普通化学（新教材）习题参考答案第一章化学反应的基本规律 (习题P50-52）16解（1）H2O( l ) == H2O(g)/ kJ⋅mol-1-285.83 -241.82∆f HθmSθ/ J⋅mol-1⋅k-1 69.91 188.83m(298k) = [-241.82-(-285.83) ] kJ⋅mol-1 = 44.01 kJ⋅mol-1∆r Hθm∆r Sθ(298k) = (188.83-69.91) J⋅mol-1⋅k-1 = 118.92 J⋅mol-1⋅k-1m( 2 ) ∵是等温等压变化(298k) ⨯ N = 44.01 kJ⋅mol-1⨯ 2mol = 88.02 kJ∴ Q p = ∆r HθmW = -P⋅∆V = -nRT = -2 ⨯ 8.315 J⋅k-1⋅mol-1 ⨯ 298k = -4955.7 J= -4.956 kJ (或-4.96kJ )∴ ∆U = Q p + W = 88.02 kJ - 4.96kJ = 83.06 kJ17解（1）N2 （g）+ 2O2（g）== 2 NO2 (g)/ kJ⋅mol-1 0 0 33.2∆f HθmSθ/ J⋅mol-1⋅k-1 191.6 205.14 240.1m(298k) = 33.2 kJ⋅mol-1 ⨯ 2 = 66.4 kJ⋅mol-1∴ ∆r Hθm∆r Sθ(298k) = ( 240.1 J⋅mol-1⋅k-1 ) ⨯ 2 -(205.14 J⋅mol-1⋅k-1 ) ⨯ 2 - 191.6 J⋅mol-1⋅k-1m= - 121.68 J⋅mol-1⋅k-1O (l) == Fe3O4 (s ) + 4 H2(g)(2) 3 Fe(s) + 4H2∆f Hθ/ kJ⋅mol-1 0 -285.83 -1118.4 0m/ J⋅mol-1⋅k-127.3 69.91 146.4 130.68Sθm∴∆r Hθ(298k) = [-1118.4 - (-285.83 ⨯ 4 ) ] kJ⋅mol-1 = 24.92 kJ⋅mol-1m∆r Sθ(298k) = [(130.68 ⨯ 4 + 146.4 ) - (27.3 ⨯ 3 + 69.91 ⨯ 4 )] J⋅mol-1⋅k-1m18. 解：2Fe2O3 (s) + 3C (s ,石墨) == 4 Fe (s) + 3 CO2 (g)∆f Hθm(298k)/ kJ⋅mol-1- 824.2Sθm(298k)/ J⋅mol-1⋅k-187.4 5.74 27.3∆f Gθm(298k)/ kJ⋅mol-1 -742.2∵∆r Gθm = ∆r Hθm- T •∆r Sθm∴ 301.32 kJ⋅mol-1= 467.87 kJ⋅mol-1 - 298 k•∆r Sθm∴∆r Sθm= 558.89 J⋅mol-1⋅k-1∴∆r Sθm = 3 Sθm( CO2(g) 298k) + 27.3 J⋅mol-1⋅k-1⨯ 4 -87.4 J⋅mol-1⋅k-1⨯ 2 - 5.74 J⋅mol-1⋅k-1⨯ 3∴Sθm( CO2(g) 298k) = 1/3 (558.89 +192.02 - 109.2 ) J⋅mol-1⋅k-1 = 213.90 J⋅mol-1⋅k-1∆f Hθm (298k, C (s ,石墨))=0 ∆f Gθm(298k, C (s ,石墨))=0∆f Hθm （298k, Fe (s)）=0 ∆f Gθm（298k, Fe (s)）=0∆r Hθm =3∆f Hθm(298k, CO2(g) ) -2∆f Hθm(298k, Fe2O3 (s) )⇒467.87 kJ⋅mol-1 =3∆f Hθm(298k, CO2(g) ) -2 ⨯ (- 824.2 kJ⋅mol-1)∴∆f Hθm(298k, CO2(g) ) = 1/3 (467.87-1648.4) kJ⋅mol-1 = -393.51 kJ⋅mol-1同理∆r Gθm =3∆f Gθm(298k, CO2(g) ) -2∆f Gθm(298k, Fe2O3 (s) )⇒301.32 kJ⋅mol-1 = 3∆f Gθm(298k, CO2(g) ) -2 ⨯ (-742.2 kJ⋅mol-1 )∴∆f Gθm(298k, CO2(g) ) = 1/3 (301.32 - 1484.4 ) kJ⋅mol-1 = -394.36 kJ⋅mol-119.解6CO2(g) + 6H2O（l）== C6H12O6 (s) + 6O2（g）∆f Gθm(298k)/ kJ⋅mol-1-394.36 -237.18 902.9 0∴∆r Gθm(298k) = [ 902.9 - (-237.18 ⨯ 6 ) - (-394.36 ⨯ 6 ) ] kJ⋅mol-1 = 4692.14 kJ⋅mol-1 >020．解(1) 4NH3(g) + 5O2(g) == 4NO(g) + 6H2O(l) ∆f Gθm(298k) /kJ⋅mol-1-16.4 0 86.57 -237.18∴∆r Gθm(298k) =[ (-237.18) ⨯6 + 86.57⨯ 4 - (-16.4) ⨯4 ] kJ⋅mol-1 = -1011.2 kJ⋅mol-1<0 ∴此反应能自发进行。

(2) 2SO3(g) == 2SO2(g) + O2(g)∆f Gθm(298k) / kJ⋅mol-1-371.1 -300.19 0∴∆r Gθm(298k) = [(-300.19) ⨯2 - (-371.1) ⨯ 2] kJ⋅mol-1= 141.82 kJ⋅mol-1 > 0 ∴此反应不能自发进行。

21．解（1）MgCO3(s) == MgO(s) + CO2(g)∆f Hθm(298k)/ kJ⋅mol-1 -1111.88 -601.6 -393.51Sθm(298k)/ J⋅mol-1⋅k-165.6 27.0 213.8∆f Gθm(298k) / kJ⋅mol-1-1028.28 -569.3 -394.36∴∆r Hθm(298k) = [ -601.6 + (-393.51) - (-1111.88)] = 116.77 kJ⋅mol-1∆r Sθm(298k) = [ 213.8+ 27.0 - 65.6] = 175.2 J⋅mol-1⋅k-1∆r Gθm(298K) = [ (-394.36) +(-569.3)-(-1028.28)] = 64.62 kJ⋅mol-1(2) ∆r Gθm (1123K) = ∆r Hθm(298k)-T⋅∆r Sθm(298k) = 116.77 kJ⋅mol-1- 1123k ⨯175.2 J⋅mol-1⋅k-1 = 116.77 kJ⋅mol-1-196.75 kJ⋅mol-1 = -79.98 kJ⋅mol-1又∵ RT ln Kθ(1123k)= -∆r Gθm(1123k)∴ 8.315 J⋅mol-1⋅k-1⨯1123 k⋅ln Kθ(1123k) = -(-79.98) kJ⋅mol-1∴ K θ(1123k ) = 5.25 ⨯ 103(3) ∵ 刚刚分解时 ∆r G θm (T) =∆r H θm (298k)-T ⋅ ∆r S θm (298k) =0∴ 分解温度T 可求: k kmol J mol kJ k S k H T m r m r 5.6662.17577.116)298()298(111=⋅⋅⋅=∆∆≈---θθ∴ 分解最低温度为666.5 k22.解法一: K θ (298k)=5.0 ⨯ 1016∴∆r G θm (298k ) = -RT ln K θ(298k )= -8.315 J ⋅mol -1⋅k -1⨯298k ⋅ln(5.0 ⨯ 1016) = -95.26 kJ ⋅mol -1∵∆r G θm (298k) = ∆r H θm (298k)-298k ⋅ ∆r S θm (298k)∴-95.26 kJ ⋅mol -1 = -92.31 kJ ⋅mol -1-298k ⋅∆r S θm (298k)∴∆r S θm (298k) =9.90 J ⋅mol -1⋅k-1 ∴∆r G θm (500k) = ∆r H θm (298k)-500k ⋅ ∆r S θm (298k)= -92.31 kJ ⋅mol -1-500k ⨯9.90 J ⋅mol -1⋅k -1= -97.26 kJ ⋅mol -1而 ∆r G θm (500k) = -RT ln K θ(500k )= -8.315 J ⋅mol -1⋅k -1⨯ 500k ⋅ln K θ(500k )∴ ln K θ(500k )= -RT k G m r )500(θ∆ = )500(315.81026.971113k k mol J mol J ⋅⋅⋅⋅⨯---= 23.40 ∴ K θ(500k ) = 1.45 ⨯ 1010 解法二：∵ ln )298()500(k K k K θθ = ()298(⋅∆-R k H m r θ5001)2981-=(315.8)1031.92(1113⨯⋅⋅⋅⨯-----k mol J mol J k )298500202⨯-= -15.05 ∴ )298()500(k K k K θθ = 2.9 ⨯ 10-7∴ K θ(500k ) =2.9 ⨯ 10-7 ⨯ K θ(298k ) = 2.9 ⨯ 10-7 ⨯ ( 5.0 ⨯ 1016 ) = 1.45 ⨯ 101023.解： N 2(g ) + 3H 2(g ) == 2NH 3(g )∆f H θm (298k)/ kJ ⋅mol -1 0 0 -45.9S θm (298k)/ J ⋅mol -1⋅k -1 191.6 130.68 192.8∴ ∆r H θm (298k) = 2⨯(-45.9) kJ ⋅mol -1 = -91.8 kJ ⋅mol -1S θm (298k) = (2⨯192.8 -191.6 -3⨯130.68 ) J ⋅mol -1⋅k -1= -198.04 J ⋅mol -1⋅k -1∆r G θm (T) = ∆r H θm (298k) -T ⋅ ∆r S θm (298k) =0= -91.8 kJ ⋅mol -1 -T ⋅ (-198.04 J ⋅mol -1⋅k -1 ) =0∴ T = 111304.198108.91---⋅⋅-⋅⨯-k mol J mol J = 463.5 k∴ T>463.5 k 时 反应能自发进行。