210 Chapter 5: Gibbs and Helmholtz Energies and Their Applications 5.27 Calculate the number of components present in each of the following situations: (a) Water, including autodissociation into H+ and OH~ ions. (b) Consider the following reaction in a closed container: 2NH3(g)-N2(g) + 3H2(g) (i) All three gases were present initially in arbitrary amounts, but the temperature is too low for the reaction to occur. (ii) Same as (i), but the temperature is raised sufficiently to allow for the equilibrium to be established. (iii) Initially only NH3 was present. The system is then allowed to reach equilibrium. Additional Problems 5.28 Give the conditions under which each of the following equations may be applied. (a) dA < 0 (for equilibrium and spontaneity) (b) dG < 0 (for equilibrium and spontaneity) M 1n*2 = Ag(r2-7j) wmP1 R TXT2 (d) AG = nRT In ^ 5.29 When ammonium nitrate is dissolved in water, the solution becomes colder. What conclusion can you draw about A.S° for the process? 5.30 Protein molecules are polypeptide chains made up of amino acids. In their physiologically functioning or native state, these chains fold in a unique manner such that the nonpolar groups of the amino acids are usually buried in the interior region of the proteins, where there is little or no contact with water. When a protein denatures, the chain unfolds so that these nonpolar groups are exposed to water. A useful estimate of the changes of the thermodynamic quantities as a result of denaturation is to consider the transfer of a hydrocarbon such as methane (a non-polar substance) from an inert solvent (such as benzene or carbon tetrachloride) to the aqueous environment: (a) CH4(inert solvent) -> CH4(g) (b) CHtfe) - CH4(a /if", dG > 0, and the process will be spontaneous from B to A. As we shall see later, the transfer can be from one phase to another or from one state of chemical combination to another. The transfer can be transport by diffusion, evaporation, sublimation, condensation, crystallization, solution formation, or chemical reaction. Regardless of the nature of the process, in each case the transfer proceeds from 218 Chapter 6: Nonelectrolyte Solutions a higher ft value to a lower ft value. This characteristic explains the name chemical potential. In mechanics, the direction of spontaneous change always takes the system from a higher potential-energy state to a lower one. In thermodynamics, the situation is not quite so simple because we have to consider both energy and entropy factors. Nevertheless, we know that at constant temperature and pressure, the direction of a spontaneous change is always toward a decrease in the system's Gibbs energy. Thus, the role Gibbs energy plays in thermodynamics is analogous to that of potential energy in mechanics. This is the reason that molar Gibbs energy or, more commonly, partial molar Gibbs energy, is called the chemical potential. 6.3 Thermodynamics of Mixing The formation of solutions is governed by the principles of thermodynamics. In this section, we shall discuss the changes in thermodynamic quantities that result from mixing. In particular, we shall focus on gases. Equation 6.9 gives the dependence of the Gibbs energy of a system on its composition. The spontaneous mixing of gases is accompanied by a change in composition; consequently, the system's Gibbs energy decreases. In Section 5.4, we obtained an expression for the molar Gibbs energy of an ideal gas (Equation 5.18): G = G° + RTlnTT— In a mixture of ideal gases, the chemical potential of the z'th component is given by ft = /!}> + *T In y^j bar (6.10) For simplicity, we omit the term "1 bar." Note that the resulting P values are dimensionless. where Pi is the partial pressure of component z in the mixture and fi° is the standard chemical potential of component i when its partial pressure is 1 bar. Now consider the mixing of nx moles of gas 1 at temperature T and pressure P with n2 moles of gas 2 at the same T and P. Before mixing, the total Gibbs energy of the system is given by Equation 6.9, where chemical potentials are the same as molar Gibbs energies, G = nxGx + n2G2 = nxfix + n2fi2 initial = "M + RTlnP) + n2(fi + RTIn P) After mixing, the gases exert partial pressures Px and P2, where Pl + P2 = P, and the Gibbs energy is* Gfi„ai = «i(ft° + BT In + nln\ + RT\nP2) * Note that Px + P2 = P only if there is no change in volume as a result of mixing; that is, hmi*V = 0. This condition holds for ideal solutions. 6.3 Thermodynamics of Mixing 219 The Gibbs energy of mixing, AmixG, is given by AmixC? = Gflnal — ^initial = nxRT \ny+ n2RT In y = nxRT\nxx + ^^In^ where Px = xxP and P2 = x2P, and xx and x2 are the mole fractions of 1 and 2, respectively. (The standard chemical potential, is the same in the pure state and in the mixture.) Further, from the relations x, = —:-= — and x7 = —;-= — 1 nx + n2 n L nx + n2 n where n is the total number of moles, we have AmixG = nRT(xx]n xx + x2ln x2) (6.11) Because both jcj and x2 are less than unity, In xx and In x2 are negative quantities, and hence so is AmixG. This result is consistent with our expectation that the mixing of gases is a spontaneous process at constant T and P. Now we can calculate other thermodynamic quantities of mixing. From Equation 5.10, we see that at constant pressure dT 8G\ = -S ip Thus, the entropy of mixing is obtained by differentiating Equation 6.11 with respect to temperature at constant pressure: dAmixG 8T Ip nR(xx In xx + x2 In x2) = -AmixS or &mixS = —nR(xx lnxx + x2 InjCj) (6-12) This result is equivalent to Equation 4.17. The minus sign in Equation 6.12 makes AmixS a positive quantity, in accord with a spontaneous process. The enthalpy of mixing is given by rearranging Equation 5.3: Amix# = AmixG + TAmixS 220 Chapter 6: Nonelectrolyte Solutions Figure 6.4 Plots of 7"AmixS, AmixH, and AmixG as a function of composition for the mixing of two components to form an ideal solution. to CD 3 0 o -3 + 0.0 0.2 0.4 0.6 0.8 1.0 This result is not surprising, because molecules of ideal gases do not interact with one another, so no heat is absorbed or produced as a result of mixing. Figure 6.4 shows the plots of AmixG, TAmixS, and Amix/f for a two-component system as a function of composition. Note that both the maximum (for TAmixS) and the minimum (for AmixG) occur at xx = 0.5. This result means that we achieve maximum disorder by mixing equimolar amounts of gases and that the Gibbs energy of mixing reaches a minimum at this point (see Problem 6.57). Reversing the process for a two-component solution of equal mole fractions leads to an increase in Gibbs energy and a decrease in entropy of the system, so energy must be supplied to the system from the surroundings. Initially, at xx ~ x2, the AminG and TAmixS curves are fairly flat (see Figure 6.4), and separation can be carried out easily. However, as the solution becomes progressively richer in one component, say 1, the curves become very steep. Then, a considerable amount of energy input is needed to separate component 2 from 1. This difficulty is encountered, for example, in the attempt to clean up a lake contaminated by small amounts of undesirable chemicals. The same consideration applies to the purification of compounds. Preparing most compounds in 95% purity is relatively easy, but much more effort is required to attain 99% or higher purity, which is needed, for example, for the silicon crystals used in solid-state electronics. As another example, let us explore the possibility of mining gold from the oceans. Estimates are that there is approximately 4 x 10-12 g of gold/mL of seawater. This amount may not seem like much, but when we multiply it by the total volume of ocean water, 1.5 x 1021 L, we find the amount of gold present to be 6 x 1012 g or 7 million tons, which should satisfy anybody. Unfortunately, not only is the concentration of gold in seawater very low, but gold is also just one of some 60 different elements in the ocean. Separating one pure component initially present in a very low concentration in seawater (i.e., starting at the steep portions of the curves in Figure 6.4) would be a very formidable (and expensive) undertaking indeed. 6.4 Binary Mixtures of Volatile Liquids EXAMPLE 6.1 Calculate the Gibbs energy and entropy of mixing 1.6 moles of argon at 1 atm and 25°C with 2.6 moles of nitrogen at 1 atm and 25°C. Assume ideal behavior. ANSWER The mole fractions of argon and neon are 1.6 _ 2.6 *Ar ~ 1.6 + 2.6 ~ °-38 X^ - 1.6 + 2.6 ~ 0,62 From Equation 6.11, AmixG = nRT(xx lnxj + x2 lnx2) = (4.2 mol)(8.314 JK"1 mol"1)(298 K)[(0.38) In 0.38 + (0.62) In 0.62] = -6.9 kJ Because AmixS = AmixGJT, we write -6.9 x 103 J 298 K 23 J K"1 COMMENT In this example, the gases are at the same temperature and pressure when they are mixed. If the initial pressures of the gases differ, then there will be two contributions to AmixG: the mixing itself and the changes in pressure. Problem 6.58 illustrates this situation. 6.4 Binary Mixtures of Volatile Liquids The results obtained in Section 6.3 for mixtures of gases also apply to ideal liquid solutions. For the study of solutions, we need to know how to express the chemical potential of each component. We shall consider a solution containing two volatile liquids, that is, liquids with easily measurable vapor pressures. Let us start with a liquid in equilibrium with its vapor in a closed container. Because the system is at equilibrium, the chemical potentials of the liquid phase and the vapor phase must be the same, that is, 222 Chapter 6: Nonelectrolyte Solutions where the asterisk denotes a pure component. Further, from the expression for fj*(g) for an ideal gas, we can write'!' where p°(g) is the standard chemical potential at P* = 1 bar. For a two-component solution at equilibrium with its vapor, the chemical potential of each component is still the same in the two phases. Thus, for component 1 we write f*i(l) = M = M°(g) + RTl*^ (6-14) where Pr is the partial pressure. Because ft°(g) = M°(g), we can combine the previous two equations to get M = fife) + *T]n^ = ri(/) + *rin£ (6.15) " l Thus, the chemical potential of component 1 in solution is expressed in terms of the chemical potential of the liquid in the pure state and the vapor pressures of the liquid in solution and in the pure state. Raoult's Law The French chemist Francois Marie Raoult (1830-1901) found that for some solutions, the ratio P\fP{ in Equation 6.15 is equal to the mole fraction of component 1, that is, P\ Xl or Px = xxP\ (6.16) Equation 6.16 is known as Raoult's law, which states that the vapor pressure of a component of a solution is equal to the product of its mole fraction and the vapor pressure of the pure liquid. Substituting Equation 6.16 into Equation 6.15, we obtain fix{l) = ft\{l) + RT\nXl (6.17) t This equation follows from Equation 5.18. For a pure component, the chemical potential is equal to the molar Gibbs energy. 6.4 Binary Mixtures of Volatile Liquids We see that in a pure liquid (xx = 1 and lnxl = 0), nx{l) = Solutions that obey Raoult's law are called ideal solutions. An example of a nearly ideal solution is the benzene-toluene system. Figure 6.5 shows a plot of the vapor pressures versus the mole fraction of benzene. Benzene Toluene I — Pc.h. + Pc. 600 - CD X E E 400 200 0.0 0.2 0.4 0.6 *CßHB Figure 6.5 Total vapor pressure of the benzene-toluene mixture as a function of the benzene mole fraction at 80.1 °C. 0.8 1.0 EXAMPLE 6.2 Liquids A and B form an ideal solution. At 45°C, the vapor pressures of pure A and pure B are 66 torr and 88 torr, respectively. Calculate the composition of the vapor in equilibrium with a solution containing 36 mole percent A at this temperature. ANSWER Because xA = 0.36 and xB = 1 — 0.36 = 0.64, we have, according to Raoult's law PA = xpfl = 0.36(66 torr) = 23.8 torr PB = xBPB = 0.64(88 torr) = 56.3 torr The total vapor pressure, PT, is given by 224 Chapter 6: Nonelectrolyte Solutions Finally, the mole fractions of A and B in the vapor phase, x\ and x\, are given by XA Pt 23.8 torr _ = on. . - 0.30 80.1 torr and XB p 56.3 torr = „ni( - 0.70 80.1 torr In an ideal solution, all intermolecular forces are equal, whether the molecules are alike or not. The benzene-toluene system approximates this requirement because benzene and toluene molecules have similar shapes and electronic structures. For an ideal solution, we have both AmixH = 0 and AmixF = 0. Most solutions do not behave ideally, however. Figure 6.6 shows the positive and negative deviations from Raoult's law. The positive deviation (Figure 6.6a) corresponds to the case in which the intermolecular forces between unlike molecules are weaker than those between like molecules, and *CS2 XCHCI; (a) (b) Figure 6.6 Deviations from Raoult's law by nonideal solutions, (a) Positive deviation from Raoult's law: carbon disulfide-acetone system at 35.2°C. (b) Negative deviation: chloroform-acetone system at 25.2°C. (From Hildebrand, J. and R. Scott, The Solubility of Nonelectrolytes © 1950, Litton Educational Publishing. Reprinted by permission of Van Nostrand Reinhold Company.) 6.4 Binary Mixtures of Volatile Liquids 225 there is a greater tendency for these molecules to leave the solution than in the case of an ideal solution. Consequently, the vapor pressure of the solution is greater than the sum of the vapor pressures for an ideal solution. Just the opposite holds for a negative deviation from Raoult's law (Figure 6.6b). In this case, unlike molecules attract each other more strongly than they do their own kind, and the vapor pressure of the solution is less than the sum of the vapor pressures for an ideal solution. Henry's Law When one solution component is present in excess (this component is called the solvent), its vapor pressure is quite accurately described by Equation 6.16. The regions where Raoult's law is applicable are shown for the carbon disulfide-acetone system in Figure 6.7. In contrast, the vapor pressure of the component present in a small amount (this component is called the solute) does not vary with the composition of the solution, as predicted by Equation 6.16. Still, the vapor pressure of the solute varies with concentration in a linear manner: There is no sharp distinction between solvent and solute. Where applicable, we shall call component 1 the solvent and component 2 the solute. P2 = Kx2 (6.18) Equation 6.18 is known as Henry's law (after the English chemist William Henry, 1775-1836), where K, the Henry's law constant, has units of pressure. Henry's law relates the mole fraction of the solute to its partial (vapor) pressure. Alternatively, Henry's law can be expressed as P2 = K'm (6.19) 226 Chapter 6: Nonelectrolyte Solutions Table 6.1 Henry's Law Constants for Some Gases in Water at 298 K Gas K/torr KVatm moM kg H20 H2 5.54 x 107 1311 He 1.12 x 108 2649 Ar 2.80 x 107 662 N2 6.80 x 107 1610 02 3.27 x 107 773 co2 1.24 x 106 29.3 H2S 4.27 x 105 10.1 where m is the molality of the solution and the constant K' now has the units atm mol-1 kg of the solvent. Table 6.1 lists the values of K and K' for several gases in water at 298 K. Henry's law is normally associated with solutions of gases in liquids, although it is equally applicable to solutions containing nongaseous volatile solutes. It has great practical importance in chemical and biological systems and therefore merits further discussion. The effervescence observed when a soft drink or champagne bottle is opened is a nice demonstration of the decrease in gas—mostly C02—solubility as its partial pressure is lowered. The emboli (gas bubbles in the bloodstream) suffered by deep-sea divers who rise to the surface too rapidly also illustrate Henry's law. At a point some 40 m below the surface of seawater, the total pressure is about 6 atm. The solubility of nitrogen in the blood plasma is then about 0.8 x 6 atm/1610 atm mol"1 kg H20, or 3.0 x 10~3 mol (kg H20)_1, which is six times the solubility at sea level. If the diver swims upward too rapidly, dissolved nitrogen gas will start boiling off. The mildest result is dizziness; the most serious, death* Because helium is less soluble in the blood plasma than nitrogen is, it is the preferred gas for diluting oxygen gas for use in deep-sea diving tanks. There are several types of deviations from Henry's law. First, as mentioned earlier, the law holds only for dilute solutions. Second, if the dissolved gas interacts chemically with the solvent, then the solubility can be greatly enhanced. Gases such as C02, H2S, NH3, and HC1 all have high solubilities in water because they react with the solvent. The third type of deviation is illustrated by the dissolution of oxygen in blood. Normally, oxygen is only sparingly soluble in water (see Table 6.1), but its solubility increases dramatically if the solution contains hemoglobin or myoglobin. * For other interesting illustrations of Henry's law, see T. C. Loose, J. Chem. Educ. 48,154 (1971), W. J. Ebel, J. Chem. Educ. 50, 559 (1973), and D. R. Kimbrough, J. Chem. Educ. 76, 1509 (1999). 6.4 Binary Mixtures of Volatile Liquids 227 EXAMPLE 6.3 Calculate the molal solubility of carbon dioxide in water at 298 K and a C02 pressure of 3.3 x 10"4 atm, which corresponds to the partial pressure of C02 in air. ANSWER The mole fraction of solute (carbon dioxide) is given by Equation 6.18: *co2 — co2 K Because the number of moles of C02 dissolved in 1000 g of water is small, we can approximate the mole fraction as follows: fco, vco, ~ C°2 WC02 + WH2Q WH2Q so that fco, Ax)2WH20 K Finally, looking up the lvalue of C02 in Table 6.1 we write 1000 g 1 nco = (3.3 x 10-4 x 760) torr x rr~:-^-pr x , ^ 1A6+— LU2 v ' 18.01 g mol 1 1.24 x 10b torr = 1.12 x 10"5 mol m Because this is the number of moles of C02 in 1000 g or 1 kg of H20, the molality is 1.12 x 10"5 mol (kg H20)_1. Alternatively, we can use Equation 6.19 and proceed as follows: COMMENT Carbon dioxide dissolved in water is converted to carbonic acid, which causes water that is exposed to air for a long period of time to become acidic. ' co2 3.3 x 10~4 atm 29.3 atm mol-1 kg H20 1.12 x 10"5 mol (kg HjO)"1 228 Chapter 6: Nonelectrolyte Solutions 6.5 Real Solutions As pointed out in Section 6.4, most solutions do not behave ideally. One problem that immediately arises in dealing with nonideal solutions is how to write the chemical potentials for the solvent and solute components. The Solvent Component Let us look at the solvent component first. As we saw earlier, the chemical potential of the solvent in an ideal solution is given by (see Equation 6.17) /"i(0 = Pi® + RT In xx where xx = Px/Pi* and P* is the equilibrium vapor pressure of pure component 1 at T. The standard state is the pure liquid and is attained when xx = 1. For a nonideal solution, we write /ix(l) = /i[(t) + RT In ax (6.20) where ax is the activity of the solvent. Nonideality is the consequence of unequal inter-molecular forces between solvent-solvent and solvent-solute molecules. Therefore, the extent of nonideality depends on the composition of solution, and the activity of the solvent plays the role of "effective" concentration. The solvent's activity can be expressed in terms of vapor pressure as «i = (6-21) Table 6.2 Activity of Water in Water-Urea Solutions at 273 Ka Molality of urea, m2 Mole fraction of water, Vapor pressure of water, P^atm Activity of water, Activity coefficient of water, yx 0 1.000 6.025 x 10"3 1.000 1.000 1 0.982 5.933 x 10"3 0.985 1.003 2 0.965 5.846 x 10"3 0.970 1.005 4 0.933 5.672 x 10-3 0.942 1.010 6 0.902 5.501 x 10"3 0.913 1.012 10 0.847 5.163 x IQ"3 0.857 1.012 a Data from National Research Council. International Critical Tables of Numerical Data: Physics, Chemistry, and Technology, Vol. 3. © 1928, McGraw-Hill. Used by permission of McGraw-Hill. Note that the solute (urea) is nonvolatile. 6.5 Real Solutions 229 where Pl is the partial vapor pressure of component 1 over the (nonideal) solution. Activity is related to concentration (mole fraction) as follows: «i = V\x\ (6-22) where yx is the activity coefficient. Equation 6.20 can now be written as fix{l) = fi\{l) + RT In ft + RT In xt (6.23) The value of yl is a measure of the deviation from ideality. In the limiting case, where Xj —> 1, y1 —>■ 1 and activity and the mole fraction are identical. This condition also holds for an ideal solution at all concentrations. Equation 6.21 provides a way of obtaining the activity of the solvent. By measuring i5! of the solvent vapor over a range of concentrations, we can calculate the value of ax at each concentration if P\ is known.? Table 6.2 shows the activity of water in various water-urea solutions. The Solute Component We now come to the solute. Ideal solutions in which both components obey Raoult's law over the entire concentration range are rare. In dilute, nonideal solutions in which there is no chemical interaction, the solvent obeys Raoult's law, and the solute obeys Henry's law.* Such solutions are sometimes called "ideal dilute solutions." If the solution were ideal, the chemical potential of the solute is also given by Raoult's law M = + RTlnx2 = lf2{l) + RT\n^ In an ideal dilute solution, Henry's law applies. That is P2 = Kx2, so that M = fffl) + RTln£+ RT In x2 = H°2(l) + RTIn x2 (6.24) where ix2(l) = /i*2(l) + RT In (KjP*2). Although Equation 6.24 seems to take the same form as Equation 6.17, there is an important difference, which lies in the choice of standard state. According to Equation 6.24, the standard state is defined as the pure solute, attained by setting x2 = 1. Equation 6.24 holds only for dilute solutions, however. How can these two conditions be met simultaneously? The simple way out of this dilemma is to recognize that standard states are often hypothetical states, not physically realizable. Thus, the standard state of solute defined by Equation 6.24 is f To obtain the value of P±, we must measure the total pressure, P, and also analyze the composition of the mixture. Then we can calculate partial pressure Pl using Equation 1.6; that is, = x\P, where x\ is the mole fraction of the solvent in the vapor phase. * For ideal solutions, Raoult's law and Henry's law become identical; that is, P2 = Kx2 = P2X2- 230 Chapter 6: Nonelectrolyte Solutions the hypothetical pure component 2 with a vapor pressure equal to K (when x2 = 1, P2 = K). In a sense, this is an "infinite dilution state of unit mole fraction"; that is, it is infinitely dilute with respect to component 1, the solvent, with the solute at unit mole fraction. For nonideal solutions in general (beyond the dilute solution limit), Equation 6.24 is modified to fi2(l) = + RT In a2 (6.25) where a2 is the activity of the solute. As in the case of the solvent component, we have a2 = y2x2, where y2 is the activity coefficient of the solute. Here, we have a2 —*■ x2 or y2 —» 1 as x2 —> 0. Henry's law is now given by Ka, (6.26) Concentrations are usually expressed in molalities (or molarities) instead of mole fractions. In molality, Equation 6.24 takes the form /i2(I) = H2{1) + RT In m° (6.27) where m° = 1 mol kg"1 so that the ratio m2\m° is dimensionless. Here, the standard state is defined as a state at unit molality but in which the solution is behaving ideally. Again, this standard state is a hypothetical state, not attainable in practice (Figure 6.8). For nonideal solutions, Equation 6.27 is rewritten as ^2{l) = til{l)+RT\na2 (6.28) where a2 = ^("W"*0)-1*1 tne limiting case of m2 —> 0, we have a2 —> m2jm° or y2 —* 1 (see Figure 6.8b). Ideal solution In 012 (a) m2 m~° (b) Figure 6.8 (a) Chemical potential of a solute plotted against the logarithm of molality for a nonideal solution, (b) Activity of a solute as a function of molality for a nonideal solution. The standard state is at m2/m° = 1. 6.6 Phase Equilibria of Two-Component Systems Keep in mind that although Equations 6.24 and 6.27 were derived using Henry's law, they are applicable to any solute, whether or not it is volatile. These expressions are useful in discussing the colligative properties of solutions (see Section 6.7), and, as we shall see in Chapter 8, in deriving the equilibrium constant. 6.6 Phase Equilibria of Two-Component Systems Before we study colligative properties, let us first apply the phase diagram and the Gibbs phase rule to the properties of solutions containing two components. Distillation The separation of two volatile liquid components is usually accomplished by fractional distillation, which has many applications in the laboratory and in industrial processes. To use this procedure, we must understand how pressure and temperature affect the vapor-liquid equilibrium of binary liquid mixtures. Pressure-Composition Diagram. We begin by constructing phase diagrams that show the vapor pressure of a solution as a function of mole fraction and as a function of the composition of the vapor in equilibrium with the solution. Using the ideal benzene-toluene solution as an example, we can express the vapor pressures of both components in terms of Raoult's law: where xb and xt are the mole fractions of benzene and toluene in solution, respectively, and the asterisk denotes a pure component. The total pressure, P, is given by Figure 6.9a shows a plot of P versus xb, which is a straight line. Next we want to express P in terms of x\, which is the mole fraction of benzene in the vapor phase. According to Equation 1.6, the mole fraction of benzene in the vapor phase is given by Pb = xbPb and Pt = xtft* P = Pb + Pt - xbPb + xtPt = xbp*b + (l - xb)p; = p; + (p*b-p;)Xb (6.29) xb^b p; + (p;-p;)Xb This equation can be solved for xb to obtain the following expression for the mole fraction of benzene in solution that corresponds to its mole fraction xb in the vapor phase at equilibrium: 232 Chapter 6: Nonelectrolyte Solutions 0.0 0.2 0.4 0.6 0.8 1.0 xb or xvb (b) (c) Figure 6.9 Pressure versus composition liquid (L)-vapor (V) phase diagram for the benzene-toluene solution at 23°C. (a) Plot of vapor pressure versus the mole fraction of a benzene in solution, (b) Plot of vapor pressure versus the mole fraction of benzene in the vapor phase, (c) Combined plots of (a) and (b). Above the straight line, the system is entirely in the liquid state; below the curve, the system is entirely in the vapor state; both liquid and vapor are present in the enclosed area. At point c, the liquid composition isxb = 0.30 and the vapor composition isxjj = 0.80. Again, referring to Equation 1.6 and Raoult's law, Pb = A>P = xb^b and using Equation 6.30, we write p = V£ =-KP[- /fi 31x xi p;-(p;-p;)xi ^ Figure 6.9b shows a plot of P versus x\. The combined plots of Equations 6.29 and 6.31 are shown in Figure 6.9c. Above the straight line, the system is in the liquid state. (This result is consistent with our expectation that at high pressures, liquid is the more stable phase.) Below the curve, that is, at low pressures, the system exists as a vapor. At a point between the two lines (i.e., a point in the enclosed area) both liquid and vapor are present. 6.6 Phase Equilibria of Two-Component Systems Figure 6.10 The lever rule gives the ratio of moles present in each phase of a two-phase system as nL(L = nv€v, where (L and €v are the distances from the point corresponding to the system's overall mole fraction to the endpoints of the tie line. Now suppose we begin with the benzene-toluene system initially in the liquid phase at point a in Figure 6.9c. According to the Gibbs phase rule (see p. 196), we have a two-component, one-phase system, so that the degrees of freedom are f=c-p+2 =2-1+2=3 At constant temperature and a fixed mole fraction, we are left with one degree of freedom. Therefore, we can lower the pressure at constant composition and temperature (from point a) until point b is reached. At this point, the liquid can exist in equilibrium with its vapor. (This is the point at which the liquid just begins to vaporize.) Further decreasing the pressure takes us to point c. The horizontal "tie line" indicates that the composition (in mole fraction) of the liquid is /and that of the vapor is g at this point. The relative amounts of liquid and vapor present in equilibrium are given by the lever rule (Figure 6.10): »L€L = «v€v (6.32) where «L and «y are the number of moles of liquid and vapor, respectively, and €L and €v are the distances defined in Figure 6.10. At point c, we have a two-component, two-phase system so that the total degrees of freedom are /= 2 - 2 + 2 = 2 At constant temperature, then, we have only one degree of freedom, so that if we select a particular pressure, the compositions of the liquid and vapor must be fixed (as shown by the tie line). Figure 6.11 suggests a way to separate benzene from toluene at a fixed temperature. We can lower the pressure on a solution until it begins to vaporize (point a). At this point, the mole fractions are xb = 0.2 and xt = 1 — 0.2 = 0.8, respectively. The composition of the vapor in equilibrium with the solution (point b) has the following values: jc^ = 0.5 and x\ = 0.5. Thus, the vapor phase is richer in benzene than is the liquid phase. Now, if we condense the vapor (b —> c) and re-evaporate the liquid (c —> d), the mole fraction of benzene will be even higher in the vapor phase. Repeating the process at constant temperature will eventually lead to a quantitative separation of benzene and toluene. 234 Chapter 6: Nonelectrolyte Solutions 0.0 0.2 0.4 0.6 0.8 1.0 xh or xl Figure 6.11 Pressure-composition diagram of the benzene-toluene system at 23°C. Temperature-Composition Diagram. In practice, distillation is carried out more conveniently at constant pressure than at constant temperature; therefore, we need to examine the temperature-composition or boiling-point diagram. The relationship between temperature and composition is complex and is usually determined experimentally. Referring again to the benzene-toluene system and comparing Figure 6.12 with Figure 6.9, we see that the liquid and vapor regions are inverted and that neither the liquid nor the vapor curve is a straight line. The more volatile component, benzene, has a higher vapor pressure and hence a lower boiling point. Figure 6.9 shows that at a constant temperature, liquid is the stable phase at high pressures. Similarly, Figure 6.12 tells us that at a constant pressure, the stable phase at low temperatures is also the liquid. During fractional distillation, the solution is heated until it starts to evaporate (a—* b). The vapor, which is richer in benzene, is condensed (b —*■ c) and then evaporated (c^d). Repeating the process will eventually separate the two components completely. This procedure is called fractional distillation. Each vaporization and condensation step is called a theoretical plate. In the laboratory, chemists use an apparatus like that shown in Figure 6.13 to separate volatile liquids. The round-bottomed flask containing the benzene-toluene solution is fitted with a long column packed with small glass beads. When the solution boils, the vapor condenses on the beads in the lower portion of the column, and the liquid falls back into the distilling flask. As time goes on, the beads gradually heat up, allowing the vapor to move upward slowly. In essence, the packing material gives the column many theoretical plates and causes the benzene-toluene mixture to be subjected continuously to numerous vaporization-condensation steps. At each step, the composition of the vapor in the column will be richer in the more volatile, or lower boiling-point, component (in this case, benzene). The vapor that rises to the top 6.6 Phase Equilibria of Two-Component Systems of the column is essentially pure benzene, which is then condensed and collected in a receiving flask. Petroleum refining employs a similar approach. Crude oil is a complex mixture of thousands of compounds. By heating and condensing the crude oil up a distillation column that may be 80 meters high and contain hundreds of theoretical plates, workers can separate fractions according to their boiling-point ranges. 236 Chapter 6: Nonelectrolyte Solutions Azeotropes. Because most solutions are nonideal, their experimentally determined temperature-composition diagrams are more complicated than the one shown in Figure 6.12. If the system exhibits a positive deviation from Raoult's law, the curve will show a minimum boiling point. Conversely, a negative deviation from Raoult's law will result in a maximum boiling point (Figure 6.14). Examples of the former are acetone-carbon disulfide, ethanol-water, and «-propanol-water. Systems that show a maximum boiling point are less common; among the known examples are acetone-chloroform and hydrochloric acid-water. In every case, the mixture cannot be completely separated into pure components by simple fractional distillation. Consider the following steps, shown in Figure 6.14a. The solution is heated at a certain composition denoted by point a. The condensed vapor (b —> c) becomes richer in component 1, while the solution remaining in the pot becomes richer in component 2. Consequently, the point representing the composition of the solution in the pot on the liquid curve will move toward the left as the distillation proceeds and the boiling point rises. Boiling the solution, condensing the vapor, and again boiling the condensed vapor results in a distillate that has the same composition as the solution in the pot. Such a distillate is known as an azeotrope (from the Greek word meaning "to boil unchanged"). The boiling point of the solution remaining in the pot will eventually reach that of pure component 2 (i.e., Tj). Once the azeotrope distillate has been produced, further distillation will result in no further separation, and it will boil at a constant temperature. If we started at point a' and went through the same evaporation-condensation steps, the vapor would become richer in component 2 until it formed the same azeotrope, while the boiling point of the solution would approach that of component 1. The maximum boiling point system (Figure 6.14b) can be similarly explained except that a pure component appears in the distillate, and the azeotrope appears in the pot. 6.6 Phase Equilibria of Two-Component Systems 237 Table 6.3 Variation of Boiling Point and Composition of the HCI-Water Azeotrope with Pressure3 Composition (percent P/torr by weight HCl) Boiling point/°C 760 20.222 108.584 700 20.360 106.424 500 20.916 97.578 a W. D. Bonner and R. E. Wallace, J. Am. Chem. Soc. 52,1747 (1930). Although an azeotrope behaves in distillation as if it were a single component, that can easily be demonstrated not to be the case. As the data in Table 6.3 show, the composition of the azeotrope depends on the pressure. Solid-Liquid Equilibria If a liquid solution of two substances is cooled to a sufficiently low temperature, a solid will form. This temperature is the solution's freezing point, which depends on the composition of the solution. As we shall see in the next section, the freezing point of a solution is always lower than that of the solvent. Let us consider a two-component system made of antimony (Sb) and lead (Pb). The solid-liquid phase diagram for this system is shown in Figure 6.15. This phase diagram is constructed by measuring the melting points of a series of solutions of 0 1 238 Chapter 6: Nonelectrolyte Solutions different compositions at constant pressure. The asymmetric, V-shaped curve is the freezing point curve, above which the system is liquid. (The melting points of Pb and Sb are 328°C and 631°C, respectively.) Consider what happens when the solution at point a is cooled at constant pressure. When point b is reached, the solution begins to freeze, and the solid that separates from the solution is pure Sb. As the temperature is lowered further, more Sb freezes, and the solution becomes progressively richer in Pb. The composition of the solution at point c, for example, is given by drawing the tie line gch. As this point, the composition of the solution is given by the projection of the vertical line from point h onto the x axis. If we continue to lower the temperature of the solution, eventually we reach point d. At this temperature, the composition of the solution is given by point e. Now suppose we start to cool the solution at point i. At point j, the solution begins to freeze, and solid Pb forms. On further cooling, eventually we reach point k. At this point, the composition of the solution is again given by point e; therefore, this is the point at which the liquid is in equilibrium with both solids. Point e is called the The eutectic point eutectic point. The eutectic point has the following significance: (l) It represents the is at 246°C. lowest temperature at which a liquid solution can exist, and (2) at the eutectic point, the solid that separates from the solution has the same composition as the solution. In this respect, a solution that has eutectic composition behaves like a pure compound. Because the composition of a eutectic mixture depends on external pressure, however, we can readily distinguish its behavior from that of a pure liquid by studying the freezing phenomenon at different pressures. At the eutectic point, we have two components (Pb and Sb) and three phases (solid Pb, solid Sb, and solution), so the degrees of freedom,/ is given by f=c-p+2 =2-3+2 = 1 This single degree of freedom, however, is used to specify the pressure. Consequently, at the eutectic point, neither the temperature nor the composition can vary. A familiar eutectic mixture is the solder used in constructing electronic circuit boards. Solder is about 33% lead and 67% tin and melts at 183°C. (Tin melts at 232°C.) 6.7 Colligative Properties General properties of solutions include vapor-pressure lowering, boiling-point elevation, freezing-point depression, and osmotic pressure. These properties are commonly referred to as colligative, or collective, properties because they are bound together through their common origin. Colligative properties depend only on the number of solute molecules present, not on the size or molar mass of the molecules. To derive equations describing these phenomena, we shall make three important assumptions: (l) The solutions are ideal dilute, so that the solvent obeys Raoult's law; (2) the solutions are dilute; and (3) the solutions contain nonelectrolytes. As usual, we shall consider only a two-component system. 6.7 Colligative Properties 239 Vapor-Pressure Lowering Consider a solution that contains a solvent 1 and a nonvolatile solute 2, such as a solution of sucrose in water. Because the solution is ideal dilute, Raoult's law applies: P = xxP\ Because xl = \ — x2> the equation above becomes />! = (1 - X2)P{ Rearranging this equation gives P\ -Px = AP = x2P\ (6.33) where AP, the decrease in vapor pressure from that of the pure solvent, is directly proportional to the mole fraction of the solute. Why does the vapor pressure of a solution fall in the presence of a solute? We may be tempted to suggest that it does because of the modification of intermolecular forces. But this idea is not right because vapor-pressure lowering occurs even in ideal solutions, in which there is no difference between solute-solvent and solvent-solvent interactions. A more convincing explanation is provided by the entropy effect. When a solvent evaporates, the entropy of the universe increases because the entropy of any substance in the gaseous state is greater than that in the liquid state (at the same temperature). As we saw in Section 6.3, the solution process itself is accompanied by an increase in entropy. This result means that there is an extra degree of randomness, or disorder, in a solution that was not present in the pure solvent. Therefore, the evaporation of solvent from a solution will result in a smaller increase in entropy. Consequently, the solvent has less of a tendency to leave the solution, and the solution will have a lower vapor pressure than the pure solvent (see Problem 6.40). Boiling-Point Elevation The boiling point of a solution is the temperature at which its vapor pressure is equal to the external pressure. The previous discussion might lead you to expect that because the addition of a nonvolatile solute lowers the vapor pressure, it should also raise the boiling point of a solution. This effect is indeed the case. For a solution containing a nonvolatile solute, the boiling-point elevation originates in the change in the chemical potential of the solvent due to the presence of the solute. From Equation 6.17, we can see that the chemical potential of the solvent in a solution is less than the chemical potential of the pure solvent by an amount equal to RT In Xj. How this change affects the boiling point of the solution can be seen from Figure 6.16. The solid lines refer to the pure solvent. Because the solute is nonvolatile, it does not vaporize; therefore, the curve for the vapor phase is the same as that for the pure vapor. On the other hand, because the liquid contains a solute, the chemical potential of the solvent decreases (see the dashed curve). The points where the curve for the vapor intersects the curves for the liquids (pure and solution) correspond to the boiling points of the pure solvent and the solution, respectively. We see that the boiling point of the solution (7^) is higher than that of the pure solvent (7"b). Chapter 6: Nonelectrolyte Solutions Figure 6.16 Plot of chemical potentials versus temperature to illustrate colligative properties. The dashed red line denotes the solution phase. 7b and 7^ are the boiling points of the solvent and the solution, and Tf and 7/ are the freezing points of the solvent and solution, respectively. Temperature We now turn to a quantitative treatment of the boiling-point elevation phenomenon. At the boiling point, the solvent vapor is in equilibrium with the solvent in solution, so that or A^ = ^(g)- »\{l) = RT\nxl (6.34) where A/j1 is the Gibbs energy change associated with the evaporation of 1 mole of solvent from the solution at temperature T, its boiling point. Thus, we can write A^Wj = AvapG. Dividing Equation 6.34 by T, we obtain Avapg _ A*i(g) - Al(0 = R In xx T T From the Gibbs-Helmholz equation (Equation 5.15), we write 4AG/T) AH . —^— = ~~f2 (at constant or dT T2 dT where AyspH is the molar enthalpy of vaporization of the solvent from the solution. Because the solution is dilute, AvapH is taken to be the same as the molar enthalpy of vaporization of the pure solvent. Rearranging the last equation gives dXnx^-^^dT (6.35) 6.7 Colligative Properties 241 To find the relationship between jtj and T, we integrate Equation 6.35 between the limits Tb and Tb, the boiling points of the solution and pure solvent, respectively. Because the mole fraction of the solvent is xx at Tb and 1 at Tb, we write in*, rTl - ah dlnx^ lnl ■"vap-1 RT2 dT or lnx, = _ Avap/// l___i_ R \Tb Th -Avap///^ - Tb r \ nTb -Avap# AT r Tl (6.36) where AT = Tb — Tb. Two assumptions were used to obtain Equation 6.36, both of which are based on the fact that Tb and Tb differ only by a small amount (a few degrees). First, we assumed \apH to be temperature independent and second, Tb ~ Tb, sothatr;rb = rb2. Equation 6.36 gives the boiling-point elevation, AT, in terms of the concentration of the solvent (jcj). By custom, however, we express the concentration in terms of the amount of solute present, so we write , , /1 \ ~Avap^ Ar In Xj = ln(l - x2) = - where* R Tl 2 3 x2 x2 ln(l -x2) = -x2----~ —x2 (x2 >w2) * This expansion is known as Maclaurin's series. You can verify this relationship by employing a small numerical value for x2 (s 0.1). 242 Chapter 6: Nonelectrolyte Solutions where wx is the mass of the solvent in kg and Mx is the molar mass of the solvent in kg mol-1, respectively. Because n2/wx gives the molality of the solution, m2, it follows that x2 = Mxm2 and thus AT=—^-m2 (6.37) Note that all the quantities in the first term on the right of Equation 6.37 are constants for a given solvent, and so we have (6'38) where Kb is called the molal boiling-point-elevation constant. The units of Kb are K mol-1 kg. Finally, AT = Khm2 (6.39) The advantage of using molality, as mentioned in Section 6.1, is that it is independent of temperature and thus is suitable for boiling-point elevation studies. Figure 6.17 shows the phase diagrams of pure water and an aqueous solution. Upon the addition of a nonvolatile solute, the vapor pressure of the solution decreases at every temperature. Consequently, the boiling point of the solution at 1 atm will be greater than 373.15 K. Figure 6.17 Phase diagrams of pure water (solid red lines) and of water in an aqueous solution containing a nonvolatile solid (dashed red lines). (Note that the axes are not drawn to scale.) 6.7 Colligative Properties 243 Freezing-Point Depression A nonchemist may be forever unaware of the boiling-point-elevation phenomenon, but any casual observer living in a cold climate witnesses an illustration of freezing-point depression: ice on winter roads and sidewalks melts readily when sprinkled with salt* This method of thawing depresses the freezing point of water. The thermodynamic analysis of freezing-point depression is similar to that of boiling-point elevation. If we assume that when a solution freezes, the solid that separates from the solution contains only the solvent, then the curve for the chemical potential of the solid does not change (see Figure 6.16). Consequently, the solid curve for the solid and the dashed curve for the solvent in solution now intersect at a point (Tf) below the freezing point of the pure solvent (rf). By following the same procedure as that for the boiling-point elevation, we can show that the drop in freezing point AT (i.e., Tf — Tf, where Tf and T'{ are the freezing points of the pure solvent and solution, respectively) is AT = Kfm2 (6.40) where Kf is the molal freezing-point-depression constant given by RT\MX where Afusi7 is the molar enthalpy of fusion of the solvent. The freezing-point-depression phenomenon can also be understood by studying Figure 6.17. At 1 atm, the freezing point of solution lies at the intersection point of the dashed curve (between the solid and liquid phases) and the horizontal line at 1 atm. It is interesting that whereas the solute must be nonvolatile in the boiling-point-elevation case, no such restriction applies to lowering the freezing point. A proof of this statement is the use of ethanol (b.p. = 351.65 K) as an antifreeze. Either Equation 6.39 or 6.40 can be used to determine the molar mass of a solute. In general, the freezing-point-depression experiment is much easier to carry out. It is commonly employed in measuring the molar mass of compounds. Table 6.4 lists the Kb and Rvalues for several common solvents. The freezing-point-depression phenomenon has many examples in everyday life and in biological systems. As mentioned above, salts, such as sodium chloride and calcium chloride, are used to melt ice on roads and sidewalks. The organic compound ethylene glycol [CH2(OH)CH2(OH)] is the common automobile antifreeze. It is also employed to de-ice airplanes. In recent years, there has been much interest in understanding how certain species of fish manage to survive in the ice-cold waters of the polar oceans. The freezing point of seawater is approximately —1.9°C, which is the temperature of seawater surrounding an iceberg. A depression in freezing point of 1.9 degrees corresponds to a concentration of one molal, which is much too high for proper physiological function; for example, it alters osmotic balance (see section below on Osmotic Pressure). Besides dissolved salts and other substances that can (6.41) * The salt employed is usually sodium chloride, which attacks cement and is harmful to many plants. Also see "Freezing Ice Cream and Making Caramel Topping," J. O. Olson and L. H. Bowman, J. Chem. Educ. 53, 49 (1976). 244 Chapter 6: Nonelectrolyte Solutions Table 6.4 Molal Boiling-Point-Elevation and Molal Freezing-Point-Depression Constants of Some Common Solvents Solvent Kb/K mol"1 kg Kf/K mol"1 kg h20 0.51 1.86 C2H5OH 1.22 1.99 c6a. 2.53 5.12 CHC13 3.63 4.68 CH3COOH 2.93 3.90 CC14 5.03 29.8 lower the freezing point colligatively, a special class of proteins resides in the blood of polar fishes that has some kind of protective effect. These proteins contain both amino acid and sugar units and are called glycoprotetins. The concentration of glycoproteins in the fishes' blood is quite low (approximately 4 x 10"4 m), so their action cannot be explained by colligative properties. The belief is that the glycoproteins have the ability to adsorb onto the surface of each tiny ice crystal as soon as it begins to form, thus preventing it from growing to a size that would cause biological damage. Consequently, the freezing point of blood in these fishes is below —2°C. EXAMPLE 6.4 For a solution of 45.20 g of sucrose (C^H^Ojj) in 316.0 g of water, calculate (a) the boiling point, and (b) the freezing point. m-, = = 0.418 mol kg"1 ANSWER (a) Boiling point: Kb = 0.51 K mol-1 kg, and the molality of the solution is given by (45.20 g)(l000g/l kg) Frc Th 2 (342.3 g mor1) (316.0 g) From Equation 6.39 AT=Kbm2 = (0.51 K mol"1 kg) (0.418 mol kg"1) = 0.21 K Thus, the solution will boil at (373.15 + 0.21) K, or 373.36 K. 6.7 Colligative Properties 245 (b) Freezing point: From Equation 6.40, = Kfm2 = (1.86 K mor1 kg) (0.418 mol kg"1) = 0.78 K Thu Thus, the solution will freeze at (273.15 - 0.78) K, or 272.37 K. COMMENT For aqueous solutions of equal concentrations, the depression in freezing point is always greater than the corresponding elevation in boiling point. The reason can be seen by comparing the following two expressions from Equations 6.38 and 6.41: RT\MX Avap# = Kf = rt\mx Although rb > Tf, Avap/7 for water is 40.79 kJ mol"1, whereas AtusH is only 6.01 kJ mol"1. The large value of Avap// in the denominator is what causes Kh and hence AT to be smaller. Osmotic Pressure The phenomenon of osmosis is illustrated in Figure 6.18. The left compartment of the apparatus contains pure solvent; the right compartment contains a solution. The two compartments are separated by a semipermeable membrane (e.g., a cellophane membrane), one that permits the solvent molecules to pass through but does not permit the movement of solute molecules from right to left. Practically speaking, then, this system has two different phases. At equilibrium, the height of the solution in the tube on the right is greater than that of the pure solvent in the left tube by h. This excess hydrostatic pressure is called the osmotic pressure. We can now derive an expression for osmotic pressure as follows. Let ^ and /if be the chemical potential of the solvent in the left and right compartments, respectively. Initially, before equilibrium is established, we have li{ = n\ + RT\nxx = /t\ (r! = 1) and ft = //* + RT\nxx (xi $ = /J* + RT\nxx Note that fi\ is the same as the standard chemical potential for the pure solvent, p\, and the inequality sign denotes that RT In xx is a negative quantity. Consequently, more solvent molecules, on the average, will move from left to right across the membrane. The process is spontaneous because the dilution of the solution in the right compartment by solvent leads to a decrease in the Gibbs energy and an increase in entropy. Equilibrium is finally reached when the flow of solvent is exactly balanced by the hydrostatic pressure difference in the two side tubes. This extra pressure increases the chemical potential of the solvent in solution, fif. From Equation 5.16, we know that We can write a similar equation for the variation of the chemical potential with pressure at constant temperature. Thus, for the solvent component in the right compartment, where Vl is the partial molar volume of the solvent. For a dilute solution, Vx is approximately equal to V, the molar volume of the pure solvent. The increase in the chemical potential of the solvent in the solution compartment (A//f) when the pressure increases from P, the external atmospheric pressure, to (P + II) is given by ■p+n vdp = vn p 6.7 Colligative Properties 247 Note that Vis treated as a constant because the volume of a liquid changes little with pressure. The Greek letter 77 represents the osmotic pressure. The term osmotic pressure of a solution refers to the pressure that must be applied to the solution to increase the chemical potential of the solvent to the value of its pure liquid under atmospheric pressure. At equilibrium, the following relations must hold: pt{ = ft = n\ + RTinxl + nv Because $ = fi[, we have nv= -RT\nxx (6.43) To relate 77 to the concentration of the solute, we take the following steps. From the procedure employed for boiling-point elevation: -In Xj = —In (1 — x2) = x2 (x2 <3C 1) Furthermore, x2 «2 n2 nx + n2 (wx » n2) where nx and n2 are the number of moles of solvent and solute, respectively. Equation 6.43 now becomes nV= RTx2 Substituting V = V\nx into Equation 6.44, we get IIV= n2RT If Fis in liters, then (6.44) (6.45) 77: -fRT MRT (6.46) where Mis the molarity of the solution. Note that molarity is a convenient concentration unit here, because osmotic pressure measurements are normally made at constant temperature. Alternatively, we can rewrite Equation 6.46 as 77: _C2 RT (6.47) Equation 6.46 also applies to two similar solutions that have different concentrations. 248 Chapter 6: Nonelectrolyte Solutions M>2 c2 Figure 6.19 Determination of the molar mass of a solute by osmotic pressure measurements for an ideal and a nonideal solution. The intercept on they axis (as c2 —► 0) gives the correct value for the molar mass. or l-l <-> where c2 is the concentration of the solute in g L_1 of the solution and m2 is the molar mass of the solute. Equation 6.48 provides a way to determine molar masses of compounds from osmotic pressure measurements. Equation 6.48 is derived by assuming ideal behavior, so it is desirable to measure II at several different concentrations and extrapolate to zero concentration for molar mass determination (Figure 6.19). For a nonideal solution, the osmotic pressure at any concentration, c2, is given by Compare Equation 6.49 U RT, „ , „ 2 , r» 3 , \ it <«\ ^Equation 1.10 V2 = ^ + Bc> + Cc> + Dc* + -> ^ where B, C, and D are called the second, third, and fourth virial coefficients, respectively. The magnitude of the virial coefficients is such that B » C » D. In dilute solutions, we need be concerned only with the second virial coefficient. For an ideal solution, the second and higher virial coefficients are all equal to zero, so Equation 6.49 reduces to Equation 6.48. Even though osmosis is a well-studied phenomenon, the mechanism involved is not always clearly understood. In some cases, a semipermeable membrane may act as a molecular sieve, allowing smaller solvent molecules to pass through while blocking larger solute molecules. In other cases, osmosis may be caused by the higher solubility of the solvent in the membrane than the solute. Each system must be studied individually. The previous discussion illustrates both the usefulness and limitation of thermodynamics. We have derived a convenient equation relating the molar mass of the solute to an experimentally measurable quantity—the osmotic pressure—simply in terms of the chemical potential difference. Because thermodynamics is not based on any specific model, however, Equation 6.47 tells us nothing about the mechanism of osmosis. 6.7 Colligative Properties 249 EXAMPLE 6.5 Consider the following arrangement, in which a solution containing 20 g of hemoglobin in 1 liter of the solution is placed in the right compartment, and pure water is placed in the left compartment (see Figure 6.18). At equilibrium, the height of the water in the right column is 77.8 mm in excess of the height of the solvent in the left column. What is the molar mass of hemoglobin? The temperature of the system is constant at 298 K. ANSWER To determine the molar mass of hemoglobin, we first need to calculate the osmotic pressure of the solution. We start by writing pressure force area Ahpg = hpg where A is the cross-sectional area of the tube, h the excess liquid height in the right column, p the density of the solution, and g the acceleration due to gravity. The parameters are h = 0.0778 m p = 1 x 103 kg nr3 g = 9.81 m s-2 (We have assumed that the density of the dilute solution is the same as that of water.) The osmotic pressure in pascals (N m-2) is given by n = 0.0778 m x 1 x 103 kg nr3 x 9.81 m s"2 = 763 kg nr1 s-2 = 763 Nm"2 From Equation 6.47, Ji7 = -±RT z n (20 kg nr3) (8.314 J K"1 mol"1) (298 K) 1 N m 250 Chapter 6: Nonelectrolyte Solutions (a) (b) (c) Figure 6.20 A cell in (a) an isotonic solution, (b) a hypotonic solution, and (c) a hypertonic solution. The cell remains unchanged in (a), swells in (b), and shrinks in (c). High-resolution mass spectrometry is now a common and convenient method for determining the molar mass of macromolecules. Example 6.5 shows that osmotic pressure measurement is a more sensitive method to determine molar mass than the boiling-point-elevation and freezing-point-depression techniques, because 7.8 cm is an easily measurable height. On the other hand, the same solution will lead to an elevation in boiling point of approximately 1.6 x 10"4 °C and a depression in freezing point of 5.8 x 10~4 °C, which are too small to measure accurately. Most proteins are less soluble than hemoglobin. Nevertheless, their molar masses can often be determined by osmotic pressure measurements. A disadvantage of osmotic pressure measurements is that the time to reach equilibrium may be hours or days. Many examples of the osmotic-pressure phenomenon are found in chemical and biological systems. If two solutions are of equal concentration, and hence have the same osmotic pressure, they are said to be isotonic. For two solutions of unequal osmotic pressures, the more concentrated solution is said to be hypertonic, and the less concentrated solution is said to be hypotonic (Figure 6.20). To study the contents of red blood cells, which are protected from the outside environment by a semipermeable membrane, biochemists employ a technique called hemolysis. They place the red blood cells in a hypotonic solution, which causes water to move into the cell. The cells swell and eventually burst, releasing hemoglobin and other protein molecules. When a cell is placed in a hypertonic solution, on the other hand, the intracellular water tends to move out of the cell by osmosis to the more concentrated, surrounding solution. This process, known as crenation, causes the cell to shrink and eventually cease functioning. The mammalian kidney is a particularly effective osmotic device. Its main function is to remove metabolic waste products and other impurities from the bloodstream by osmosis to the more concentrated urine outside through a semipermeable membrane. Biologically important ions (such as Na+ and CI") lost in this manner are then actively pumped back into the blood through the same membrane. The loss of water through the kidney is controlled by the antidiuretic hormone (ADH), which is secreted into the blood by the hypothalamus and posterior pituitary gland. When little or no ADH is secreted, large amounts of water (perhaps 10 times normal) pass into the urine each day. On the other hand, when large quantities of ADH are present in the blood, 6.7 CoUigative Properties 251 the permeability of water through the membrane decreases so that the volume of urine formed may be as little as one-half the normal amount. Thus, the kidney-ADH combination controls the rate of loss of both water and other small waste molecules. The chemical potential of water within the body fluids of freshwater fishes is lower than that in their environment, so they are able to draw in water by osmosis through their gill membranes. Surplus water is excreted as urine. An opposite process occurs for the marine teleost fishes. They lose body water to the more concentrated environment by osmosis across the gill membranes. To balance the loss, they drink seawater. Osmotic pressure is also the major mechanism for water rising upward in plants. The leaves of trees constantly lose water to their surroundings, a process called transpiration, so the solute concentrations in leaf fluids increase. Water is then pushed up through the trunks and branches by osmotic pressure, which, to reach the tops of the tallest trees, can be as high as 10 to 15 atm* Leaf movement is an interesting phenomenon that may also be related to osmotic pressure. The belief is that some processes can increase salt concentration in leaf cells in the presence of light. Osmotic pressure rises and cells become enlarged and turgid, causing the leaves to orient toward light. Reverse Osmosis. A related phenomenon to osmosis is called reverse osmosis. If we apply pressure greater than the equilibrium osmotic pressure to the solution compartment shown in Figure 6.18, pure solvent will flow from the solution to the solvent compartment. This reversal of the osmotic process results in the unmixing of the solution components. An important application of reverse osmosis is the desalination of water. Several techniques discussed in this chapter are suitable, at least in principle, for obtaining pure water from the sea. For example, either distilling or freezing seawater would achieve the goal. However, these processes involve a phase change from liquid to vapor or liquid to solid and so would require considerable energy input to maintain. Reverse osmosis is more appealing, for it does not involve a phase change and is economically sound for large amounts of water.t Seawater, which is approximately 0.7 M in NaCl, has an estimated osmotic pressure of 30 atm. For a 50% recovery of pure water from the sea, an additional 60 atm would have to be applied on the seawater-side compartment to cause reverse osmosis. The success of large-scale desalination depends on the selection of a suitable membrane that is permeable to water but not to dissolved salts and that can withstand the high pressure over long periods of time. * See "Entropy Makes Water Run Uphill—in Trees," P. E. Steveson, J. Chem. Educ. 48, 837 (1971). t See "Desalination of Water by Reverse Osmosis," C. E. Hecht, J. Chem. Educ. 44, 53 (1967). 252 Chapter 6: Nonelectrolyte Solutions Key Equations V = nxVx + n2V2 (Volume of a solution in terms of partial molar volumes) (6.6) G, = Hi = (tt^) (Definition of chemical potential) (6.7) \öni/T,P,nj AmixG = nRT{xx \nxx + x2 lnx2) (Gibbs energy of mixing) (6.11) Amix 0, AmixH< 0, AmixF> 0, AmixV< 0. 6.10 Calculate the changes in entropy for the following processes: (a) mixing of 1 mole of nitrogen and 1 mole of oxygen, and (b) mixing of 2 moles of argon, 1 mole of helium, and 3 moles of hydrogen. Both (a) and (b) are carried out under conditions of constant temperature (298 K) and constant pressure. Assume ideal behavior. 6.11 At 25°C and 1 atm pressure, the absolute third-law entropies of methane and ethane are 186.19 J K"1 mol"1 and 229.49 J K"1 mol"1, respectively, in the gas phase. Calculate the absolute third-law entropy of a "solution" containing 1 mole of each gas. Assume ideal behavior. Henry's Law 6.12 Prove the statement that an alternative way to express Henry's law of gas solubility is to say that the volume of gas that dissolves in a fixed volume of solution is independent of pressure at a given temperature. Chapter 6: Nonelectrolyte Solutions 6.13 A miner working 900 ft below the surface had a soft drink beverage during the lunch break. To his surprise, the drink seemed very flat (i.e., not much effervescence was observed upon removing the cap). Shortly after lunch, he took the elevator up to the surface. During the trip up, he felt a great urge to belch. Explain. 6.14 The Henry's law constant of oxygen in water at 25°C is 773 atm mol-1 kg of water. Calculate the molality of oxygen in water under a partial pressure of 0.20 atm. Assuming that the solubility of oxygen in blood at 37°C is roughly the same as that in water at 25°C, comment on the prospect for our survival without hemoglobin molecules. (The total volume of blood in the human body is about 5 L.) 6.15 The solubility of N2 in blood at 37°C and a partial pressure of 0.80 atm is 5.6 x 10~4 mol L_1. A deep-sea diver breathes compressed air with a partial pressure of N2 equal to 4.0 atm. Assume that the total volume of blood in the body is 5.0 L. Calculate the amount of N2 gas released (in liters) when the diver returns to the surface of water, where the partial pressure of N2 is 0.80 atm. Chemical Potential and Activity 6.16 Which of the following has a higher chemical potential? If neither, answer "same." (a) H20(s) or H20(/) at water's normal melting point, (b) H20(s) at —5°C and 1 bar or H20(/) at —5°C and 1 bar, (c) benzene at 25°C and 1 bar or benzene in a 0.1 M toluene solution in benzene at 25°C and 1 bar. 6.17 A solution of ethanol and w-propanol behaves ideally. Calculate the chemical potential of ethanol in solution relative to that of pure ethanol when its mole fraction is 0.40 at its boiling point (78.3°C). 6.18 Derive the Gibbs phase rule (Equation 5.23) in terms of chemical potentials. 6.19 The following data give the pressures for carbon disulfide-acetone solutions at 35.2°C. Calculate the activity coefficients of both components based on deviations from Raoult's law and Henry's law. (Hint: First determine Henry's law constants graphically.) xCs2 0 0.20 0.45 0.67 0.83 1.00 Pcsj/torr 0 272 390 438 465 512 ^CHso/torr 344 291 250 217 180 0 6.20 A solution is made up by dissolving 73 g of glucose (C6H12Oe; molar mass 180.2 g) in 966 g of water. Calculate the activity coefficient of glucose in this solution if the solution freezes at -0.66°C. 6.21 A certain dilute solution has an osmotic pressure of 12.2 atm at 20°C. Calculate the difference between the chemical potential of the solvent in the solution and that of pure water. Assume that the density is the same as that of water. (Hint: Express the chemical potential in terms of mole fraction, jcl5 and rewrite the osmotic pressure equation as nv = n2RT, where n2 is the number of moles of the solute and V= 1 L.) 6.22 At 45°C, the vapor pressure of water is 65.76 mmHg for a glucose solution in which the mole fraction of glucose is 0.080. Calculate the activity and activity coefficient of the water in the solution. The vapor pressure of pure water at 45°C is 71.88 mmHg. 6.23 Consider a binary liquid mixture of A and B, where A is volatile and B is nonvolatile. The composition of the solution in terms of mole fraction is xA = 0.045 and xB = 0.955. The vapor pressure of A from the mixture is 5.60 mmHg, and that of pure A is 196.4 mmHg at the same temperature. Calculate the activity coefficient of A at this concentration. Problems 257 Colligative Properties 6.24 List the important assumptions in the derivation of Equation 6.39. 6.25 Liquids A (b.p. = 7^) and B (b.p. = Tg) form an ideal solution. Predict the range of boiling points of solutions formed by mixing different amounts of A and B. 6.26 A mixture of ethanol and w-propanol behaves ideally at 36.4°C. (a) Determine graphically the mole fraction of K-propanol in a mixture of ethanol and «-propanol that boils at 36.4°C and 72 mmHg. (b) What is the total vapor pressure over the mixture at 36.4°C when the mole fraction of n-propanol is 0.60? (c) Calculate the composition of the vapor in (b). (The equilibrium vapor pressures of ethanol and w-propanol at 36.4°C are 108 mmHg and 40.0 mmHg, respectively.) 6.27 Two beakers, 1 and 2, containing 50 mL of 0.10 Murea and 50 mL of 0.20 Murea, respectively, are placed under a tightly sealed bell jar at 298 K. Calculate the mole fraction of urea in the solutions at equilibrium. Assume ideal behavior. (Hint: Use Raoult's law and note that at equilibrium, the mole fraction of urea is the same in both solutions.) 6.28 At 298 K, the vapor pressure of pure water is 23.76 mmHg and that of seawater is 22.98 mmHg. Assuming that seawater contains only NaCl, estimate its concentration. (Hint: Sodium chloride is a strong electrolyte.) 6.29 Trees in cold climates may be subjected to temperatures as low as —60°C. Estimate the concentration of an aqueous solution in the body of the tree that would remain unfrozen at this temperature. Is this a reasonable concentration? Comment on your result. 6.30 Explain why jams and honey can each be stored under atmospheric conditions for long periods of time without spoilage. 6.31 Provide a molecular interpretation for the positive and negative deviations in the boiling-point curves and the formation of azeotropes. 6.32 The freezing-point-depression measurement of benzoic acid in acetone yields a molar mass of 122 g; the same measurement in benzene gives a value of 244 g. Account for this discrepancy. (Hint: Consider solvent-solute and solute-solute interactions.) 6.33 A common antifreeze for car radiators is ethylene glycol, CH2(OH)CH2(OH). How many milliliters of this substance would you add to 6.5 L of water in the radiator if the coldest day in winter is —20°C? Would you keep this substance in the radiator in the summer to prevent the water from boiling? (The density and boiling point of ethylene glycol are 1.11 g cm-3 and 470 K, respectively.) 6.34 For intravenous injections, great care is taken to ensure that the concentration of solutions to be injected is comparable to that of blood plasma. Why? 6.35 The tallest trees known are the redwoods in California. Assuming the height of a redwood to be 105 m (about 350 ft), estimate the osmotic pressure required to push water up from the roots to the treetop. 6.36 A mixture of liquids A and B exhibits ideal behavior. At 84°C, the total vapor pressure of a solution containing 1.2 moles of A and 2.3 moles of B is 331 mmHg. Upon the addition of another mole of B to the solution, the vapor pressure increases to 347 mmHg. Calculate the vapor pressures of pure A and B at 84°C. 6.37 Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be 0.20 atm and 0.80 atm, respectively, calculate the mole fractions of oxygen and nitrogen in water at 298 K. Comment on your results. Chapter 6: Nonelectrolyte Solutions 6.38 Liquids A (molar mass 100 g mol-1) and B (molar mass 110 g mol-1) form an ideal solution. At 55°C, A has a vapor pressure of 95 mmHg and B a vapor pressure of 42 mmHg. A solution is prepared by mixing equal weights of A and B. (a) Calculate the mole fraction of each component in the solution, (b) Calculate the partial pressures of A and B over the solution at 55°C. (c) Suppose that some of the vapor described in (b) is condensed to a liquid. Calculate the mole fraction of each component in this liquid and the vapor pressure of each component above this liquid at 55°C. 6.39 Lysozyme extracted from chicken egg white has a molar mass of 13,930 g mol-1. Exactly 0.1 g of this protein is dissolved in 50 g of water at 298 K. Calculate the vapor pressure lowering, the depression in freezing point, the elevation of boiling point, and the osmotic pressure of this solution. The vapor pressure of pure water at 298 K is 23.76 mmHg. 6.40 The following argument is frequently used to explain the fact that the vapor pressure of the solvent is lower over a solution than over the pure solvent and that lowering is proportional to the concentration. A dynamic equilibrium exists in both cases, so that the rate at which molecules of solvent evaporate from the liquid is always equal to that at which they condense. The rate of condensation is proportional to the partial pressure of the vapor, whereas that of evaporation is unimpaired in the pure solvent but is impaired by solute molecules in the surface of the solution. Hence the rate of escape is reduced in proportion to the concentration of the solute, and maintenance of equilibrium requires a corresponding lowering of the rate of condensation and therefore of the partial pressure of the vapor phase. Explain why this argument is incorrect. [Source: K. J. Mysels, J. Chem. Educ. 32, 179 (1955).] 6.41 A compound weighing 0.458 g is dissolved in 30.0 g of acetic acid. The freezing point of the solution is found to be 1.50 K below that of the pure solvent. Calculate the molar mass of the compound. 6.42 Two aqueous urea solutions have osmotic pressures of 2.4 atm and 4.6 atm, respectively, at a certain temperature. What is the osmotic pressure of a solution prepared by mixing equal volumes of these two solutions at the same temperature? 6.43 A forensic chemist is given a white powder for analysis. She dissolves 0.50 g of the substance in 8.0 g of benzene. The solution freezes at 3.9°C. Can the chemist conclude that the compound is cocaine (C17H21N04)? What assumptions are made in the analysis? The freezing point of benzene is 5.5°C. 6.44 "Time-release" drugs have the advantage of releasing the drug to the body at a constant rate so that the drug concentration at any time is not high enough to have harmful side effects or so low as to be ineffective. A schematic diagram of a pill that works on this basis is shown below. Explain how it works. 6.45 A nonvolatile organic compound, Z, was used to make up two solutions. Solution A contains 5.00 g of Z dissolved in 100 g of water, and solution B contains 2.31 g of Z dissolved in 100 g of Rigid wall containing tiny holes Elastic impermeable membrane Problems benzene. Solution A has a vapor pressure of 754.5 mmHg at the normal boiling point of water, and solution B has the same vapor pressure at the normal boiling point of benzene. Calculate the molar mass of Z in solutions A and B, and account for the difference. 6.46 Acetic acid is a polar molecule that can form hydrogen bonds with water molecules. Therefore, it has a high solubility in water. Yet acetic acid is also soluble in benzene (CgHg), a nonpolar solvent that lacks the ability to form hydrogen bonds. A solution of 3.8 g of CH3COOH in 80 g C6Hg has a freezing point of 3.5°C. Calculate the molar mass of the solute, and suggest what its structure might be. {Hint: Acetic acid molecules can form hydrogen bonds among themselves.) 6.47 At 85°C, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate the composition of a mixture of A and B that boils at 85°C when the pressure is 0.60 atm. Also, calculate the composition of the vapor mixture. Assume ideal behavior. 6.48 Comment on whether each of the following statements is true or false, and briefly explain your answer: (a) If one component of a solution obeys Raoult's law, then the other component must also obey the same law. (b) Intermolecular forces are small in ideal solutions, (c) When 15.0 mL of an aqueous 3.0 Methanol solution is mixed with 55.0 mL of an aqueous 3.0 Methanol solution, the total volume is 70.0 mL. 6.49 Liquids A and B form an ideal solution at a certain temperature. The vapor pressures of pure A and B are 450 torr and 732 torr, respectively, at this temperature, (a) A sample of the solution's vapor is condensed. Given that the original solution contains 3.3 moles of A and 8.7 moles of B, calculate the composition of the condensate in mole fractions, (b) Suggest a method for measuring the partial pressures of A and B at equilibrium. 6.50 Nonideal solutions are the result of unequal intermolecular forces between components. Based on this knowledge, comment on whether a racemic mixture of a liquid compound would behave as an ideal solution. 6.51 Calculate the molal boiling-point elevation constant (Kb) for water. The molar enthalpy of vaporization of water is 40.79 kJ mol-1 at 100°C. 6.52 Explain the following phenomena, (a) A cucumber placed in concentrated brine (saltwater) shrivels into a pickle, (b) A carrot placed in fresh water swells in volume. Additional Problems 6.53 Calculate the change in the Gibbs energy at 37°C when the human kidneys secrete 0.275 mole of urea per kilogram of water from blood plasma to urine if the molal concentrations of urea in blood plasma and urine are 0.005 mol kg"1 and 0.326 mol kg"1, respectively. 6.54 (a) Which of the following expressions is incorrect as a representation of the partial molar volume of component A in a two-component solution? Why? How would you correct it? (b) Given that the molar volume of this mixture (Vm) of A and B is given by Vm = [0.34 + 3.6xAxB + 0.4jcb(1 - jca)] L mol-1 derive an expression for the partial molar volume for A and calculate its value at xA = 0.20. Chapter 6: Nonelectrolyte Solutions 6.55 The partial molar volumes for a benzene-carbon tetrachloride solution at 25°C at a mole fraction of 0.5 are: Vh = 0.106 L mol-1 and Vc = 0.100 L mol-1, respectively, where the subscripts b and c denote CsHg and CC14. (a) What is the volume of a solution made up of one mole of each? (b) Given that the molar volumes are: CgHg = 0.089 L mol"1 and CC14 = 0.097 L mol"1, what is the change in volume on mixing 1 mole each of C6Hg and CC14? (c) What can you deduce about the nature of intermolecular forces between C6Hg and CC14? 6.56 The osmotic pressure of poly(methyl methacrylate) in toluene has been measured at a series of concentrations at 298 K. Determine graphically the molar mass of the polymer. /7/atm 8.40X10"4 1.72 x 10"3 2.52 x 10"3 3.23 x 10"3 7.75 x 10"3 c/gL"1 8.10 12.31 15.00 18.17 28.05 6.57 Benzene and toluene form an ideal solution. Prove that to achieve the maximum entropy of mixing, the mole fraction of each component must be 0.5. 6.58 Suppose 2.6 moles of He at 0.80 atm and 25°C are mixed with 4.1 moles of Ne at 2.7 atm and 25°C. Calculate the Gibbs energy change for the process. Assume ideal behavior. 6.59 Two beakers are placed in a closed container. Beaker A initially contains 0.15 mole of naphthalene (C10Hg) in 100 g of benzene (CgFQ and beaker B initially contains 31 g of an unknown compound dissolved in 100 g of benzene. At equilibrium, beaker A is found to have lost 7.0 g. Assuming ideal behavior, calculate the molar mass of the unknown compound. State any assumptions made. 6.60 As an after dinner party trick, the host brings out a glass of water with an ice cube floating on top and a thread. Then he asks the guests to remove the ice cube with the thread, but they are not allowed to tie a loop around the ice cube. Describe how the guests might accomplish this task. 6.61 A student carried out the following procedure to measure the pressure of carbon dioxide in a carbonated soft drink bottle. First, she weighed the bottle (853.5 g). Next, she carefully removed the cap to let the C02 gas escape. Finally, she measured the volume of the soft drink (452.4 mL). Given that the Henry's law constant for C02 in water at 25°C is 3.4 x 10~2 mol L_1 atm-1, calculate the pressure of C02 in the original bottle. List the sources of errors. 6.62 (a) Derive the equation relating the molality (m) of a solution to its molarity (M) M 1000 where d is the density of the solution (g/mL) and m is the molar mass of the solute (g/mol). (Hint: Start by expressing the solvent in kilograms in terms of the difference between the mass of the solution and the mass of the solute.) (b) Show that, for dilute aqueous solutions, m is approximately equal to M. 6.63 At 298 K, the osmotic pressure of a glucose solution is 10.50 atm. Calculate the freezing point of the solution. The density of the solution is 1.16 g/mL. 6.64 The mole fractions of dry air are approximately 21% 02 and 79% N2. Calculate the masses of these two gases dissolved in 1000 g of water at 25°C and 1 atm. Copyrighted Materials Copyright©2014 University Science Books Retrieved from www.knovel.com CHAPTER 7 Electrolyte Solutions And when the rain has wet the kite and twine, so that it can conduct the electric fire freely, you will find it stream out plentifully from the key on the approach of your knuckle. —Benjamin Franklin* All biological and many chemical systems are aqueous solutions that contain various ions. The rates of many reactions are highly dependent on the type and concentration of ions present. It is important to have a clear understanding of the behavior of ions in solution. An electrolyte is a substance that, when dissolved in a solvent (usually water), produces a solution that will conduct electricity. An electrolyte can be an acid, a base, or a salt. In this chapter, we shall consider ionic conductance, ionic dissociation, the thermodynamics of ions in solution, and the theory and colligative properties of electrolyte solutions. 7.1 Electrical Conduction in Solution Some Basic Definitions The ability of an electrolyte to conduct electricity provides us with a simple and direct means of studying ionic behavior in solution. We begin with a few basic definitions. Ohm's Law. Ohm's law (after the German physicist George Simon Ohm, 1787-1854) states that the current (/) flowing through a particular medium is directly proportional to the voltage or the electrical potential difference (V) across the medium and indirectly proportional to the resistance (R) of the medium. Thus, /-| (7.D where lis in amperes (A), Vin volts (V), and R in ohms (Q). * Labaree. L. W., et al., Eds., The Papers of Benjamin Franklin, Yale University Press, New Haven, CT, 1961. Vol. 4, p.367. Used by permission. 261 262 Chapter 7: Electrolyte Solutions Resistance (R). The resistance across a particular medium depends on the geometry of the medium; it is directly proportional to the length (/) and inversely proportional to the cross section area (A) of the medium. Thus, Rot I Figure 7.1 A conductance cell. The electrodes are made of platinum. The historical unit of electric conductance was the mho ("ohm" backwards). PA (7.2) where the proportionality constant p is called the specific resistance, or resistivity. Because the units of R are ohms (Q), of / centimeters or meters, and of A square centimeters or square meters, the units of p are Q cm or Q m. Resistivity is a property characteristic of the material comprising the medium. Conductance (C). Conductance is the reciprocal of resistance, that is, R ]_A P I Kl (7.3) where k is the specific conductance, or conductivity equal to \\p. The SI unit for conductance is Siemens (S) (after the German physicist Werner von Siemens, 1816-1892), where 1 S = 1 Q"1. Conductivity has the units Q"1 cm-1 or Q"1 m"1. A typical conductance cell* is shown in Figure 7.1. The conductance is given by Equation 7.3. The ratio IJA, called the cell constant, is the same for all solutions. Here, A is the area of the electrodes and / the distance of separation between the electrodes. In practice, the cell constant is calibrated by measuring the conductance of a standard solution of potassium chloride with a known value of k. * It is customary to quote the conductance rather than the resistance of an electrolyte solution. EXAMPLE 7.1 The conductance of a solution is 0.689 Q i. Calculate the specific conductance if the cell constant is 0.255 cm"1. ANSWER From Equation 7.3, k = C I = 0.689 Q"1 x 0.255 cm"1 = 0.176 Q-1 cm"1 7.1 Electrical Conduction in Solution Although the specific conductance can be measured easily (from the known cell constant and the experimentally determined conductance), it is not the best value to use in studying the conduction process in electrolyte solutions. The specific conductances of solutions of different concentrations, for example, will differ simply because a given volume of the solutions will contain different numbers of ions. For this reason, expressing the conductance as a molar quantity is preferable. We define the molar conductance (A) as A = ~c (7-4) where c is the molar concentration of the solution. The SI units for A are Q_1 mol-1 m2, although it is often more conveniently expressed as Q_1 mol-1 cm2. EXAMPLE 7.2 The conductance of a cell containing an aqueous 0.0560 MKC1 solution is 0.0239 Q_1. When the same cell is filled with an aqueous 0.0836 MNaCl solution, its conductance is 0.0285 Q_1. Given that the molar conductance of KC1 is 1.345 x 102 Q_1 mol-1 cm2, calculate the molar conductance of the NaCl solution. ANSWER We need the cell constant. Our first step is to calculate the specific conductance of the KC1 solution. From Equation 7.4, k = Ac ^9~i ii 9 0.0560 mol 1.345 x 102 Q_1 mol"1 cm2 x-—-x Next, from Equation 7.3 we obtain the cell constant / k 7.53 x 10"3 Q"1 cm"1 Th( 1L 1000 cm3 l_ A~ C~ 0.0239 Q"1 = 0.315 cm"1 The specific conductance of the NaCl solution is obtained by rearranging Equation 7.3: k = ~tC A = (0.315 cm"1) (0.0285 Q-1) = 8.98 x 10-3 Q"1 cm"1 264 Chapter 7: Electrolyte Solutions Finally, the molar conductance of the NaCl solution is given by (see Equation 7.4) k = 8.98 x 10~3 Q-1 cm-1 1000 cm3 ~ c ~ 0.0836 mol L"1 X 1 L 1.07 x 102 Q"1 mor1 cm2 Strong electrolytes are substances that are completely dissociated into ions in solution. Looking at Equation 7.4, we might expect A to be independent of the concentration of the solution, (k is directly proportional to concentration, but k\c should be constant for a given substance.) Careful measurements show that this is not the case, however. The German chemist Friedrich Wilhelm Georg Kohlrausch (1840-1910) discovered the following relationship between molar conductance and concentration for strong electrolytes at a particular temperature: A = An - BVc (7.5) where A0 is the molar conductance at infinite dilution; that is, A —> A0 as c —> 0, and B is a positive constant for a given electrolyte. Thus, A0 can be readily obtained by plotting A versus Vc and extrapolating to zero concentration (Figure 7.2). This method is unsatisfactory for weak electrolytes because of the steepness of their curves at low concentrations (see plot for CH3COOH). Table 7.1 shows the values of A0 for several electrolytes. An interesting pattern emerges when we examine the difference in A0 for two electrolytes containing the same cation or anion. For example, A™ - ANaCi = 23.4 Q"1 mol"1 cm2 akno3 _ ANaN03 = 23.4 Q"1 mol"1 cm2 This same difference and similar observations led Kohlrausch to suggest that molar conductance at infinite dilution can be broken down into two contributions, one from the anion and the other from the cation: A0 = v+X% + v_l0 (7.6) Note that these are all 1:1 electrolytes, so v+ = v_ = 1. where 1$ and Aq are the molar ionic conductances at infinite dilution, and v+ and v_ are the number of cations and anions in the formula. Equation 7.6 is known as Kohlrausch's law of independent migration. It means that molar conductance at infinite dilution is made up of independent contributions from the cationic and anionic species. We can now see why the same value was obtained in the example above, because AKCl _ ANaCl — 5K+ + ;cr _ lNa+ _ 5C1" — ;K+ _ ;Na and AKN03 ANaN03 2Na+ lNa+ A0 Table 7.2 lists the molar conductances of a number of ions. 7.1 Electrical Conduction in Solution 265 Table 7.1 Molar Conductance at Infinite Dilution for Some Electrolytes in Water at 298 Ka Electrolyte A0/O"1 mol"1 cm2 HCl 426.16 CH3COOH 390.71 LiCl 115.03 NaCl 126.45 AgCl 137.20 KCl 149.85 L1NO3 110.14 NaNC-3 121.56 KNO3 144.96 C11SO4 267.24 CH3COONa 91.00 a To express A0 as £2 1 mol 1 m2, multiply each number by 10-4. Thus, A0 for HC1 is 426.16 Or1 mol"1 cm2 or 4.2616 x 10"2 ST1 mol"1 m2. 266 Chapter 7: Electrolyte Solutions Table 7.2 Molar Ionic Conductance and Ionic Mobility of Some Common Ions at 298 K Ion fi-1 moM cm2 Ionic mobility13 10-4cm2 s-1 V-1 Ionic radius/Ä H+ 349.81 36.3 Li+ 38.68 4.01 0.60 Na+ 50.10 5.19 0.95 K+ 73.50 7.62 1.33 Rb+ 77.81 7.92 1.48 Cs+ 77.26 7.96 1.69 NH4 73.5 7.62 Mg2+ 106.1 5.50 0.65 Ca2+ 119.0 6.17 0.99 Ba2+ 127.3 6.59 1.35 Cu2+ 107.2 5.56 0.72 OH" 198.3 20.50 F" 55.4 5.74 1.36 ci- 76.35 7.91 1.81 Br 78.14 8.10 1.95 r 76.88 7.95 2.16 N03 71.46 7.41 HC03 44.50 4.61 CH3COO- 40.90 4.24 so^- 160.0 8.29 a From Robinson, R. A., and R. H. Stokes, Electrolyte Solutions, Academic Press, New York, 1959. Used by permission. b From Adamson, A. W. A Textbook of Physical Chemistry, Academic Press, New York, 1973. Used by permission. Degree of Dissociation At a certain concentration, an electrolyte may be only partially dissociated. At infinite dilution, any electrolyte, weak or strong, is completely dissociated. In 1887, the Swedish chemist Svante August Arrhenius (1859-1927) suggested that the degree of dissociation (a) of an electrolyte can be calculated by the simple relation A a = -7-A0 (7.7) 7.1 Electrical Conduction in Solution 267 where A is the molar conductance at a particular concentration to which a refers. Using Equation 7.7, the German chemist Wilhelm Ostwald (1853-1932) showed how one can measure the dissociation constant of an acid. Consider a weak acid HA of concentration c (mol L_1). At equilibrium, we have HA ^ H+ + A" c(l — a) ca ca where a is the fraction of HA dissociated. The dissociation constant, Ka, is given by [H+][A1 c2a2 ca' [HA] c(l - a) (1 - a) Using the expression for a in Equation 7.7, we obtain cA2 a Ao(Ao-A) Equation 7.8 can be rearranged to give 1 [ [(Ac) + 1 A *aA2^' 1 A0 (7.8) (7.9) Equation 7.9 is known as the Ostwald dilution law. Thus, the value of Ka can be obtained either directly from Equation 7.8 or more accurately from Equation 7.9 by plotting 1 /A versus Ac (Figure 7.3). Note that A0 must be known to apply Equation 7.8 alone. Figure 7.3 Graphical determination of Ka according to Equation 7.9. EXAMPLE 7.3 The molar conductance of an aqueous acetic acid solution at concentration 0.10 Mis 5.2 Q"1 mol"1 cm2 at 298 K. Calculate the dissociation constant of acetic acid. ANSWER From Table 7.1, Aq = 390.71 Q"1 mol"1 cm2, and from Equation 7.8 we write (0.10 mol L-1)^ Q"1 mor1 cm2)2_ a ~ (390.71 Q"1 mol"1 cm2) (390.71 - 5.2)0-! mol"1 cm2 = 1.8 x 10"5 mol L"1 268 Chapter 7: Electrolyte Solutions Ionic Mobility The molar conductance of a solution depends on the ease of ionic movement. Ionic velocity is not a constant, however, because it depends on the strength of the electric field (E).* On the other hand, ionic mobility (u), defined as the ionic velocity per unit electric field, is a constant. Thus, the ionic mobility of a cation, w+, is given by «+ = f (7.10) where v+ is the velocity of the cation moving in an electric field of strength E. Ionic mobility has the units (cm s_1)/V cm-1, or cm2 s_1 V-1; it is related to the molar ionic conductance at infinite dilution as follows: u+ = — and u_ = — (7.11) where F is the Faraday constant (after the English chemist and physicist Michael Faraday, 1791-1867).t Table 7.2 lists the ionic mobilities of various ions at 298 K. The ionic mobilities of H+ and OH" ions are much higher than those of other ions. These high values are due to hydrogen bonding. In water, the proton is hydrated, and its movement can be represented as follows: H+ H H H H H+ \ \ / \ / / » O—H.....O—H.....O -> O.....H—O.....H—O -» / ^ ^ \ / \ H H H H Similarly, the movement of the hydroxide ion is H H H H H H \ / / \ \ / <- _0.....H—O.....H—O <- O—H—O—H—O- <- In each case, the ion can move along an extended hydrogen-bond network, resulting in very high mobility. Ionic mobility is utilized in electrophoresis, a technique for purifying and identifying proteins and nucleic acids. * For example, if the potential drop between two electrodes separated by 2.0 cm in an electrolytic cell is 5.0 V, then the electric field is 5.0 V/2.0 cm. or 2.5 V cm-1. t The charge carried by 1 mole of electrons is called the faraday (F), given by 96,485 coulombs. In most calculations, we shall round off the quantity to 96,500 C mol-1. 7.1 Electrical Conduction in Solution EXAMPLE 7.4 The mobility of a chloride ion in water at 25°C is 7.91 x 10-4 cm2 s_1 V-1. (a) Calculate the molar conductance of the ion at infinite dilution, (b) How long will it take for the ion to travel between two electrodes separated by 4.0 cm if the electric field is 20 V cm"1? ANSWER From Equation 7.11, 1q = Fu_ = (96500 C mol"1) (7.91 x 10"4 cm2 s~l V"1) = 76.3 C s"1 V"1 mol"1 cm2 = 76.3 Q_1 mol"1 cm2 because 1 C s"1 = 1 A and A/V = Q_1 (Ohm's law), (b) First, we calculate the ionic velocity, given by Equation 7.10, for an anion u_ = Eu_ = (20 V cm"1) (7.91 x 10"4 cm2 s"1 V"1) = 1.58 x 10"2 cm s"1 Next, we write time distance velocity 4.0 cm 1.58 x 10-2 cm s"1 2.5 x 102 s = 4.2 min Applications of Conductance Measurements Accurate conductance measurements are easy to make and have many applications. Two examples are described below. Acid-Base Titration. As mentioned earlier, the conductances of H+ and OH" are considerably higher than those of other cations and anions. By plotting the conductance of a HC1 solution as a function of NaOH solution added, we obtain a titration 270 Chapter 7: Electrolyte Solutions ^Equivalence point Volume of NaOH added Figure 7.4 Acid-base titration monitored by conductance measurements. Note the difference between a strong acid-base titration (HCI vs. NaOH) and a weak acid-strong base titration (CHgCOOH vs. NaOH). curve such as the one shown in Figure 7.4. Initially, the conductance of the solution falls, because H+ ions are replaced by Na+ ions, which have a lower ionic conductance. This trend continues until the equivalence point is reached. Beyond this point, the conductance begins to rise due to the excess of OH" ions. If the acid is a weak electrolyte, such as acetic acid, the slope of the first part of the curve is much less steep—the conductance actually increases right from the beginning—and there is more uncertainty in determining the equivalence point. Solubility Determination. We have seen how the dissociation constant of acetic acid can be obtained from conductance measurements. The same procedure can be applied to determine the solubility of a sparingly soluble salt. Suppose that we are interested in the molar solubility (mol L"1) and the solubility product of AgCl (a 1:1 electrolyte) in water at 298 K. From Equation 7.4, we write where S is the molar solubility in mol L 1. Because AgCl is an insoluble salt, the concentration of the solution is low, so we can assume that A ~ A0. Thus, S"A~Ao Experimentally, the specific conductance of a saturated AgCl solution is 1.86 x 10"6 Q"1 cm-1. Because water is a weak electrolyte, however, we must take out the contribution due to water itself (k for water is 6.0 x 10"8 Q"1 cm"1). Thus, «(AgCl) = (1.86 x 10"6) - (6.0 x 10"8) = 1.8 x 10"6 Q"1 cm"1 7.2 A Molecular View of the Solution Process According to Table 7.1, A0 = 137.2 Q 1 mol 1 cm2 for AgCl. Finally, we have 7.2 A Molecular View of the Solution Process Why does NaCl dissolve in water and not in benzene? We know that NaCl is a stable compound in which the Na+ and CI- ions are held by electrostatic forces in the crystal lattice. For NaCl to enter the aqueous environment, the strong attractive forces must somehow be overcome. The dissolution of NaCl in water presents two questions: How do the ions interact with water molecules, and how do they interact with one another? Water is a good solvent for ionic compounds because it is a polar molecule and therefore can stabilize the ions through ion-dipole interaction that results in hydration. Generally, smaller ions can be hydrated more effectively than larger ions. A small ion contains a more concentrated charge, which leads to greater electrostatic interaction with the polar water molecules* Figure 7.5 shows a schematic diagram of hydration. Because a different number of water molecules surrounds each type of ion, we speak of the hydration number of an ion. This number is directly proportional to the charge and inversely proportional to the size of the ion. Note that water in the "hydration sphere" and bulk water molecules have different properties^ which can be distinguished by spectroscopic techniques such as nuclear magnetic resonance. There is a dynamic equilibrium between the two types of molecules. Depending on the ion, the mean lifetime of a H20 molecule in the hydration sphere can vary tremendously. For example, consider the mean lifetime of HzO in the hydration sphere for the following ions: Br~, 10-11 s; Na+, 10"9 s; Cu2+, 10"7 s; Fe24,1(T5 s; Al3+, 7 s; and Cr34, 1.5 x 105 s, or 42 h. * According to electrostatic theory, the electric field at the surface of a charged sphere of radius r is proportional to ze/r2, where z is the number of charges and e is the electronic charge. t Water molecules in the hydration sphere of an ion do not exhibit individual translational motion. They move with the ion as a whole. k_ _ 1.8 x 10~6 Q"1 cm"1 1000 cm3 A0 ~ 137.2 Q"1 mol"1 cm2 X 1 L = 1.3 x 10"5 mol L_1 The solubility product, K , for AgCl is given by *sp = [Ag+][C1-] = (1.3 x 10"5)(1.3 x IQ"5) = 1.7 x 10"10 Figure 7.5 Hydration of a cation and an anion. In general, each cation and each anion has a specific number of water molecules associated with it in the hydration sphere. 272 Chapter 7: Electrolyte Solutions The ion-dipole interaction (see Chapter 17) between dissolved ions and water molecules can affect several bulk properties of water. Small and/or multicharged ions such as Li+, Na+, Mg2+, Al3+, Er34, OH", and F" are often called structure-making ions. The strong electric fields exerted by these ions can polarize water molecules, producing additional order beyond the first hydration layer. This interaction increases the solution's viscosity. On the other hand, large monovalent ions such as K+, Rb+, Cs+, NHJ, Cl~, NOj, and CIO4 are structure-breaking ions. Because of their diffuse surface charges and hence weak electric fields, these ions are unable to polarize water molecules beyond the first layer of hydration. Consequently, the viscosities of solutions containing these ions are usually lower than that of pure water. The effective radii of hydrated ions in solution can be appreciably greater than their crystal or ionic radii. For example, the radii of the hydrated Li+, Na+, and K+ ions are estimated to be 3.66 A, 2.80 A, and 1.87 A, respectively, although the ionic radii actually increase from Li+ to K+. We might expect the mobility of an ion to be inversely proportional to its hydrated radius. Table 7.2 nicely demonstrates the truth of this expectation. The high ionic mobility of the proton, which we would expect to be strongly hydrated because of its very small size, stems from the rapid movement of the H+ ions via hydrogen bonds. We now turn to the other question raised earlier: How do ions interact with one another? According to Coulomb's law (after the French physicist Charles Augustin de Coulomb, 1736-1806), the force (F) between Na+ and Cl~ ions in a vacuum is given by F = ^f (7.12) where s0 is the permittivity of the vacuum (8.854 x 10"12 C2 N"1 m"2), qm+ and qa- are the charges on the ions, and r is the distance of separation. The factor 4jt£0 is present as a result of using SI units. In the polar medium of water, as Figure 7.6 shows, the dipolar molecules align themselves with their positive ends facing the negative charge and the negative ends facing the positive charge. This arrangement reduces the effective charge at the positive and negative charge centers by a factor of lis, where s is The dielectric constant of the dielectric constant of the medium (see Appendix 7.1 on p. 295). Therefore, in any vacuum is 1. medium other than a vacuum, Equation 7.12 takes the form \%E(frL Table 7.3 lists the dielectric constants of several solvents. Keep in mind that £ always decreases with increasing temperature. For example, at 343 K, the dielectric constant Figure 7.6 (a) Separation of a cation and an anion in a vacuum, (b) Separation of the same ions in water. The alignment of polar water molecules is exaggerated. Because of thermal motion, the polar molecules are only partially aligned. Nevertheless, this arrangement reduces the electric field and hence the attractive force between the ions. (a) (b) 7.2 A Molecular View of the Solution Process Table 7.3 Dielectric Constants of Some Pure Liquids at 298 K Liquid Dielectric constant, e H2S04 101 H20 78.54 (CH3)2SO (dimethylsulfoxide) 49 C3Hg03 (glycerol) 42.5 CH3N02 (nitromethane) 38.6 HOCH2CH2OH (ethylene glycol) 37.7 CH3CN (acetonitrile) 36.2 CH3OH 32.6 C2H5OH 24.3 CH3COCH3 (dimethyl ether) 20.7 CH3COOH 6.2 4.6 C2H5OC2H5 (diethyl ether) 4.3 CS2 2.6 The dielectric constant is a dimensionless quantity. of water is reduced to about 64. Water's large dielectric constant is what reduces the attractive force between the Na+ and CI" ions and enables them to separate in solution. The dielectric constant of a solvent also determines the "structure" of ions in solution. To maintain electrical neutrality in solution, an anion must be near a cation, and vice versa. Depending on the proximity of these two ions, we can think of them either as "free" ions or as "ion pairs." Each free ion is surrounded by at least one and perhaps several layers of water molecules. In an ion pair, the cation and anion are close to each other, and few or no solvent molecules are between them. Generally, free ions and ion pairs are thermodynamically distinguishable species that have quite different chemical reactivities. For dilute 1:1 aqueous electrolyte solutions, such as NaCl, ions are believed to be in the free-ion form. In higher-valence electrolytes, such as CaCl2 and NajSC^, the formation of ion pairs is indicated by conductance measurements, for a neutral ion pair cannot conduct electricity. Two opposing factors determine whether we have free ions or ion pairs in solution: the potential energy of attraction between the cation and anion, and the kinetic or thermal energy, of the order of kBT, for individual ions. We can now understand easily why NaCl does not dissolve in benzene. A non-polar molecule, benzene does not solvate Na+ and CI" ions effectively. In addition, its small dielectric constant means that the cations and anions will have little tendency to enter the solution as separate ions. 274 Chapter 7: Electrolyte Solutions EXAMPLE 7.5 Calculate the force in newtons between a pair of Na+ and Cl~" ions separated by exactly 1 nm (10 A) in (a) a vacuum and (b) water at 25°C. The charges on the Na+ and CI" ions are 1.602 x 10"19 C and -1.602 x 10"19 C, respectively. ANSWER From Equations 7.12 and 7.13 and Table 7.3, we proceed as follows. (1.602 x 10-19 C) (-1.602 x 10~19 C) (a) (b) F = 4tz:(8.854 x 10"12 C2 N"1 m"2)(l)(l x 10"9 m)2 = -2.31 x 10"10N F = (1.602 x 10~19 C) (-1.602 x 10~19 C) 4tt(8.854 x 10"12 C2 N"1 m"2)(78.54)(1 x 10"9 m)2 = -2.94 x 10"12 N COMMENT As expected, the attractive force between the ions is reduced by a factor of about 80 from vacuum to the aqueous environment. The negative sign convention for F denotes attraction. 7.3 Thermodynamics of Ions in Solution In this section, we shall briefly examine the thermodynamic parameters of the solution process involving ionic compounds and the thermodynamic functions of the formation of ions in aqueous solution. The constant-pressure dissolution of NaCl can be represented by Na+(gr) + Cr(g) I2 Na+(a"NaCl + ^ln »jNa+ff2cl- 7.4 Ionic Activity 279 where /"°NaCl = i"Na+ + i"Cl" In general, a salt with the formula Mv+Xv dissociates as follows: MVXV ^ v+Mz+ + v_Xz- where v+ and v_ are the numbers of cations and anions per unit and z+ and z_ are the numbers of charges on the cation and anion, respectively. For NaCl, v+ = v_ = 1, z+ = +1, z_ = —1. For CaCl2, v+ = 1, v_ = 2, z+ = +2, andz_ = —1. The chemical potential is given by H = v+fi+ + v_n_ (7.15) where n , , Recall that each m term is fi+ = v+ + RTIn m+ divided by and H_ = fil + RT\nm_ The molalities of the cation and anion are related to the molality of the salt originally dissolved in solution, m, as follows: m+ = v_m m_ = v_m Substituting the expressions for fi+ and ju_ into Equation 7.15 yields p= (v+fii°+ + v_jj°_) + RTlnml+mv_- (7.16) We define mean ionic molality (m±) as a geometric mean (see Appendix A) of the individual ionic molalities m± = (ml+ml-)l/v (7.17) where v = v+ + v_, and Equation 7.16 becomes p = (v+fi°+ + v_/j°_) + vRT In m± (7.18) Mean ionic molality can also be expressed in terms of the molality of the solution, m. Because m+ = v+m and m_ = v_m, we have m± = [(v+»j)v+(v_7w)v-]1/v = m[(vl+)(vl-)Y/v (7.19) 280 Chapter 7: Electrolyte Solutions EXAMPLE 7.7 Write the expression for the chemical potential of Mg3(P04)2 in terms of the molality of the solution. ANSWER For Mg3(P04)2, we have v+ = 3, v_ = 2, and v = 5. The mean ionic molality is Wl± = («7>2)l/5 and the chemical potential is given by //Mg3(po4)2 = AMg3(po4)2 + 5RT\nm± From Equation 7.19, m± = m(33 x22y/5 = 2.55m so that ^Mg3(Po4)2 = AMg3(Po4)2 + 5*7/In 2.55m Unlike nonelectrolyte solutions, most electrolyte solutions behave nonideally. The reason is as follows. The intermolecular forces between uncharged species generally depend on ljr7, where r is the distance of separation; a OA-m nonelectrolyte solution is considered ideal for most practical purposes. But Coulomb's law has a 1/r2 dependence (Figure 7.7). This dependence means that even in quite dilute solutions (e.g., 0.05 m), the electrostatic forces exerted by ions on one another are enough to cause a deviation from ideal behavior. Thus, in the vast majority of cases, we must replace molality with activity. By analogy to the mean ionic molality, we define the mean ionic activity (a+) as a± = {al+al-yiv (7.20) where a+ and a_ are the activities of the cation and anion, respectively. The mean ionic activity and mean ionic molality are related by the mean ionic activity coefficient, y+; that is, a± = y+ffz.. (7.21) where (7.22) 7.4 Ionic Activity 281 Distance/arbitrary units Figure 7.7 Comparison of dependence of attractive forces on distance, r. electrostatic forces (1/r2) between ions and van der Waals forces (1/r7) between molecules. The chemical potential of a nonideal electrolyte solution is given by H = (v+fjl + v_/*°) + vRT In a± = (v+^° + v_[f_) + RT In al = (v+p° + v_n°_) + RT In a (7.23) where the activity of the electrolyte, a, is related to its mean ionic activity by a = a+ Experimental values of y+ can be obtained from freezing-point depression and osmotic-pressure measurements* or electrochemical studies (see Chapter 9). Hence, the value of a± can be calculated from Equation 7.21. In the limiting case of infinite dilution (m —> 0), we have Lim y+ = 1 m->0 _ Figure 7.8 shows the plots of y+ versus m for several electrolytes. At very low concentrations, y+ approaches unity for all types of electrolytes. As the concentrations of electrolytes increase, deviations from ideality occur. The variation of y+ with concentration for dilute solutions can be explained by the Debye-Hiickel theory, which is discussed next. * Interested readers should consult the standard physical chemistry texts listed in Chapter 1 for details of y± measurements. Chapter 7: Electrolyte Solutions Figure 7.8 Plots of mean activity coefficient, y± versus molality, m, for several electrolytes. At infinite dilution (m —* 0), the mean activity coefficient approaches unity. KCI CaCI2 EXAMPLE 7.8 Write expressions for the activities (a) of KCI, Na2Cr04, and A12(S04)3 in terms of their molalities and mean ionic activity coefficients. ANSWER We need the relations a = a\ = (y±m±)v. KCI: v = 1 + 1 = 2; m± = (m2) 1/2 m Therefore, aKC1 = m2yl N^CrCv v = 2 + 1 = 3; m± = [(2m)2(m)]m = 4mm Therefore, %a2cro4 = ^yl A12(S04)3: v = 2 + 3 = 5;w± = [(2tw)2(3t«)3]1/5 = 1081/5tw Therefore, aAi2(so4)3 = 108m5y| 7.5 Debye-Huckel Theory of Electrolytes Our treatment of deviations from ideality by electrolyte solutions has been empirical: Using the ionic activities obtained from the activity coefficient and the known concentration, we calculate chemical potential, the equilibrium constant, and other properties. Missing in this approach is a physical interpretation of ionic behavior in solution. In 1923, Debye and the German chemist Erich Htickel (1896-1980) put forward a quantitative theory that has greatly advanced our knowledge of electrolyte solutions. 7.5 Debye-Huckel Theory of Electrolytes 283 Based on a rather simple model, the Debye-Huckel theory enables us to calculate the value of y+ from the properties of the solution. The mathematical details of Debye's and HiickePs treatment are too complex to present here. (Interested readers should consult the standard physical chemistry texts listed in Chapter 1.) Instead, we shall discuss the underlying assumptions and final results. Debye and Huckel began by assuming the following: (l) electrolytes are completely dissociated into ions in solution; (2) the solutions are dilute, with a concentration of 0.01 m or lower; and (3) on average, each ion is surrounded by ions of opposite charge, forming an ionic atmosphere (Figure 7.9a). Working from these assumptions, Debye and Huckel calculated the average electric potential at each ion caused by the presence of other ions in the ionic atmosphere. The Gibbs energy of the ions was then related to the activity coefficient of the individual ion. Because neither y+ nor y_ could be measured directly, the final result is expressed in terms of the mean ionic activity coefficient of the electrolyte as follows: logy± = -0.509|z+z_|v7 (7.24) where the | | signs denote the magnitude but not the sign of the product z+z_. Thus, for CuS04, we have z+ = 2 and z_ = —2, but |z+z_| = 4. The quantity I, called the ionic strength, is defined as follows: (7.25) where mt and z, are the molality and the charge of the /th ion in the electrolyte, respectively. This quantity was first introduced by the American chemist Gilbert Newton Lewis (1875-1946), who noted that nonideality observed in electrolye solutions primarily stems from the total concentration of charges present rather than from the chemical nature of the individual ionic species. Equation 7.25 enables us to express the ionic concentrations for all types of electrolytes on a common basis so that we need not sort out the charges on the individual ions. Equation 7.24 is known as the Debye-Huckel limiting law. As written, it applies to an aqueous electrolyte solution at 298 K. Note that the right side of Equation 7.24 is dimensionless. Thus, the / term is assumed to be divided by m° [1 mol (kg H20)-1]. The word "limiting" means that the law applies only to solutions in the limit of dilute concentrations. To cathode Figure 7.9 (a) Simplified presentation of an ionic atmosphere surrounding a cation in solution, (b) In a conductance measurement, the movement of a cation toward the cathode is retarded by the electric field exerted by the ionic atmosphere left behind. 284 Chapter 7: Electrolyte Solutions EXAMPLE 7.9 Calculate the mean activity coefficient (y+) of a 0.010 m aqueous solution of CuS04 at 298 K. in ANSWER The ionic strength of the solution is given by Equation 7.25: / = ^[(0.010m)x22 + (0.010m) x (-2)2] = 0.040 m log y± = -0.509(2 x 2)VO040 = -0.407 From Equation 7.26, - y± = 0.392 Experimentally, y+ is found to be 0.41 at the same concentration. Two points are worth noting in applying Equation 7.24. First, in a solution containing several electrolytes, all the ions in solution contribute to the ionic strength, but z+ and z_ refer only to the ionic charges of the particular electrolyte for which y+ is being calculated. Second, Equation 7.24 can be used to calculate the ionic activity coefficient of individual cations or anions. Thus, for the fth ion, we write log y, = -0.509z2v7 (7.26) where zt is the charge of the ion. The y+ and y_ values calculated this way are related to y+ according to Equation 7.22. Figure 7.10 shows calculated and measured values of log y+ at various ionic strengths. We can see Equation 7.24 holds quite well for dilute solutions but must be modified to account for the drastic deviations that occur at high concentrations of electrolytes. Several improvements and modifications have been applied to this equation for treating more concentrated solutions. The generally good agreement between experimentally determined y+ values and those calculated using the Debye-Htickel theory provides strong support for the existence of an ionic atmosphere in solution. The model can be tested by taking a conductance measurement in a very strong electric field. In reality, ions do not move in a straight line toward the electrodes in a conductance cell but move along 7.5 Debye-Hiickel Theory of Electrolytes 0.0 0.2 0.4 0.6 0.8 1.0 1.2 Figure 7.10 Plots of log y± versus V7 for several electrolytes. The straight lines are those predicted by Equation 7.24. a zigzag path. Microscopically, the solvent is not a continuous medium. Each ion actually "jumps" from one solvent hole to another, and as the ion moves across the solution, its ionic atmosphere is repeatedly being destroyed and formed again. The formation of an ionic atmosphere does not occur instantaneously but requires a finite amount of time, called the relaxation time, which is approximately 10-7 s in a 0.01-m solution. Under normal conditions of conductance measurement, the velocity of an ion is sufficiently slow so that the electrostatic force exerted by the atmosphere on the ion tends to retard its motion and hence to decrease the conductance (see Figure 7.9b). If the conductance measurement were carried out at a very strong electric field (approximately 2 x 105 V cm-1), the ionic velocity would be approximately 10 cm s"1. The radius of the ionic atmosphere in a 0.01-tm solution is approximately 5 A, or 5 x 10"8 cm, so that the time required for the ion to move out of the atmosphere is 5 x 10~8 cm/(l0 cm s"1), or 5 x 10"9 s, which is considerably shorter than the relaxation time. Consequently, the ion can move through the solution free of the retarding influence of the ionic atmosphere. The free movement leads to a marked increase in conductance. This phenomenon is called the Wien effect, after the German physicist Wilhelm Wien (1864-1928), who first performed the experiment in 1927. The Wien effect is one of the strongest pieces of evidence for the existence of an ionic atmosphere. 286 Chapter 7: Electrolyte Solutions The Salting-In and Salting-Out Effects The Debye-Hiickel limiting law can be applied to study the solubility of proteins. The solubility of a protein in an aqueous solution depends on the temperature, pH, dielectric constant, ionic strength, and other characteristics of the medium. In this section, however, we shall focus on the influence of ionic strength. Let us first investigate the effect of ionic strength on the solubility of an inorganic compound, AgCl. The solubility equilibrium is AgCl(s) ^ Ag+(a<7) + C\-{aq) The thermodynamic solubility product for the process, k\v, is given by k°p = «Ag+«Cl" The ionic activities are related to ionic concentrations as follows: a+ = y+m+ and a_ = y_m_ so that = rAg^Ag^Cl-^Cl- = 7Ae+7a- -^sp where ksp = mAg+ma- is the apparent solubility product. The difference between the thermodynamic and apparent solubility products is as follows. As we can see, the apparent solubility product is expressed in molalities (or some other concentration unit). We can readily calculate this quantity if we know the amount of AgCl dissolved in a known amount of water to produce a saturated solution. Because of electrostatic forces, however, the dissolved ions are under the influence of their immediate neighbors. Consequently, the actual or effective number of ions is not the same as that calculated from the concentration of the solution. For example, if a cation forms a tight ion pair with an anion, then the actual number of species in solution, from a thermodynamic perspective, is one and not, as we would expect, two. This is the reason for replacing concentration with activity, which is the effective concentration. Thus, the thermodynamic solubility product represents the true value of the solubility product, which generally differs from the apparent solubility product. Because we write kzp = y2±ksv Taking the logarithm of both sides and rearranging, we obtain Ik \m -logy± = log(^j = 0.509|z+z_|Vf 7.5 Debye-Huckel Theory of Electrolytes 287 The last equality in the above equation is the Debye-Huckel limiting law. The solubility product can be directly related to the molar solubility (S); for a 1:1 electrolyte, {Kj/2 = S and (K°J<2 = S° where S and S° are the apparent and thermodynamic molar solubilities in mol L"1. Finally, we obtain the following equation relating the solubility of an electrolyte to the ionic strength of the solution: log J; = 0.509|z+z_|Vf (7.27) Note that the value of S° can be determined by plotting log S versus VZ The intercept on the log S axis (/ = 0) gives log S°, and hence S°. If AgCl is dissolved in water, its solubility (S) is 1.3 x 10"5 mol L"1. If it is dissolved in a KN03 solution, according to Equation 7.27, its solubility is greater because of the solution's increase in ionic strength. In a KN03 solution, the ionic strength is a sum of two concentrations, one from AgCl and the other from KN03. The increase in solubility, caused by the increase in ionic strength, is called the salting-in effect. Equation 7.27 holds up only to a certain value of ionic strength. As the ionic strength of a solution increases further, it must be replaced by the following expression: log J; = -K'l (7.28) where K' is a positive constant whose value depends on the nature of the solute and on the electrolyte present. The larger the solute molecule, the greater the value of K' is. Equation 7.28 tells us that the ratio of the solubilities in the region of high ionic strength actually decreases with I (note the negative sign). The decrease in solubility with increasing ionic strength of the solution is called the salting-out effect. This phenomenon can be explained in terms of hydration. Recall that hydration is the process that stabilizes ions in solution. At high salt concentrations, the availability of water molecules decreases, and so the solubility of ionic compounds also decreases. The salting-out effect is particularly noticeable with proteins, whose solubility in water is sensitive to ionic strength because of their large surface areas. Combining Equations 7.27 and 7.28, we have the approximate equation log = 0.509|z+z_| Vf - K'l (7.29) Equation 7.29 is applicable over a wider range of ionic strengths. Figure 7.11 shows how the ionic strength of various inorganic salts affects the solubility of horse hemoglobin. As we can see, the protein exhibits a salting-in region at low ionic strengths* As I increases, the curve goes through a maximum and eventually the slope becomes negative, indicating that the solubility decreases with increasing ionic strength. In this region, the second term in Equation 7.29 predominates. This trend is most pronounced for salts such as Na2S04 and (NFL^SCv * When I is less than unity, V7 > I. Thus, at low ionic strengths, the first term in Equation 7.29 predominates. 288 Chapter 7: Electrolyte Solutions (NH4)2S04 -2.0 0 2 3 4 5 6 Ionic strength Figure 7.11 Plots of log(S/S°) versus ionic strength for horse hemoglobin in the presence of various inorganic salts. Note that when / = 0, all the curves converge to the same point on the log(S/S°) axis at zero and S = S°. [From Cohn, E., and J. Edsall, Proteins, Amino Acids and Peptides, © Litton Educational Publishing, 1943. Reprinted by permission of Van Nostrand Reinhold, New York.] The practical value of the salting-out effect is that it enables us to precipitate proteins from solutions. In addition, the effect can also be used to purify proteins. Figure 7.12 shows the range of the salting-out phenomenon for several proteins in the presence of ammonium sulfate. Although the solubility of proteins is sensitive to the degree of hydration, the strength of binding of water molecules is not the same for all proteins. The relative solubility of different proteins at a particular ionic strength provides a means for selective precipitation. The point is that although higher ionic strengths are needed to salt out proteins, precipitation occurs over a small range of ionic strength, providing sharp separations. 7.6 Colligative Properties of Electrolyte Solutions The colligative properties of an electrolyte solution are influenced by the number of ions present in solution. For example, we expect the aqueous freezing-point depression caused by a 0.01-7« solution of NaCl to be twice that effected by a 0.01-zw sucrose solution, assuming complete dissociation of the former. For incompletely dissociated salts, the relationship is more complicated, but understanding it provides us with another way to measure an electrolyte's degree of dissociation. Let us define a factor /, called the van't Hoff factor (after the Dutch chemist Jacobus Hendricus van't Hoff, 1852-1911), as follows: actual number of particles in solution at equilibrium (7.30) i number of particles in solution before dissociation 7.6 Colligative Properties of Electrolyte Solutions 289 0 2 4 6 8 10 Ionic strength Figure 7.12 Plots of (S/S°) versus ionic strength for several proteins in aqueous ammonium sulfate, demonstrating the salting-out effect. [From Cohn, E. J., Chem. Rev. 19, 241 (1936). Used by permission of Williams & Wilkins, Baltimore.] If a solution contains A7 units of an electrolyte, and if a is the degree of dissociation, Mv Xv ^ v+Mz+ + v_Xz-N{\ - a) Nv+a Nv_a there will be N(l — a) undissociated units and (Nv+a + Nv_a), or Nva ions in solution at equilibrium, where v = v+ + v_. We can now write the van't Hoff factor as N(l - a) + Nva i =-—-= 1 — a + va and (7.31) For strong electrolytes, / is approximately equal to the number of ions formed from each unit of the electrolyte; for example, /« 2 for NaCl and CuS04; /« 3 for K2S04 and BaCl2, and so on. The value of i decreases with increasing concentration of the solution, which is attributed to the formation of ion pairs. The presence of ion pairs will also affect the colligative properties because it decreases the number of free particles in solution. In general, formation of ion pairs is most pronounced between highly charged cations and anions and in media of low dielectric constants. In an aqueous solution of Ca(N03)2, for example, the Ca2+ and NO3 ions form ion pairs as follows: Ca2+(aq) + NO^aq) ^ Ca(N03)+(a^) 290 Chapter 7: Electrolyte Solutions The equilibrium constants for such ion pairing are not accurately known, however, thus making calculations of colligative properties of electrolyte solutions difficult. A recent study* showed that deviation from colligative properties for many electrolyte solutions is not the result of ion pair formation, but rather of hydration. In an electrolyte solution the cations and anions can tie up large numbers of water molecules in their hydration spheres, thus reducing the number of free water molecules in the bulk solvent. Deviations disappear when the correct number of water molecules in the hydration sphere is subtracted from the total water solvent molecules in calculating the concentrations (molality or molarity) of the solution. EXAMPLE 7.10 The osmotic pressures of a 0.01 m solution of CaCl2 and a 0.01 m sucrose solution at 298 K are 0.605 atm and 0.224 atm, respectively. Calculate the van't Hoff factor and the degree of dissociation for CaCl2. Assume ideal behavior. ANSWER As far as osmotic pressure measurements are concerned, the main difference between calcium chloride and sucrose is that only CaCl2 can dissociate into ions (Ca2+ and CI"). Otherwise, equal concentrations of CaCl2 and sucrose solutions would have the same osmotic pressure. Because the osmotic pressure of a solution is directly proportional to the number of particles present, we can calculate the van't Hoff factor for the CaCl2 solution as follows. From Equation 7.30, 0.605 atm i = TTTönzr = 2.70 0.224 atm Because for CaCl2, v+ = 1, and v_ = 2, we have v = 3 and 2.70 -1 3 - 1 0.85 Finally, we note that the equations used to determine the colligative properties of nonelectrolyte solutions (Equations 6.39,6.40, and 6.46) must be modified for electrolyte solutions as follows: AT = Kb(im2) (7.32) AT = K{(imJ (7.33) n = iMRT (7.34) Ideal behavior is assumed for the electrolyte solutions, so we use concentrations instead of activities. * A. A. Zavitsas, J. Phys. Chem. 105, 7805 (2001). 7.6 Colligative Properties of Electrolyte Solutions The Donnan Effect The Donnan effect (after the British chemist Frederick George Donnan, 1870-1956) has its starting point in the treatment of osmotic pressure. It describes the equilibrium distribution of small diffusible ions on the two sides of a membrane that is freely permeable to these ions but impermeable to macromolecular ions, in the presence of a macromolecular electrolyte on one side of the membrane. Suppose that a cell is separated into two parts by a semipermeable membrane that allows the diffusion of water and small ions but not protein molecules. Let us consider the following three cases. Case 1. The protein solution is placed in the left compartment; water is placed in the right compartment. We assume that the protein molecules are neutral species* Let the concentration of the protein solution be c (mol L_1) so that the osmotic pressure of the solution, according to Equation 6.46, is given by nx = cRT Thus, from a measurement of the osmotic pressure, we can readily determine the molar mass of the protein molecule. Case 2. In this case, the protein is the anion of the sodium salt, Na+P", which we assume to be a strong electrolyte. Again, the protein solution of concentration c is placed in the left compartment, and pure water is in the right compartment. To maintain electrical neutrality, all of the Na+ ions remain in the left compartment; the osmotic pressure of the solution now becomes TI2 = (c + c)RT = 2cRT Because 772 = 2nu it follows that the molar mass determined in this case will be only half of the true molar mass. (From Equation 6.47 we have M2 = c2RT/II, so that doubling II would decrease the value of M2 by half.) In practice, the situation is actually much worse, because the protein ion may bear as many as 20 or 30 net negative (or positive) charges. In the early days of protein-molar-mass determination by osmotic pressure, disastrously poor results were obtained before the dissociation process was recognized and efforts made to correct for it (see Appendix 7.2 on p. 298). Case 3. We start with an arrangement similar to that discussed in Case 2 and then add NaCl (of concentration b in mol L"1) to the right compartment (Figure 7.13a). At equilibrium, a certain amount, x (mol L"1), of Na+ and CI" ions has diffused through the membrane from right to left, creating a final state shown in Figure 7.13b. Both sides of the membrane must be electrically neutral: in each compartment the number of cations equals the number of anions. The condition of equilibrium enables us to equate the chemical potentials of NaCl in the two compartments as follows (see Equation 7.23): (/"NaCl)L = (j"NaCl)R * Proteins are ampholytes, that is, they possess both acidic and basic properties. Depending on the pH of the medium, a protein can exist as an anion, a cation, or a neutral species. 292 Chapter 7: Electrolyte Solutions (a) (b) Figure 7.13 Schematic representation of the Donnan effect, (a) Before diffusion has begun, (b) At equilibrium. The membrane separating the left and right compartments is permeable to all but the P- ions. The volumes in the two compartments are equal and assumed to remain constant. or (ff + 2RTlna±)^cl = {f + 2tf7/lna±)*aC1 Because n°, the standard chemical potential, is the same on both sides, we obtain (fl±)NaCl = (a±)NaCl From Equation 7.20, (%a+aCl-)L = (%a+«Cl-)R If the solutions are dilute, then the ionic activities may be replaced by the corresponding concentrations, that is, %a+ = [Na+] and acr = [CI-]. Hence, ([Na+][C1-])L = ([Na+][C1-])K or (c + x)x = (b — x) (b — x) Table 7.6 The Donnan Effect and Osmotic Pressure Initial concentration Equilibrium concentration Left compartment Right compartment Left compartment c = [Na+] = [P-] b = [Na+] = [CI"] (c + x) = [Na+] c = [P-] x = [C1-] 0.1 0 0.1 0.1 0 0.1 0.01 0.1008 0.1 0.00083 0.1 0.1 0.1333 0.1 0.0333 0.1 1.0 0.576 0.1 0.476 0.1 10.0 5.075 0.1 4.975 [P] = 0.1 0 0 [P] = 0.1 0 7.6 Colligative Properties of Electrolyte Solutions 293 Solving for x, we obtain b2 x ~ c + 2b (7.35) Equation 7.35 says that the amount of NaCl, x, that diffuses from right to left is inversely proportional to the concentration of the nondiffusible ion (P~), c, in the left compartment. This unequal distribution of the diffusible ions (Na+ and CI-) in the two compartments is the result of the Donnan effect. In this case, then, the osmotic pressure of the protein solution is determined by the difference between the number of particles in the left compartment and the number in the right compartment. We write n3 = [(c + c + x + x)-2(b- x)]RT = (2c + 4x - 2b)RT left compartment right compartment From Equation 7.35, Two limiting cases may be applied to the equation above. If b , which is the potential energy of a unit positive charge in the electric field. Its units are J/C or V (1 J = 1 C x 1 V). A unit positive charge in an electric field E experiences a force equal in magnitude to E. When the charge is moved through a certain distance, dr, the potential energy change is equal to qEdr or Edr because q = 1 C. Because the repulsive potential energy increases as the unit positive charge comes closer to the positive charge qA that generates the electric field, the change in potential energy, d, is — Edr. (The negative sign ensures that as dr decreases, — Edr is a positive quantity, signifying the increase in repulsion between the two positive charges.) The electric potential at a certain point at a distance r from the charge qA is the potential energy change that occurs in bringing the unit positive charge from infinity to distance r from the charge: Edr = - 1a , Ansnr2 dr = 1a AmcT (4) 295 296 Chapter 7: Electrolyte Solutions (a) (b) Figure 7.14 (a) The charge separation of a capacitor. The electric field, E, is directed from the positive plate to the negative plate, separated by distance d. With a vacuum between the plates, the dielectric constant is e0. (b) Orientation of the dipoles of a dielectric in a capacitor. The degree of orientation is exaggerated. The dielectric material decreases the electric field between the capacitor plates. The dielectric constant of the medium is e. Note that (j> = 0 at r = oo. From Equation 4, we can define the electric potential difference between points 1 and 2 in an electric field as the work done in bringing a unit charge from 1 to 2; that is, W = fa ~ fa (5) This difference is commonly referred to as the voltage between points 1 and 2. Dielectric Constant (s) and Capacitance (C) When a nonconducting substance (called a dielectric) is placed between two flat, parallel metal plates with opposite charges that are equal in magnitude (called a capacitor), the substance becomes polarized. The reason is that the electric field of the plates either orients the permanent dipoles of the dielectric or induces dipole moments, as shown in Figure 7.14. The dielectric constant of the substance is defined as where E0 and E are the electric fields in the space between the plates of the capacitor in the absence of a dielectric (a vacuum) and presence of a dielectric, respectively. Keep in mind that the orientation of the dipoles (or induced dipoles) reduces the electric field between the capacitor plates so that E 1. For ions in aqueous solution, this decrease in electric field reduces the attraction between the cation and the anion (see Figure 7.6). The capacitance (C) of a capacitor measures its ability to hold charges for a given electric potential difference between the plates; that is, it is given by the ratio of charge to potential difference. The capacitances of a capacitor, when the space between the plates is filled with a dielectric (C) and with a vacuum (C0), are given respectively by Appendix 7.1: Notes on Electrostatics 297 and Because & = Ed, where d is the distance between the plates, Equation 6 can also be written as (A^0/d) (0CO) _ C_ (Q, e (At/d) (QIC) C0 W Capacitance is an experimentally measurable quantity, so the dielectric constant of a substance can be determined. It has the unit farad (F), where 1 F = 1 C/V. Note that the ratio C/C0 in Equation 9 makes e a dimensionless quantity. APPENDIX 7.2 The Donnan Effect Involving Proteins Bearing Multiple Charges For a protein at a pH other than its isoelectric point, at which it will possess either a net positive or net negative charge, an additional factor must be considered; that is, the counterions needed to maintain charge neutrality. In the chapter, we considered the simple case where the protein bears only one negative charge. Here we deal with the situation in which a protein bears a number of negative charges (z). We assume that the protein, Na+P2-, is a strong electrolyte so that Na+Pz" -> zNa+ + Pz" Referring to Figure 7.13, we shall consider two cases. Case 1 The protein solution is placed in the left compartment and water in the right compartment. The osmotic pressure of the solution (TJj) is given by n1 = (z+ \)cRT where c is the concentration (molarity) of the protein solution. Because z is typically of the order of 30, using this arrangement to determine the molar mass of the protein yields a value that is only 55 of the true value. Case 2 Again the protein solution is placed in the left compartment, but a NaCl solution is placed in the right compartment. The requirement that the chemical potential of a component be the same throughout the system applies to the NaCl as well as to the water. To attain equilibrium. NaCl will move from the right to the left compartment. We can calculate the actual amount of NaCl that is transported. The initial molar concentration of Na+Pz~ is c, and that of NaCl is b. At equilibrium, the concentrations are [Pz"]L = c [Na+]L = (zc + x) [Cl-]L = x and [Na+]R = {b-x) [C1-]R =(b-x) where x is the amount of NaCl transported from right to left. Because (/*NaCi)L = (^Naci)11 at equilibrium and for dilute solutions, we replace activities with concentrations so that 298 ([Na1[Cl-])L = ([Na+][Cl-])R Appendix 7.2: The Donnan Effect Involving Proteins Bearing Multiple Charges or (zc + x) (x) = (b — x)(b — x) b2 x = zc + 2b The osmotic pressure (772), which is proportional to the difference in solute concentration between the two sides, is now given by n2 = [(c + zc + x + x) - (b - x + b - x)]RT left compartment right compartment or n2 = (c + zc - 2b + Ax)RT Substituting for x, we obtain 77, = lc + zc - 2b + ^..W \ zc + 2b/ _ zc2 + 2cb + z2c2 „^ ~ zVTYb RT (1) Equation 1 was derived assuming no change in either the pH or the volume of the solutions. Two limiting cases follow. If b S> z2c (the salt concentration is much greater than the protein concentration),* then n2 = ^RT=cRT (2) In this limiting case, the osmotic pressure approaches that of the pure isoelectric protein. In effect, the added salt reduces (and at high enough salt concentrations, eliminates) the Donnan effect. Under these conditions, the molar mass determined by osmotic pressure measurement would correspond closely to the true value. * In practice, c£lx 10-4 M,z£ 30, so that z2c S 0.1 M. Thus, for this limiting case to hold, the concentration of the added salt should be about 1 M 300 Chapter 7: Electrolyte Solutions Suggestions for Further Reading BOOKS Hunt, J. P., Metal Ions in Aqueous Solution, W. A. Benjamin, Menlo Park, CA, 1963. Pass, G., Ions in Solution, Clarendon Press, Oxford, 1973. Robinson, R. A., and R. H. Stokes, Electrolyte Solutions, 2nd ed. Academic Press, New York, 1959. Tombs, M. P., and A. R. Peacocke, The Osmotic Pressure of Biological Macromolecules, Clarendon Press, New York, 1975. Wright, M. R., An Introduction to Aqueous Electrolyte Solutions, John Wiley & Sons, New York, 2007 ARTICLES "Ion Pairs and Complexes: Free Energies, Enthalpies, and Entropies," J. E. Prue, J. Chem. Educ. 46,12 (1969). "On Squid Axons, Frog Skins, and the Amazing Uses of Thermodynamics," W. H. Cropper, J. Chem. Educ. 48, 182 (1971). "Osmotic Pressure in the Physics Course for Students of the Life Sciences," R. K. Hobbie, Am. J. Phys. 42,188 (1974). "Electrolyte Theory and SI Units," R. I. Holliday, J. Chem. Educ. S3,21 (1976). "The Motion of Ions in Solution Under the Influence of an Electric Field," C. A. Vincent, J. Chem. Educ. 53, 490 (1976). "Thermodynamics of Ion Solvation and Its Significance in Various Systems," C. M. Criss andM. Salomon, J. Chem. Educ. 53, 763 (1976). "The Donnan Equilibrium and Osmotic Pressure," R. Chang and L. J. Kaplan, J. Chem. Educ. 54, 218 (1977). "Ionic Hydration Enthalpies," D. W. Smith, J Chem. Educ. 54, 540 (1977). "Paradox of the Activity Coefficient y+," E-I Ochiai, J. Chem. Educ. 67,489 (1990). "Standard Enthalpies of Formation of Ions in Solution," T. Solomon, J. Chem. Educ. 68,41 (1991). "Determination of the Thermodynamic Solubility Product, K^p, of Pbl2 Assuming Nonideal Behavior," D. B. Green, G. Rechtsteiner, and A. Honodel, J. Chem. Educ. 73,789 (1996). "Properties of Water Solutions of Electrolytes and Nonelectrolytes," A. A. Zavitsas, J. Phys. Chem. B, 105, 7805 (2001). "The Definition and Unit of Ionic Strength," T. Solomon, J. Chem. Educ. 78, 1691 (2001) "The Conductivity of Strong Electrolytes: A Computer Simulation in LabVIEW," A. Belletti, R. Borromei, and G. Ingletto, Chem. Educator [Online] 13, 224 (2008) DOI 10.1333/ s00897082144a. Problems Problems Ionic Conductance 7.1 The resistance of a 0.010 MNaCl solution is 172 Q. If the molar conductance of the solution is 153 Q_1 mol-1 cm2, what is the cell constant? 7.2 Using the cell described in Problem 7.1, a student determined the resistance of a 0.086 MKC1 solution to be 20.4 Q. Calculate the molar conductance of this solution. 7.3 The cell constant (ifA) of a conductance cell is 388.1 m_1. At 25°C, the resistance of a 4.8 x 10-4 mol L"1 aqueous solution of sodium chloride is 6.4 x 104 Q and that of a sample of water is 7.4 x 106 Q. Calculate the molar conductance of the NaCl in solution at this concentration. 7.4 Given that the measurement of A0 for weak electrolytes is generally difficult to obtain, how would you deduce the value of A0 for CH3COOH from the data listed in Table 7.1? (Hint: Consider CH3COONa, HC1, and NaCl.) 7.5 A simple way to determine the salinity of water is to measure its conductivity and assume that the conductivity is entirely due to sodium chloride. In a particular experiment, the resistance of a sample solution is found to be 254 Q. The resistance of a 0.050 MKC1 solution measured in the same cell is 467 Q. Estimate the concentration of NaCl in the solution. (Hint: First derive an equation relating R to A and c and then use A0 values for A.) 7.6 A conductance cell consists of two electrodes, each with an area of 4.2 x 10~4 m2, separated by 0.020 m. The resistance of the cell when filled with a 6.3 x 10"4 M KN03 solution is 26.7 Q. What is the molar conductance of the solution? 7.7 Referring to Figure 7.4, explain why the slope of conductance versus volume of NaOH added rises right at the start if the acid employed in the titration is weak. Solubility 7.8 Calculate the solubility of BaS04 (in g L"1) in (a) water and (b) a 6.5 x 10"5 MMgS04 solution. The solubility product of BaS04 is 1.1 x 10~10. Assume ideal behavior. 7.9 The thermodynamic solubility product of AgCl is 1.6 x lO-10. What is [Ag+] in (a) a 0.020 M KN03 solution and (b) a 0.020 MKC1 solution? 7.10 Referring to Problem 7.9, calculate the value of AG° for the process AgClfc) ^ Ag+(ao) + Cl-(aa) to yield a saturated solution at 298 K. (Hint: Use the well-known equation AG° = — RT In K.) 7.11 The apparent solubility products of CdS and CaF2 at 25°C are 3.8 x 10"29 and 4.0 x 10"11, respectively. Calculate the solubility (g/100 g of solution) of these compounds. 7.12 Oxalic acid, (COOH)2, is a poisonous compound present in many plants and vegetables, including spinach. Calcium oxalate is only slightly soluble in water (Ksp = 3.0 x 10"9 at 25°C) and its ingestion can result in kidney stones. Calculate (a) the apparent and thermodynamic solubility of calcium oxalate in water, and (b) the concentrations of calcium and oxalate ions in a 0.010 M Ca(N03)2 solution. Assume ideal behavior in (b). Chapter 7: Electrolyte Solutions Ionic Activity 7.13 Express the mean activity, mean activity coefficient, and mean molality in terms of the individual ionic quantities (a+, a_, y+, y_, m+, and m_) for the following electrolytes: KI, SrS04, CaCl2, Li2C03, K3Fe(CN)6, and K4Fe(CN)6. [Hint: Fe(CN)4" is a complex ion.] 7.14 Calculate the ionic strength and the mean activity coefficient for the following solutions at 298 K: (a) 0.10 m NaCl, (b) 0.010 m MgCl2, and (c) 0.10 m Kfie(Ctf)6. 7.15 The mean activity coefficient of a 0.010 m H2S04 solution is 0.544. What is its mean ionic activity? 7.16 A 0.20 m Mg(N03)2 solution has a mean ionic activity coefficient of 0.13 at 25°C. Calculate the mean molality, the mean ionic activity, and the activity of the compound. Debye-Htickel Limiting Law 7.17 The Debye-Htickel limiting law is more reliable for 1:1 electrolytes than for 2:2 electrolytes. Explain. 7.18 The size of the ionic atmosphere, called the Debye radius, is 1/k , where k is given by where e is the electronic charge, NA Avogadro's constant, s0 the permittivity of vacuum (8.854 x 10"12 C2 N"1 m~2), £ the dielectric constant of the solvent, kB the Boltzmann constant, Tthe absolute temperature, and I the ionic strength (see the physical chemistry texts listed in Chapter 1). Calculate the Debye radius in a 0.010 m aqueous Na2S04 solution at 25°C. 7.19 Explain why it is preferable to take the geometric mean rather than the arithmetic mean when defining mean activity, mean molality, and mean activity coefficient. Colligative Properties 7.20 The freezing-point depression of a 0.010 m acetic acid solution is 0.0193 K. Calculate the degree of dissociation for acetic acid at this concentration. 7.21 A 0.010 m aqueous solution of the ionic compound Co(NH3)5Cl3 has a freezing point depression of 0.0558 K. What can you conclude about its structure? Assume the compound is a strong electrolyte. 7.22 The osmotic pressure of blood plasma is approximately 7.5 atm at 37°C. Estimate the total concentration of dissolved species and the freezing point of blood plasma. 7.23 Calculate the ionic strength of a 0.0020 m aqueous solution of MgCl2 at 298 K. Use the Debye-Htickel limiting law to estimate (a) the activity coefficients of the Mg2+ and Q~ ions in this solution and (b) the mean ionic activity coefficients of these ions. 7.24 Referring to Figure 7.13, calculate the osmotic pressure for the following cases at 298 K: (a) The left compartment contains 200 g of hemoglobin in 1 liter of solution; the right compartment contains pure water, (b) The left compartment contains the same hemoglobin solution as in part (a), and the right compartment initially contains 6.0 g of NaCl in 1 liter of solution. Assume that the pH of the solution is such that the hemoglobin molecules are in the Na+ Hb~ form. (The molar mass of hemoglobin is 65,000 g mol-1.) Problems 303 Additional Problems 7.25 From the following data, calculate the heat of solution for KI: NaCl NaT KC1 KI Lattice energy/kJ mol 1 Heat of solution/kJ mol-1 787 3.8 -5.1 700 17.1 716 643 9 7.26 From the data in Table 7.2, determine A0 for H20. Given that the specific conductance (k) for water is 5.7 x 10-8 Q_1 cm-1, calculate the ion product (Kw) of water at 298 K. 7.27 In this chapter (see Figures 7.2 and 7.11) and in Chapter 6 (see 77measurements in Figure 6.19), we extrapolated concentration-dependent values to zero solute concentration. Explain what these extrapolated values mean physically and why they differ from the value obtained for the pure solvent. 7.28 (a) The root cells of plants contain a solution that is hypertonic in relation to water in the soil. Thus, water can move into the roots by osmosis. Explain why salts (NaCl and CaCl2) spread on roads to melt ice can be harmful to nearby trees, (b) Just before urine leaves the human body, the collecting ducts in the kidney (which contain the urine) pass through a fluid whose salt concentration is considerably greater than is found in the blood and tissues. Explain how this action helps conserve water in the body. 7.29 A very long pipe is capped at one end with a semipermeable membrane. How deep (in meters) must the pipe be immersed into the sea for fresh water to begin passing through the membrane? Assume seawater is at 20°C and treat it as a 0.70 MNaCl solution. The density of seawater is 1.03 g cm"3. 7.30 (a) Using the Debye-Hiickel limiting law, calculate the value of y+ for a 2.0 x 10-3 m Na3P04 solution at 25°C. (b) Calculate the values of y+ and y_ for the Na3P04 solution, and show that they give the same value for y+ as that obtained in (a). Copyrighted Materials Copyright©2014 University Science Books Retrieved from www.knovel.com CHAPTER Chemical Equilibrium When a system is in chemical equilibrium, a change in one of the parameters of the equilibrium produces a shift in such a direction that, were no other factors involved in this shift, it would lead to a change of opposite sign in the parameter considered. Turning from our discussion of nonelectrolyte and electrolyte solutions and physical equilibria in Chapters 6 and 7, we shall focus in this chapter on chemical equilibrium in gaseous and condensed phases. Equilibrium is a state in which there are no observable changes over time; at equilibrium, the concentrations of reactants and products in a chemical reaction remain constant. Much activity occurs at the molecular level, however, because reactant molecules continue to form product molecules while product molecules react to yield reactant molecules. This process is an example of dynamic equilibrium. The laws of thermodynamics help us predict equilibrium composition under different reaction conditions. 8.1 Chemical Equilibrium in Gaseous Systems In this section, we shall derive an expression relating the Gibbs energy change for a reaction in the gas phase to the concentrations of the reacting species and temperature. We first consider the case in which all gases exhibit ideal behavior. Ideal Gases Examples of the simplest type of chemical equilibrium, are cis-trans isomerization, racemization, and the cyclopropane-ring-opening reaction to form propene. The progress of the reaction can be monitored by the quantity £ (Greek letter xi), called the extent of reaction. When an infinitesimal amount of A is converted to B, the change in A is dnA = —dg and that in B is dnB = +dg where dn denotes the change in number of moles. The change in Gibbs energy for this transformation at constant T and P is given by —Henri-Louis Le Chatelier A(g)^B(g) dG = nAdnA + fiBdnB = -[iAd£ + [iBd£ (8.1) 305 306 Chapter 8: Chemical Equilibrium ArG = 0 0 Extent of reaction, ^ For simplicity, we use the units kJ mol-1 rather than kJ (mol reaction)-1. Figure 8.1 A plot of Gibbs energy versus extent of reaction. For a reacting system at equilibrium, the slope of the curve is zero. where /iA and ^ are the chemical potentials of A and B, respectively. Equation 8.1 can be rearranged to give The quantity (dGJd£)T P is represented by ArG, which is the change in Gibbs energy per mole of reaction; it has the units kJ mol-1. During the reaction, the chemical potentials vary with composition. The reaction proceeds in the direction of decreasing G, that is, (dGld£)T P<0. Therefore, the forward reaction (A —> B) is spontaneous when jua > juB whereas the reverse reaction (B —> A) is spontaneous when //B > [tA. At equilibrium, fiA = pB so that (8.2) Figure 8.1 shows a plot of Gibbs energy versus extent of reaction. At constant T and P, we have ArG < 0 Forward reaction is spontaneous ArG > 0 Reverse reaction is spontaneous ArG = 0 Reacting system is at equilibrium Let us now consider a more complicated case: aA(g) ^ bB(g) 8.1 Chemical Equilibrium in Gaseous Systems 307 where a and b are stoichiometric coefficients. According to Equation 6.10, the chemical potential of the /th component in a mixture, assuming ideal behavior, is given by fi^tf + RTln^ where Pt is the partial pressure of component / in the mixture, /i° is the standard chemical potential of component /, and P° = 1 bar. Therefore, we can write ftA = M°A +RT In ^ (8.3a) ^ = H\\ + RT\n^ (8.3b) The Gibbs energy change for the reaction, ArG, can be expressed as ArG = b/ta - a/iA (8.4) Substituting the expressions in Equation 8.3 into Equation 8.4, we get ArG = - ctfiA + bRT In ^ - aRT In ^ (8.5) The standard Gibbs energy change of the reaction, ArG°, is just the difference between the standard Gibbs energies of products and reactants, that is, ArG° = tyg - afi°A Therefore, we can write Equation 8.5 as ArG = ArG° + Jmn|j^ (8.6) By definition, ArG = 0 at equilibrium, so Equation 8.6 becomes 0 = ATG° + RTlnKp or ArG° = —RTlnKp (8.7) KP, the equilibrium constant (where the subscript P denotes that concentrations are expressed in pressures), is given by Kp ^r = ^po) (8-8) 308 Chapter 8: Chemical Equilibrium Equilibrium 0 1 Extent of reaction, € Figure 8.2 Total Gibbs energy versus extent of reaction for the aA(g) ^ oB(g) reaction assuming ArG° < 0. At equilibrium, product is favored over reactant. Note that the equilibrium point, which is at the minimum of Gibbs energy, is a compromise between ArG° and the Gibbs energy of mixing. Equation 8.7 is one of the most important and useful equations in chemical thermodynamics. It relates the equilibrium constant, KP, and the standard Gibbs energy change of the reaction, ArG°, in a remarkably simple fashion. Keep in mind that at a given temperature, ArG° is a constant whose value depends only on the nature of the reactants and products and temperature. In our example, ArG° is the standard Gibbs energy change when reactant A at 1 bar pressure and temperature T is converted to product B at 1 bar pressure and the same temperature per mole of reaction, as shown above. Figure 8.2 shows the Gibbs energy versus the extent of reaction for ArG° < 0. When there is no mixing of reactants with products, the Gibbs energy decreases linearly as the reaction progresses, and eventually the reactants will be completely converted to products. As Equation 6.11 shows, however, AmixG is a negative quantity; therefore, the Gibbs energy for the actual path will be lower than that for the nonmixing case. Consequently, the equilibrium point, which is at the minimum of the Gibbs energy, is a compromise between these two opposing tendencies, that is, the conversion of reactants to products and the mixing of products with reactants. Equation 8.7 tells us that if we know the value of ArG°, we can calculate the equilibrium constant KP, and vice versa. The standard Gibbs energy of a reaction is just the difference between the standard Gibbs energies of formation (AfG°) of the products and the reactants, discussed in Section 5.3. Thus, once we have defined a reaction, we can usually calculate the equilibrium constant from the Af G° values listed in Appendix B and Equation 8.7. Note that these values all refer to 298 K. Later in this chapter (Section 8.5), we shall learn how to calculate the value of KP at another temperature if its value at 298 K is known. Finally, note that the equilibrium constant is a function of temperature alone (because /x° depends only on temperature) and is dimensionless. This follows from the fact that in the expression for KP, each pressure term is divided by its standard-state value of 1 bar, which cancels the pressure unit but does not alter P numerically. 8.1 Chemical Equilibrium in Gaseous Systems EXAMPLE 8.1 From the thermodynamic data listed in Appendix B, calculate the equilibrium constant for this reaction at 298 K: N2(g) + 3H2(g) ^ 2NH3(g) ANSWER The equilibrium constant for the equation is given by (W°)2 KP To calculate the value of KP, we need Equation 8.7 and the value of ArG°. From Equation 5.12 and Appendix B, we have ArG° = 2AfG°(NH3) - AfG°(N2) - 3AfG°(H2) = (2) (-16.6 kJmol"1) - (0) - (3)(0) = -33.2k.Imoi-1 From Equation 8.7, -33,200 J mor1 = -(8.314 J Kr1 mol"1)(298 K) In KP or In KP = 13.4 KP = 6.6 x 105 COMMENT Note that if the reaction were written as |N2(g) + |H2(g)^NH3(g) the value of ArG° would be —16.6 kJ mol 1, and the equilibrium constant would be calculated as follows: (W1 3/2 8.1 x 102 Thus, whenever we multiply a balanced equation throughout by a factor n, we change the equilibrium constant KP to KP. Here n = \, so we have changed KP toK, 1/2 310 Chapter 8: Chemical Equilibrium A Closer Look at Equation 8.7 The change in standard Gibbs energy, ArG°, is generally not equal to zero. According to Equation 8.7, if ArG° is negative, the equilibrium constant must be greater than unity; in fact, the more negative ArG° is at a given temperature, the larger KP is. The reverse holds true if ArG° is a positive number. Here the equilibrium constant is less than unity. Of course, just because ArG° is positive does not mean that no reaction will take place. For example, if ArG° = 10 kJ mol-1 and T = 298 K, then KP = 0.018. While 0.018 is a small number compared to unity, an appreciable amount of products can still be obtained at equilibrium if we use large quantities of reactants for the reaction. There is also the special case in which ArG° = 0, which corresponds to a KP of unity, meaning that the products and reactants are equally favored at equilibrium. It is instructive to look at the factors that affect ArG° and hence KP. From Equation 5.3 we have ArG° = ATH° - TATS° Figure 8.3 The occupancy of energy levels in B is greater than than in A. Consequently, there are more microstates in B and the entropy of B is greater than that of A. At equilibrium, B dominates even though the A —> B reaction is endothermic. so the equilibrium constant at temperature T is governed by two terms: the change in enthalpy and temperature times the change in entropy. For many exothermic reactions (Ari7° < 0) at room temperature and below, the first term on the right side of the above equation dominates. This means that KP is greater than one, so products are favored over reactants. For an endothermic reaction (Aj//0 > 0), the equilibrium composition will favor products only if ArS° > 0 and the reaction is run at a high temperature. Consider the following reaction in which the A —»■ B step is endothermic A^B As Figure 8.3 shows, the energy levels of A are below those of B, so the conversion of A to B is energetically unfavorable. This is, in fact, the nature of all endothermic reactions. But because the energy levels of B are more closely spaced together, the Boltzmann distribution law (see Equation 2.33) tells us that the population spread of the B molecules over the energy levels is greater than that of the A molecules. Consequently Ar5° > 0 because the entropy of B is greater than that of A. At a sufficiently high temperature, then, the TATS° term will outweigh the AJP term in magnitude and ArG° will be a negative quantity. As an illustration of the relative importance of AfH0 versus TArS°, let's consider the thermal decomposition of limestone or chalk (CaC03): CaC03(s) ^ CaO(s) + C02(g) ATH° = 177.8 kJ mol"1 Using the data in Appendix B we can show that ArS° = 160.5 J Kr1 mol"1. At 298 K, we have ArG° = 177.8 kJ mol"1 - (298 K) (160.5 J Kr1 mol"1)(l kJ/1000 J) = 130.0 kJ mol"1 Because ArG° is a large positive quantity, we conclude that the reaction is not favored for product formation at 298 K. Indeed, the equilibrium pressure of C02 is so low at room temperature that it cannot be measured. In order to make ArG° negative, we first have to find the temperature at which ArG° is zero; that is, 0 = ATH° - TATS° 8.1 Chemical Equilibrium in Gaseous Systems 311 or _ ArH° ArS° (177.8 kJ mor'XlOOO J/l kJ) 160.5 J Kr1 mol"1 = 1108 K or 835°C At a temperature higher than 835°C, ATG° becomes negative, indicating that the reaction now favors the formation of CaO and C02. For example, at 840°C, or 1113 K, Two points are worth making about such a calculation. First, we used the ATH° and AjS0 values at 25°C to calculate changes that occur at a much higher temperature. Because both A^ and A^0 change with temperature, this approach will not give us an accurate value of ArG°, but it is good enough for "ballpark" estimates. Second, we should not be misled into thinking that nothing happens below 835°C and that at 835°C CaC03 suddenly begins to decompose. Far from it. The fact that ArG° is a positive value at some temperature below 835°C does not mean that no C02 is produced, but rather that the pressure of the C02 gas formed at that temperature will be below 1 bar (its standard-state value). The significance of 835°C is that this is the temperature at which the equilibrium pressure of C02 reaches 1 bar. Above 835°C, the equilibrium pressure of C02 exceeds 1 bar. A Comparison of ArG° with ArG Suppose we start a gaseous reaction with all the reactants in their standard states (i.e., all at 1 bar). As soon as the reaction starts, the standard-state condition no longer exists for the reactants or the products because their pressures are different from 1 bar. Under conditions that are not standard state, we must use ArG rather than ArG° to predict the direction of a reaction. We can use the change in standard Gibbs energy (ArG°) and Equation 8.6 to find ArG. The value of ArG is determined by two terms: ArG° and a concentration dependent term. At a given temperature the value of ArG° is fixed, but we can change the value of ArG by adjusting the partial pressures of the gases. Although the quotient composed of the pressures of reactants and products has the form of an equilibrium constant, it is not equal to the equilibrium constant unless PA and PB are the partial pressures at equilibrium. In general, we can rewrite Equation 8.6 as where Q is the reaction quotient and Q ¥= KP unless ArG = 0. The usefulness of Equation 8.6 or Equation 8.9 is that it tells us the direction of a spontaneous change if the concentrations of the reacting species are known. If ArG° is a large positive or a large negative number (say, 50 kJ mol-1 or more), then the direction of the reaction The sign of Afi and not that of Afi° determines the direction of reaction spontaneity. ArG° = Aji/0 - TATS° = 177.8 kJ mol"1 - (1113 K) (160.5 J K"1 mol"1) = -0.8 kJ mol"1 312 Chapter 8: Chemical Equilibrium (or the sign of ArG) is primarily determined by ArG° alone, unless either the reac-tants or the products are present in a much larger amount so that the RT In Q term in Equation 8.9 is comparable to ArG° in magnitude but opposite in sign. If ArG° is a small number, either positive or negative (say, 10 kJ mol"1 or less), then the reaction can go either way* * Alternatively, we can determine the direction of a reaction by comparing Q with K} From Equations 8.7 and 8.9 we can show that ArG = RT \n{QjKP). Therefore, if Q < Kj ArG is negative and the reaction will proceed in the forward direction (left to right). If Q > K} ArG is positive. Here, the reaction will proceed in the reverse direction (right to left). ■ EXAMPLE 8.2 The equilibrium constant (KP) for the reaction N204(g) ^ 2N02(g) is 0.113 at 298 K, which corresponds to a standard Gibbs energy change of 5.40 kJ mol-1. In a certain experiment, the initial pressures are PNC>2 = 0.122 bar and ^n2o4 = 0-453 bar. Calculate ArG for the reaction at these pressures and predict the direction of the net reaction. ANSWER To determine the direction of the net reaction, we need to calculate the Gibbs energy change under nonstandard-state conditions (ArG) using Equation 8.9 and the given ArG° value. Note that the partial pressures are expressed as dimensionless quantities in the reaction quotient Q because each pressure is divided by its standard-state value of 1 bar. ArG = ArG° + RT In Q = ArG° + RTIn -'no. rn204 = 5.40 x 103 J mol"1 + (8.314 J K"1 mol"1) (298 K) x In = 5.40 x 103 J mol"1 - 8.46 x 103 J mol"1 = -3.06 x 103 J mor1 = -3.06 kJ mol"1 (0.122)2 0.453 Because ArG < 0, the net reaction proceeds from left to right to reach equilibrium. COMMENT Note that although ArG° > 0, the reaction can be made initially to favor product formation by having a small concentration (pressure) of the product compared to that of the reactant.