M, 0.01 f. KI sat. with I 2 1 1 FeCl 2 (0.02 f.) 4. FeCl 8 (0.01 f.), M. Regard the ionization of all the salts as complete. The solubility of CuCl is 0.0010 formal. For the composition of iodide solutions saturated with iodine see Prob. 52. Art. 92. 137. Evaluation of Liquid-Potentials. Just as the electromotive force of a whole cell is determined by the free-energy-decrease attend- ing the change in state that takes place in the cell, and just as the electrode-potential is determined by the free-energy-decrease attend- 166 ELECTROCHEMISTRY ing the change in state that takes place at the electrode, so a liquid- potential is determined by the free-energy-decrease attending the change in state that takes place at the boundary between the two - solutions. The change in state at such a boundary is of a simple character in the case where the two solutions contain only the same solute at two different concentrations, as in the combination, NaCl(0.01 f.), NaCl(0.1f.). In this case the calculation is based on the considera- tion of the number of equivalents of the positive ion-constituent and of the negative ion-constituent that pass through the boundary in the two opposite directions per faraday passed through. It therefore involves a knowledge of the transference-numbers of the ions. It a^o involves, since the free-energy-decrease attending the transference of the ions must be evaluated, a knowledge of the ion-concentrations in the two solutions. Prob. 87. Calculate the liquid-potential at 18 of the combinations: a, NaCl(0,lf.), NaCl(O.Olt) ; 6, K 2 SO 4 (0.1 f.), KjSCMO.Ol f.). For the data needed see Arts. 53 and 56. *By a similar consideration of the ion-transference that takes place at the boundary of the two solutions an expression can be derived for* the liquid-potential of combinations of the types KC1, KOH; KC1, NaCl; K,SO 4 , Na 2 S0 4 ; ZnSO 4 , CuS0 4 , for the case that the two solutes have the same concentration and ionization. Namely, taking as a specific example the combination KCl(c formal), NaCl(c formal), it will be noted that there must be at the boundary a portion of the solution in which the two salts are present in varying proportions, the concentration of the potassium chloride decreasing continuously from left to right from c to 0, and that of the sodium chloride increas- ing continuously in the same direction from to c. Representing now by T and T the transference-numbers of the respective ions in the two solutions of the pure salts, it is evident that per faraday of electricity T equivalents of potassium-ion enter the left-hand side of the boundary-portion, and that no potassium-ion leaves the right-hand side of that portion ; that no sodium-ion enters the left-hand side, but that T equivalents of sodium-ion leave the right-hand side ; and that t93> 1 T equivalents of chloride-ion leave the left-hand side and 1 T Na "This paragraph and the problem following it may be omitted in brief courses. LIQUID-POTENTIALS 107 equivalents of chloride-ion enter the right-hand side of the boundary- portion. The resultant change in state attending the passage of one faraday is therefore the transfer of T equivalents of KC1 from the pure potassium chloride solution (where its concentration is c) into the boundary-portion (where its concentration has values varying from c to 0), and the transfer of T equivalents of NaCl from the boundary- portion (where its concentration has values varying from to c) into the pure sodium chloride solution (where its concentration is c). The free-energy-decrease attending these transfers is in this case an integral of an infinite number of infinitesimal free-energy-changes. By evaluating it and placing it equal to the electrical work that attends these transfers when they are brought about by the passage of the current, an expression for the liquid-potential is obtained. Pro ft. 38. Describe in detail the change in state that takes place when one faraday passes through the combination Cu(NO,) a (0.01 f.), Zn(NO 8 ) 2 (0.01f.). The general expression, obtained in the way just described, for the liquid-potential E between solutions of any two substances of the same ionic type having one ion in common and having equal concen- trations and ionizations is: EL = RT- A,, " ? A 02 In this expression A 01 and A 02 denote the equivalent conductances at zero concentration of the left-hand and right-hand solutes, respectively, and v denotes the number of positive charges on the ions not common to the two solutes, it being given a negative value when the charges are negative. In the derivation of this equation it is assumed that intermediate or complex ions are not present, and that the simple ions behave as perfect solutes and have the same relative conductances up to the concentration involved. Pro ft. S9. Calculate the liquid-potential at 18 of: a, KOH(0.1f.), KC1(0.1 f.) ; 6, HC1(0.01 f.). KC1(0.01 f.) ; c, ZnSO(0.1 f.), CuSO 4 (0.1 f.). Pro ft. 40. Calculate the electromotive force at 18 of the cell Ag, AgN0 3 (0.1n.), KNO s (0.1n.), KCl(0.1n.), KC1(1 n.) -f Hg,Cl 2 , Hg. Assume that the potassium chloride is 74% ionized at 1 normal and 84% ionized at 0.1 normal, and that the other two salts are 84% ionized at 0.1 normal. 168 ELECTROCHEMISTRY For combinations of solutions of two salts having a common ion but different concentrations, such as KNO 3 (0.1 n.), KCXO-Oln.), and for combinations of solutions of salts without a common-ion, such as KN"O 3 (0.1n.), NaCl(0.01n.), the calculation of the liquid-potential is very complicated. Combinations of these kinds can, however, be avoided, as illustrated by the cell of Prob. 40, by connecting the two solutions through intermediate ones so as to produce only combina- tions of the two types for which the liquid-potentials can be calculated as described above. It is to be noted that the above-derived expressions for the liquid- potentials involve the assumption that the ions behave as perfect solutes, and therefore that these expressions are exact only at small concentrations. The assumption is also involved in the case of tri- ionic salts that these are ionized only into the ultimate ions (such as K + and SO 4 = ), without formation of an appreciable proportion of the intermediate ion (such as KSO 4 ~). A liquid-potential can sometimes be experimentally determined by measuring the electromotive force of a cell involving it in which the two electrode-potentials are made substantially equal, as in the cells: Hg + Hg 2 Cl 2 , KCKO.lf.), NaCKO.lf.), Hg 2 CL + Hg. Ag, AgNO 3 (0.001 f.) + KNO 3 (0.1 f.), AgNO 3 (0.001 f.) + NaNO :! (0.1f.), Ag. ProJ). 41. a. Show that the electromotive force of the silver cell here formulated is substantially equal to the liquid-potential of KNO 3 (0.1f.), NaNO 8 (0.1f.), by specifying exactly what determines each of the three partial potentials of the cell. ft. Show why there would be a considerable error in deriving the liquid-potential of KNO 3 ( 0.1 f.), NaNO 3 ( 0.01 f.) from the electromotive force of a cell which differs from the one just considered only in the respect that the NaNO 3 (0.1f.) is replaced by NaN0 3 (0.01f.). 138. Determination of Ion-Concentrations and of Equilibrium- Constants Involving Them by Means of Electromotive-Force Meas- urements. Prol). 4%- Determination of Solubility. The electromotive force of the cell Ag + Agl, KI(0.1f.), KNO 8 (0.1f.), AgNO 3 (0.1 f. ) , Ag is 0.814 volt at 25. Calculate the solubility of silver iodide in water at 25. Assume the ionizations of all the substances to be equal and to have the average value shown by salts of the uniunivalent type (given in Art. 58). EQUILIBRIUM OF OXIDATION REACTIONS 169 Prob. 43. Determination of the lonization-Constant of Water. Calculate the ionization-constant of water (Art. 85) at 18 from the fact that the electromotive force of the cell fT 2 (latm.), HCl(0.1f.), KCl(0.1f.), KOH(0.1f.), # 2 (latm.) is 0.653 volt at 18. Make the same assumption in regard to the ionizations as in the preceding problem. Prob. 44. Determination of Complex-Constants. The electromotive force of the cell Ag, K+Ag(CN) 2 -(0.01 n.) -fKCN(ln.), KCl(ln.) t Hg 2 Cl 2 -fHg is 0.83 volt at 25. Calculate the dissociation-constant of the complex-ion Ag(CN) 2 ~ at 25. Assume the ionizatkm of the salts to be 74%, and neglect the liquid-potential, which is small in this case. Prob. 45. Determination of the Hydrolysis of Salts. When a solu- tion 0.05 formal in Na 2 HPO 4 is saturated with hydrogen at 1 atm., when a platinum electrode is placed in it, and when the half-cell HCl(0.01f.) -f- NaCl (0.1 f.), H 2 (~L atm.), is brought into contact with it, the cell thus formed is found to have at 18 an electromotive force of 0.398 volt. Calculate the hydrolysis of the salt, assuming complete ionization of the largely ionized substances, and neglecting the liquid-potential, which is made small by the addition of the sodium chloride. Prob. 46. Determination of Indicator-Constants. When the Na,HPO 4 solution of Prob. 45 is made 0.0001 formal in phenolphthalein the indi- cator is found to show a color 13% as intense as that which is pro- duced on adding to the solution an excess of sodium hydroxide. Calculate the ionization-constant of this indicator. 139. Derivation of the Equilibrium- Conditions of Oxidation Reac- tions from the Electrode-Potentials. When the concentrations of the substances involved in the two electrode-reactions are such that the two electrode-potentials are equal to each other, there is evidently no tendency for the cell to act nor for the chemical change to take place in it. In other words, the concentrations that make the two electrode-potentials equal are concentrations at which the chemical change is in equilibrium. The equilibrium-constant of the chem- ical change which is the resultant of the two electrode-reactions can therefore be calculated from the specific electrode-potentials. Prob. 4~- Calculate the concentration of copper-ion at which the reaction Zn -(- Cu++ = Cu-f Zn ++ is in equilibrium when the zinc-ion is 1 molal. Prob. 48. a. Calculate the concentration of hydrogen-ion at which the reaction Pb -f 2H+C1- = H 2 -f- Pb ++ Cl- 2 is in equilibrium at 25 when the hydrogen has a pressure of 1 atm. and the lead-ion is 0.03 molal. b. Formulate an algebraic relation between the equilibrium-constant of the corresponding ionic reaction and the specific electrode- potentials. c. Calculate the value of this equilibrium-constant. 170 ELECTROCHEMISTRY Prob. 49 ' Formulate an algebraic relation between the equilibrium- constant of the ionic reaction Ag-|- Fe* + * Ag+ -f- Fe ++ and the specific electrode-potentials involved. b. Calculate the value of this equilibrium- constant, c. Calculate the composition of the equilibrium-mixture that results when metallic silver is placed in 0.1 formal Fe(NO 3 ) 8 solution, assuming complete ionization of the salts. Prob. 50. Formulate a complete numerical expression for calculating the OH~ concentration at which 1000 times as much manganate-iou (MnO 4 =) as permanganate-ion (MnO 4 -) is present in contact with air when equilibrium is reached at 25. The specific electrode-potential of MuO 4 =, MnO 4 - has been roughly determined to be 0.61 volt at 25. Prob. 51. a. Derive an expression for the equilibrium-constant of the reaction Cu + Cu + + = 2 Cu* in terms of the specific electrode-poten- tials. I). Calculate the concentration of cuprous salt resulting when copper is shaken with a 0.1 formal CuSO 4 solution at 25, assuming complete ionization. 'c. Calculate the concentration of cupric salt in the equilibrium-mixture produced by shaking a 0.1 formal CuCl 2 solution with copper at 25. 140. Electromotive Force of Cells with Concentrated Solutions. The electromotive force of any cell in which solutes are present at large concentrations cannot be calculated from the specific electrode- potentials with the aid of the logarithmic concentration formula of Art. 134, since this holds true even approximately only when the concentration does not exceed about 0.1 normal. A knowledge of the change in state taking place in such a cell and of the reactions occurring at its electrodes is, however, of importance, since it shows qualitatively the factors which determine the magnitude of the elec* tromotive force. This is illustrated by the following problems, which relate to certain cells of technical importance. Prob. 52. One form of the Clark standard cell at 20 is repre- sented by the formula : Zn -f ZnS0 4 7H 2 0, ZnSO 4 16.8H 8 O, Hg 2 S0 4 -j- Hg. a. Specify the exact change of state that occurs when two faradays pass through the cell. b. Show that even in an actual cell, with only a finite quantity of solution, no variation of the electromotive force results when two faradays pass, provided the change in state takes place slowly enough. Prob. 58. The LeclanchS cell consists essentially of a zinc rod (amalgamated to diminish local voltaic action) dipping into a con- centrated NH 4 C1 solution containing ZnCl 2 , and a carbon rod coated with MnO 2 dipping into the same solution, a. Formulate the ion-reac- tion that takes place at each electrode and the whole reaction in the cell, taking into account the facts that the MnO 2 is reduced to Mn 2 O,, ELECTROMOTIVE FORCE OF CELLS 171 and that Zn(OH) 2 is soluble in NH 4 C1 solution. 6. Show in what direction the electromotive force of the cell would be changed by decreasing the concentration of the NH 4 C1. Note. The common "dry cell" is a Leclanche" cell to which some porous material, such as paper-pulp or sawdust, has been added to hold the liquid. Prol. 54. In the nickel-iron (Edison) storage cell, Fe, KOH(21%), Ni(OH) 3 , the main reaction is Pe -{- 2Ni(OH), Fe(OH), -f 2Ni(OH) a (the degree of hydration of the three oxides being, however, somewhat indefinite), a. Write the ion-reaction occurring at each electrode. 6. Show in what direction each of the electrode-potentials, and also the electromotive force of the whole cell, would vary with increase of the KOH concentration. 172 ELECTROCHEMISTRY VOLTAIC ACTION. ELECTROLYSIS, AND POLARIZATION 141. Concentration-Changes attending Voltaic Action and the Resulting Polarization. Throughout the foregoing considerations, as an aid in evaluating the electromotive force, it has been assumed that so large a quantity of solution is present in the cell that only infinitesimal concentration-changes are produced in it by the passage of a finite quantity of electricity. The fact that this is not the case in actual cells must be taken into account. Prob. 55. The electromotive force E at 15 of the lead storage-cell varies with the mol-fraction x of the H 2 SO 4 (for values of x up to 0.10) according to the equation E 1.855 -f- 3.80# 10# 2 . Out of a cer- tain cell which contains 1300 g. of 10 mol-percent H 2 SO 4 a steady current of 5.36 amperes is taken for 10 hours, a. Calculate the electro- motive force of the cell at the beginning and at the end. 6. Derive an integral (expressed in terms of molal composition and numerical co- efficients) by which the electrical energy producible by the outflow of any given quantity of electricity from any similar lead storage-cell could be calculated. In practice the electromotive force is often decreased much more than these considerations indicate because tho concentration-changes actually occur in the immediate neighborhood of the electrodes and are only gradually distributed by convection or diffusion through the whole body of the solution. Thus in the lead storage-cell water is produced and acid is destroyed by the electrode-reaction in the solu- tion impregnating the porous lead-peroxide electrode, and the acid can only be replenished from the main body of the solution by the slow process of diffusion. This phenomenon i? one kind of polariza- tion, sometimes called concentration-polarization; the name polar- ization being used in general to denote the production by the passage of the current of any change in the solution adjoining the electrode or in the surface of the electrode which makes its potential deviate from its normal value. Prob. 56. When a certain current is taken at 25 out of a certain cell of the form Zn, ZnSO 4 (lf.), CuSO 4 (lf.), Cu, the electromotive force soon becomes fairly constant and remains so for a time at a value 0.06 volt below its normal value, a. Show quantitatively how this might be accounted for by concentration-polarization. 6. Explain how the polarization would be affected by increasing the current ; by using ELECTROLYSIS AND POLARIZATION 173 smaller electrodes without changing the current, thereby increasing the current-density, by which is meant the current in amperes per unit- area (per square centimeter or decimeter) of electrode-surface ; and by stirring the solutions, these being separated from each other by a porous cup. 142. Electrolysis in Relation to Applied Electromotive Force. In order to produce electrolysis in any electrolytic cell there must obviously be applied from an external source an electromotive force a.t least equal to the electromotive force that is produced by the combination of solution and electrodes, considering it as a voltaic cell. The value of the electromotive force that must be applied to compensate this ~back electromotive force of the cell under the actual conditions, and thus produce appreciable electrolysis, is called the decomposition-potential. Its value, when not influenced by in- definite polarization-effects, such as are described in Art. 143, can therefore be calculated by the methods considered in Arts. 130-137. It is important to recognize this fact; for the decomposition-poten- tial is sometimes treated as if it were an essentially independent quantity. Pro&. 57. Experimental Determination of Decomposition-Potential. A 0.05 formal solution of CuSO 4 is placed at 25 in a certain electro- lytic cell between a mercury electrode covered with Hg 2 SO 4 and a cop- per electrode. An external electromotive force is applied at the mercury electrode, first of 0.1, then of 0.2, 0.3, 0.4, 0.5, and 0.6 volt. An ammeter placed in series with the cell shows that no appreciable current passes until 0.4 volt is applied, when the current is 0.8 milliamperes ; while with 0.5 volt the current is 2.8 milliamperes, and with 0.6 volt it is 4.8 milliamperes. a. Plot these values of the current as ordinates and of the potential as abscissas, and derive from the plot the value of the decomposition-potential. 6. Calculate the resistance of the solution. ProT). 58. Deposition of Metals ~by Electrolysis. A solution 0.05 formal in H 2 SO 4 and 0.05 formal in CuSO 4 is electrolyzed at 25 between a mercury anode and a platinum cathode. In the mixture the salt may be assumed to be 35% ionized and the acid to be 35% ionized into 2H+ and SO 4 = and 60% ionized into H + and HSO 4 -. a. Calculate from the electrode-potentials (taking that of Hg 4. Hg 2 S0 4 , SO 4 =(lm.) as 0.62 volt) the minimum electromotive force which would have to be applied in order to cause the copper to deposit. 6. To what value would this electromotive force have to be increased after 99% of the copper had been precipitated in order that the deposition might continue, assuming 174 ELECTROCHEMISTRY the ionizations to be the same as in the original mixture? c. What is the minimum electromotive force at which hydrogen could be continu- ously set free, assuming that it attains at the cathode an effective pressure of 1 atm.? Prod. 59. Separation of Elements by Electrolysis. A solution 0.1 normal in KC1, 0.1 normal in KBr, and 0.1 normal in KI is placed, together with a platinum electrode, in a porous cup ; and this is placed within a larger vessel containing a zinc electrode and a large quan- tity of 0.1 normal ZnCl 2 solution. Assuming complete ionization of the salts, calculate the applied electromotive force required at 25, a, to liberate 99.9% of the iodine ; 6. to set bromine free at a concentration of 0.0001 molal; c, to liberate 99.9% of the bromine (which remains in solution) ; and d, to liberate chlorine at a concentration of 0.0001 molal. 143. Electrolysis in Relation to Polarization. In electrolysis, as in voltaic action, concentration-changes are produced at the elec- trodes; and these have the effect of increasing the applied electro- motive force required. Thus in electrolyzing a solution of copper sulphate between copper electrodes (as is done in copper plating) only an infinitesimal electromotive force is required to start the deposition; but the solution around the cathode soon becomes less concentrated in copper and the solution around the anode more con- centrated in copper, producing a concentration-cell, like those of Art. 131, with an electromotive force opposite to that applied. This back electromotive force can of course be calculated for any given concentration-changes. It would evidently be diminished by decreas- ing the current-density, by agitating the whole solution, or by rotating the electrodes. When a gas, such as hydrogen or oxygen, is set free at an elec- trode, a phenomenon, known as gas-polarizaiion, is observed which does not occur in the deposition of metals. For when a gas is in- volved in the electrode-reaction its partial pressure determines the electrode-potential. Hence in a cell exposed to the air a slow, con- tinuous electrolysis may take place before the applied electromotive force equals the electromotive force of the cell when the gas-pressure is one atmosphere; for the gas-forming substance is produced on the electrode at a lower pressure and is dissolved in the solution and carried away by convection and diffusion. Moreover, a sudden in- ELECTROLYSIS AND POLARIZATION 175 crease in the rate of electrolysis does not take place when the applied electromotive force is increased beyond that corresponding to a pres- sure of one atmosphere; for the gas-forming substance then passes into the electrode-surface at this high pressure, producing a super- saturated solid solution from which the gas does not escape rapidly enough to reduce the effective pressure to one atmosphere.* There is therefore a back electromotive force produced by the cell which is larger than its electromotive force when the gas-pressure is one atmosphere. The excess of the one over the other is called the polarization or overvoltage. Proft. 60. Concentration-Polarization. Calculate the electromotive force of polarization that would result in electrolyzing at 25 a 0.01 formal CuSO 4 solution between copper electrodes if the copper-ion con- centration became 0.001 formal around the cathode and 0.1 formal around the anode, assuming equal ionization of the salt at these two concentrations. Prob. 61. Gas Polarization. Electromotive forces successively increasing in magnitude were applied at 22 to an electrolytic cell con- sisting of a large, unpolarizable, platinized platinum plate as anode, a small mercury surface as cathode, and a 0.1 normal H 2 SO 4 solution as electrolyte. Hydrogen gas at 1 atm. was bubbled steadily through the cell, and a resistance of 100,000 ohms was placed in series with it, the resistance of the cell being negligible in comparison. The current- strengths i in millionths of an ampere corresponding to various applied electromotive forces E in volts were as follows: i 0.06 0.44 1.20 2.20 3.70 4.82 E 0.32 0.48 0.62 0.77 0.95 1.08 a. Plot these current-strengths as ordinates against the electromotive forces as abscissas. 6. Calculate the back electromotive force corre- * In some cases the solid solution is probably a solution in the metal of the gas-forming substance itself or of the corresponding monatomic substance (thus of H or O in the case of H 2 or O 2 ). In other cases there is evidence that it is a solution in the metal of an unstable compound of the metal with the gas-forming substance ; thus higher oxides such as PtO 8 and NiO 3 seem to be formed when oxygen is set free on platinum or nickel. Such compounds have a high dissociation-pressure and are therefore equivalent in their effect on the potential to the gas Itself at the same high pressure, assuming equilibrium to be attained between them and the electrolyte in the solution. The slowness with which the gas escapes from these solid solutions may arise in part from the small rate at which the reactions (such as 2H = H 2 , or 2PtO 3 = 2Pt + 3O 2 ) take place within the electrode-surface. The rate of escape of the gas is, moreover, doubtless largely determined by surface-tension effects. 176 ELECTROCHEMISTRY spending to each of these applied electromotive forces, record these values, and plot the current-strengths against them on the same dia- gram, c. Calculate the effective pressure of the hydrogen at the elec- trode corresponding to the smallest of these back electromotive forces. The back electromotive force is experimentally determined by the method illustrated in Probs. 57 and 61, or by finding the smallest value of the applied electromotive force at which bubbles form slowly but continuously at the electrode surface (this giving a value corresponding practically to a minimum current-density), or by applying a definite electromotive force to the cell long enough to charge the electrodes with the decomposition-products and then short- circuiting the electrodes through a high-resistance potentiometer, the applied potential being at the same time removed. From this back electromotive force the overvoltage is obtained by subtracting the theoretical electromotive force corresponding to a gas-pressure of one atmosphere. The overvoltage is always found to increase with increase of the applied electromotive force and of the current-density; but it varies in a highly specific way with the chemical nature of the gas and with the chemical nature and physical state of the metallic electrode. Thus in certain experiments made with 1-normal H 2 SO 4 with a low applied potential and low current-density the overvoltage of the hydrogen was found to be zero on a platinized electrode, 0.03 volt on smooth platinum, 0.36 volt on lead, and 0.44 volt on mercury; while with a much higher applied potential and current-density it was found to be 0.07 volt on platinized platinum, 0.65 volt on smooth platinum, 1.23 volts on rough lead, and 1.30 volts on mercury. And in certain experiments with 2-normal KOH with a moderate cur- rent-density the overvoltage of the oxygen was found at the start to be 0.44 volt on platinized platinum, 0.84 volt on smooth platinum, and 0.50 von on iron; the value increasing to 1.46 on smooth plat- inum and to 0.59 on iron after two hours' passage of the current. The phenomenon of gas-polarization and the overvoltage attend- ing it are of great significance in technical processes. The over- voltage may greatly diminish the energy-efficiency of the process, the energy-efficiency being the ratio of the minimum electrical energy theoretically required to produce a definite quantity of some ELECTROLYSIS AND POLARIZATION 177 product of the electrolysis to the energy actually expended. Over- voltage may also make processes practicable which would otherwise not be possible; thus in charging a lead storage-cell hydrogen is not set free at the lead electrode owing to its overvoltage, although the potential of the half-cell jff 2 (latm.), H 2 SO 4 20H 2 O is about 0.4 volt less than that of the half-cell Pb -f PbS0 4 , H 2 SO 4 20H 2 0. Pro&. 62. Energy-Efficiency and Overvoltagei In a certain com- mercial alkali-chlorine cell sodium hydroxide and chlorine are produced at an iron cathode and graphite anode, respectively, by the continuous electrolysis of a 25% NaCl solution which is slowly flowed through a diaphragm from the anode to the cathode compartment (to prevent hydroxide-ions from migrating to the anode). In practice 4.5 volts are applied to the cell, whose resistance is 0.00080 ohm, yielding a current of 2000 amperes. There flow off each hour from the cathode 27,460 g. of solution containing 10% NaOH and 13% NaCl. The electromotive force of the voltaic cell # 2 (latm. ), NaOH ( 10% ) -f NaCl (13%), NaCl (25%), C7,(latm.) has been independently determined to be 2.3 volts, a. Cal- culate the current-efficiency in the production of the sodium hydroxide. &. Calculate the energy-efficiency, c. Calculate the overvoltage and the percentage loss in energy-efficiency that arises from it. 144. Review Problems. Pro&. 63. The potential of a platinum electrode in an acid solution of potassium iodate and iodine is found to depend only on the concen- trations of H+, IO 3 -, and I 2 . a. Write the electrode-reaction attending the passage of ten faradays from the electrode to the solution. 6. For- mulate an expression by which the electrode-potential E at 25 for any small concentrations (H + ), (I