CHM 1046. Chapter 17 Homework Solutions.

Problems: 12, 16, 17, 23, 34, 38a,b, 42, 48, 58, 68, 70, 72, 76

12)Entropy increases by a small amount as one goes from solid to liquid, and by a large amount as one goes from liquid to gas.

16)G = H - TS , and G < 0 for spontaneous processes.

a) H < 0 and S > 0.

In this case G < 0 at all temperatures, and so the reaction will always be spontaneous.

b) H > 0 and S < 0.

In this case G > 0 at all temperatures, and so the reaction will never be spontaneous.

c) H < 0 and S < 0.

In this case G < 0 at low temperatures, and G > 0 at high temperatures, and so the reaction will be spontaneous at low temperatures.

d) H > 0 and S > 0.

In this case G > 0 at low temperatures, and G < 0 at high temperatures, and so the reaction will be spontaneous at high temperatures.

17)The third law of thermodynamics states that the absolute entropy of a perfect crystal of a pure substance at absolute zero is exactly equal to 0 J/K. The significance of this is that it makes it possible to assign values for absolute entropy for pure substances at temperatures above absolute zero using experimental data.

23)G is the change in free energy for standard conditions (1 atm pressure for gases and 1 mol/L concentration for solutes) while G is the change in free energy for whatever conditions actually are present.

34)To do this problem recall the following

For Ssyst, when ng > 0 we expect S > 0, when ng = 0 we expect S  0, and when ng < 0 we expect S < 0.

For Ssurr, since Ssurr = - H/T, it follows that when H > 0 we expect Ssurr < 0, and when H < 0 we expect Ssurr > 0.

a)ng = -1 and so is less than zero, therefore we expect Ssyst < 0

Hrxn < 0 and so Ssurr > 0

Reaction will be spontaneous when Ssurr is large enough to make Suniv > 0, which will occur at low temperatures.

b)ng = +1 and so is greater than zero, therefore we expect Ssyst > 0

Hrxn > 0 and so Ssurr < 0

Reaction will be spontaneous when Ssurr is small enough to make Suniv > 0, which will occur at high temperatures.

c)ng = -1 and so is less than zero, therefore we expect Ssyst < 0

Hrxn < 0 and so Ssurr > 0

Reaction will be spontaneous when Ssurr is large enough to make Suniv > 0, which will occur at low temperatures.

d)ng = 0 and so is less than zero, therefore we expect Ssyst 0

Hrxn > 0 and so Ssurr < 0

Since Ssyst 0, SunivSsurr < 0, therefore we expect the reaction will never be spontaneous.

38) Recall Ssyst = Srxn, Ssurr = - Hrxn/T, and Suniv = Ssyst + Ssurr.

a) Ssyst = Srxn = - 157. J/mol.K

Ssurr = - Hrxn/T = - (- 95000. J/mol)/298. K = + 318.8 J/mol.K

Suniv = Ssyst + Ssurr = (- 157. J/mol.K) + (318.8 J/mol.K)

= 162. J/mol.K

b) Ssyst = Srxn = - 157. J/mol.K

Ssurr = - Hrxn/T = - (- 95000. J/mol)/855. K = + 111.1 J/mol.K

Suniv = Ssyst + Ssurr = (- 157. J/mol.K) + (111.1 J/mol.K)

= - 46. J/mol.K

42)Grxn = Hrxn - TSrxn, and so

Grxn = - 1269.8 kJ/mol - (298.2 K) (- 0.3646 kJ/mol) = - 1161.1 kJ/mol.

Since Grxn < 0 the reaction is spontaneous.

48)a) NaNO3(aq). Dissolving a solid in a solvent usually increases entropy.

b) CH3CH3(g). For substances in the same phase, the substance made up of the larger number of atoms usually has the higher value for entropy.

c) Br2(g). For a particular pure substance the gas phase will have a higher value for entropy than the liquid phase.

d) Br2(g). For substances in the same phase and with the same number of atoms, the heavier substance usually has the higher value for entropy. (We did not discuss this in class)

e) PCl5. For substances in the same phase, the substance made up of the larger number of atoms usually has the higher value for entropy.

f) CH3CH2CH2CH3(g). For substances in the same phase, the substance made up of the larger number of atoms usually has the higher value for entropy.

58)Data for this problem are found in Appendix II.

a) Hrxn = [Hf(C2H6(g))] - [2 Hf(CH4(g))]

= [(- 84.68)] - [2 (- 74.6)] = 64.5 kJ/mol

Srxn = [S(C2H6(g)) + S(H2(g))] - [2 S(CH4(g))]

= [(229.2) + (130.7)] - [2 (186.3)] = - 12.7 J/mol.K

Grxn = [Gf(C2H6(g))] - [2 Gf(CH4(g))]

= [(- 32.0)] - [2 (- 50.5)] = 69.0 kJ/mol

Grxn > 0, so reaction is not spontaneous. Hrxn > 0 and Srxn < 0, so the reaction is not spontaneous at any temperature.

b) Hrxn = [Hf(N2H4(g))] - [2 Hf(NH3(g))]

= [(95.4)] - [2 (- 45.9)] = 187.2 kJ/mol

Srxn = [S(N2H4(g)) + S(H2(g))] - [2 S(NH3(g))]

= [(238.5) + (130.7)] - [2 (192.8)] = - 16.4 J/mol.K

Grxn = [Gf(N2H4(g))] - [2 Gf(NH3(g))]

= [(159.4)] - [2 (- 16.4)] = 192.2 kJ/mol

Grxn > 0, so reaction is not spontaneous. Hrxn > 0 and Srxn < 0, so the reaction is not spontaneous at any temperature.

c) Hrxn = [2 Hf(NO(g))]

= [2 (91.3)] = 182.6 kJ/mol

Srxn = [2 S(NO(g)] - [S(N2(g)) + S(O2(g))]

= [2 (210.8)] - [(191.6) + (205.2)] = 24.8 J/mol.K

Grxn = [2 Gf(NO(g))]

= [2 (87.6)] = 175.2 kJ/mol

Grxn > 0, so reaction is not spontaneous. Hrxn > 0 and Srxn > 0, so the reaction will become spontaneous at high temperature.

d) Hrxn = [2 Hf(KCl(s))] - [2 Hf(KClO3(s))]

= [2 (- 436.5)] - [2 (- 397.7)] = - 77.6 kJ/mol

Srxn = [2 S(KCl(s)) + 3 S(O2(g))] - [2 S(KClO3(s))]

= [2 (82.6) + 3 (205.2)] - [2 (143.1)] = 494.6 J/mol.K

Grxn = [2 Gf(KCl(s))] - [2 Gf(KClO3(g))]

= [2 (- 408.5)] - [2 (- 296.3)] = - 224.4 kJ/mol

Grxn < 0, so reaction is spontaneous.

68)Grxn = Grxn + RT ln QQ = (pCOCl2)2 = (0.744)2 = 28.4

(pCO2) (pCCl4) (0.112) (0.174)

Grxn = [2 Gf(COCl2(g))] - [Gf(CO2(g)) + Gf(CCl4(g))]

= [2 (- 204.9)] - [(- 394.4) + (- 62.3)] = 46.9 kJ/mol

So Grxn = (46.9 kJ/mol) + (8.314 x 10-3 kJ/mol.K)(298.2 K) ln(28.4)

= 55.2 kJ/mol

70)ln K = - Grxn/RT

a) Grxn = [Gf(N2O4(g))] - [2 Gf(NO2(g))]

= [(99.8)] - [2 (51.3)] = - 2.8 kJ/mol

So ln K = - (- 2800. J/mol) = 1.13 ; K = e1.13 = 3.1

(8.314 J/mol.K) (298.2 K)

b) Grxn = [2 Gf(BrCl(g))] - [Gf(Br2(g))]

= [2 (- 1.0)] - [(3.1)] = - 5.1 kJ/mol

So ln K = - (- 5100. J/mol) = 2.06 ; K = e2.06 = 7.8

(8.314 J/mol.K) (298.2 K)

72)Grxn = - RT ln K = - (8.314 x 10-3 kJ/mol.K) (298.2 K) ln (81.9) = - 10.9 kJ/mol

Grxn = Grxn + RT ln QQ = (pICl)2

(pI2) (pCl2)

a) For standard conditions Grxn = Grxn = - 10.9 kJ/mol

b) At equilibrium Grxn = 0

c) Grxn = Grxn + RT ln Q Q = (pICl)2 = (2.55)2 = 90.5

(pI2) (pCl2) (0.325) (0.221)

Grxn = (- 10.9 kJ/mol) + (8.314 x 10-3 kJ/mol.K) (298.2 K) ln (90.5)

= 0.3 kJ/mol

76) To find Hrxn and Srxn we need to plot ln K vs 1/T

Temperature (K)Kp1/T (K-1)ln K

170.3.8 x 10-3 0.005882- 5.573

180.0.340.005556- 1.079

190. 18.40.005263 2.912

200.681. 0.005000 6.524

The data are plotted below.


Based on the plot, the best fitting line to the data has

slope = - 13710. Kintercept = 75.06

slope = - Hrxn/R, so Hrxn = - R (slope) = - (8.314 J/mol.K) (- 13710. K)

= 114.0 kJ/mol

intercept = Srxn/R, so Srxn = R (intercept) = (8.314 J/mol.K) (75.06)

= 624. J/mol.K

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