CHEM 1001 Prblem Set #3: Entry and Free Energy 19.7 (a) Negative; A liquid (mderate entry) cmbines with a slid t frm anther slid. (b)psitive; One mle f high entry gas frms where n gas was resent befre. (c) Psitive; One mle f high entry gas frms where n gas was resent befre. (d) Uncertain; Same number f mles f gaseus rducts as f gaseus reactants. (e) Negative; Tw mles f gas (and a, slid) cmbine t frm just ne mle f gas. 19.11 Trutn's rule is beyed mst clsely by liquids that d nt have a high degree f rder within the liquid. In bth HF and CH3OH, hydrgen bnds create cnsiderable rder within the liquid. In C6H5CH 3, the nly attractive frces are nn-directinal Lndn frces, which cause the mlecules t attract each ther, but have n referred rientatin as hydrgen bnds d. Thus, f the three chices, liquid C H CH wuld mst clsely fllw Trutn's rule. 6 5 3 19.19 First f all, the rcess is clearly sntaneus, and therefre G < 0. In additin, the gases are mre disrdered when they are at a lwer ressure and therefre S>0. We als cnclude that H = 0, because the gases are ideal and thus there are n frces f attractin r reulsin between them, rducing n energy f interactin. 19.21 (a) An exthermic reactin (ne that gives ff heat) may nt ccur sntaneusly if, at the same time, the system becmes mre rdered, that is, S < 0.This is articularly true at a high temerature, where the T S term dminates the G exressin. An examle f such a rcess is freezing water (clearly exthermic because the reverse rcess, melting ice, is endthermic) at temeratures abve 0 C. (b) A reactin in which S > 0 need nt be sntaneus if that rcess als is endthermic. This is articularly true at lw temeratures, where the H term dminates the G exressin. An examle is the varizatin f water (clearly an endthermic rcess, ne that requires heat, and ne that rduces a gas, s S > 0) at lw temeratures, that is, belw 100 C. 19.25 (a) S = 2S [POCl 3(l)] - 2S [PCl 3(g)] - S [O 2(g)] = 2(222.4 J/K) - 2(311. 7 J/K) - 205.1 J/K = -383.7 J/K 3 3 G = H - T S = -620.2 x 10 J - (298 K)(-383.7 J/K) = -506x10 J = -506 kj (b) The reactin rceeds sntaneusly in the frward directin when reactants and rducts are in their standard states, because the value f G is less than zer.
19.27 We cmbine the reactins in the same way as fr Hess's law calculatins. (a) N2O(g) N 2(g) + ½O 2(g) G = -½(+208.4 kj) = -104.2 kj N 2(g) + 2O 2(g) 2 NO 2(g) G = +102.6 kj Net: N2O(g) + 3/2O 2(g) 2NO 2(g) G = -104.2 + 102.6 = -1.6 kj This reactin reaches an equilibrium cnditin, a cnclusin we reach based n the relatively small abslute value f G. (b) 2N 2(g) + 6H 2(g) 4NH 3(g) G = 2(-33.0 kj) = -66.0 kj 4NH 3(g) + 5O 2(g) 4NO(g) + 6H2O(l) G =-1010.5 kj 4NO(g) 2N 2(g) + 2O 2(g) G =-2(+173.1 kj) = -346.2 kj Net: 6H 2(g) + 3O 2(g) 6H2O(I) G = -66.0 kj -1010.5 kj -346.2 kj = -1422.7 kj This reactin is three times the desired reactin, which therefre has G = -1422.7 kj 3 = -474.3 kj The large negative G value indicates that this reactin wuld tend t g t cmletin at 25 C. (c) 4NH 3(g) + 5O 2(g) 4NO(g) + 6H2O(l) G = -1010.5 kj 4NO(g) 2N 2(g) + 2O 2(g) G = -2(+173.1 kj) = -346.2 kj 2N 2(g) + O 2(g) 2N2O(g) G = +208.4 kj Net: 4NH 3(g) + 4O 2(g) 2N2O(g) + 6H2O(I) G = -1010.5 kj -346.2 kj + 208.4 kj = -1148.3kJ This reactin is twice the desired reactin, which, therefre, has G = -1148.3 kj 2 = -574.2 kj The very large negative value f G fr this reactin indicates that it will g t cmletin.
19.31 (a) S rxn = S rducts - S reactants = {1 ml x 301.2 J K ml + 2 ml x 188.8 J K ml } - {2 ml x 247.4 J K ml + 1 ml x 238.5 J K ml } = -54.5 J K -1 = -0.0545 kj K -1 (b) H rxn = {bnds brken in reactants (kj/ml)} - {bnds brken in rducts(kj/ ml)} = {4 ml x (389 kj ml ) N-H + 4 ml x (222 kj ml ) O-F} - {4 ml x (301 kj ml ) N-F + 4 ml x (464 kj ml ) O-H} = -616 kj -1 (c) G rxn = H rxn - T S rxn = -616 kj - 298 K(-0.0545 kj K ) = -600 kj Since the G rxn is negative, the reactin is sntaneus, and hence feasible (at 25 C). Because bth the entry and enthaly changes are negative, this reactin will be mre highly favred at lw temeratures (i.e. the reactin is enthaly driven). 19.33 In all three cases, K eq = K because nly gases, slids, and liquids are resent in the chemical equatins. There are n factrs fr slids and liquids in K eq exressins, and gases aear as artial ressures in atmsheres. That makes n K eq the same as K fr these three reactins. We nw recall that K = K c(rt). Hence, in these three cases we have: (a) 2SO 2(g) + O 2(g) 2SO 3(g); n gas = 2 - (2+1) = -1; K eq =K =K c(rt) -1 (b) HI(g) ½H 2(g) + ½I 2(g); n gas = 1 - (½+½) = 0; K eq = K = Kc (c) NH4HCO 3(s) NH 3(g) + CO 2(g) + H2O(I); n gas = 2 - (0) = 2; K eq =K =K c(rt) 2 ngas 19.48 (a) We knw that K = K c(rt). Fr the reactin 2 SO 2(g) + O 2(g) 2 SO 3(g), n = 2 - (2 + 1) = -1, and therefre a value f K can be btained. gas -1 2-1 K = K c(rt) = (2.8 X 10 )(0.08206 L atm ml K X 1000 K) = 3.41 = Keq We recgnize that K eq = K since all f the substances invlved in the reactin are gases. We can nw evaluate G. G = -RT In K eq = -(8.3145 J ml K )(1000 K) In (3.41) 4 = -1.02 x 10 J/ml = -10.2 kj/ml
(b) We can evaluate Q fr this situatin and cmare the value with that f K. c c Since Q c is less than K c, the reactin will shift t the right t rduce mre rducts until the reactin reaches equilibrium and these tw values are equal. 19.51 Since K = K fr this reactin, eq G = -RT In K eq = -RT In K -3 11 = -(8.3145 x 10 kj ml K )(298 K)In(6.5x10 ) = -67.4 kj/ml CO(g) + Cl 2(g) COCl 2(g) G = -67.4kJ/ml C(grahite) + ½O 2(g) CO(g) G f = -137.2 kj/ml C(grahite) + ½O (g) + CI (g) COCl (g) G = -204.6 kj/ml 2 2 2 f G f f COCI 2(g) given in Aendix D is -204.6 kj/ml which is in excellent agreement with this calculated value. 19.60 (a) H = H f [CO 2(g)] + H f [H 2(g)] - H f [CO(g)] - H f [H2O(l)] = {-393.5 + 0.00 - (-110.5 l) - (-241.8)} kj/ml = -41.2 kj/ml S = S f [CO 2(g)] + S f [H 2(g)] - S f [CO(g)] - S f [H2O(l)] = {213.6 + 130.6-197.6-188.7} J ml K = -42.1 J ml K G = H - T S -3 = -41.2 kj/ml - (298.15 K)(-42.1 X 10 kj ml K ) = -28.6 kj/ml (b) G = H - T S -3 = -41.2 kj/ml - (875 K)(-42.1 X 10 kj ml K ) = -4.4 kj/ml Since K eq = K fr this reactin, G = -RT In K eq = -RT In K 3 In K = - G /RT = (-4.4 X 10 J/ml)/{(8.3145 J ml K )(875 K)} = 0.60 K = ex(0.60) = 1.8
19.67 (a) We cmute G fr the given reactin in the fllwing manner: H = H f [TiCl 4(l)] + H f [O 2(g)] - H f [TiO 2(s)] - 2 H f [Cl 2(g)] = {-804.2 + 0.00 - (-944. 7) - 2(0.00)} kj/ml = +140.5 kj/ml S = S f [TiCl 4(l)] + S f [O 2(g)] - S f [TiO 2(s)] - 2 S f [Cl 2(g)] = {252.3 + 205.1 - (50.33) - 2(223.1)} J ml K = -39.1 J ml K G = H - T S -3 = +140.5 kj/ml - (298K)(-39.1x10 kj ml K ) = + 140.5 kj/ml + 11.6 kj/ml = + 152.1 kj/ml This reactin is nnsntaneus at 250 C. (We als culd have used values f G t calculate G.) f (b) Fr the cited reactin, G = 2 G f[co 2 (g)]- 2 G f[ CO (g)] - G f [O 2 (g)] G = {2(-394.4) - 2(-137.2) - 0.00 } kj/ml = -514.4 kj/ml Then we cule the tw reactins. TiO 2 (s) + 2Cl 2 (g) TiCl 4 (l) + O 2(g) 2CO(g) + O (g) 2CO (g) 2 2 G = +152.1 kj/ml G = -514.4 kj/ml TiO 2 (s) + 2Cl 2 (g) + 2CO(g) TiCl 4 (l) + 2CO 2(g) G = -362.3 kj/ml The culed reactin has G < 0, and, therefre, is sntaneus.