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） H 2O( l ) == H 2O(g)∆f H θm / kJ ⋅mol -1-285.83 -241.82 S θm / J ⋅mol -1⋅k-1 69.91 188.83 ∆r H θm (298k) = [-241.82-(-285.83) ] kJ ⋅mol -1 = 44.01 kJ ⋅mol -1 ∆r S θm (298k) = (188.83-69.91) J ⋅mol -1⋅k-1 = 118.92 J ⋅mol -1⋅k -1( 2 ) ∵是等温等压变化∴ Q p = ∆r H θm (298k) ⨯ N = 44.01 kJ ⋅mol -1⨯ 2mol = 88.02 kJW = -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） N 2 （g ）+ 2O 2 （g ） == 2 NO 2 (g)∆f H θm / kJ ⋅mol -10 0 33.2 S θm / J ⋅mol -1⋅k -1 191.6 205.14 240.1 ∴ ∆r H θm (298k) = 33.2 kJ ⋅mol-1 ⨯ 2 = 66.4 kJ ⋅mol -1 ∆r S θm (298k) = ( 240.1 J ⋅mol -1⋅k -1 ) ⨯ 2 -(205.14 J ⋅mol -1⋅k -1 ) ⨯ 2 - 191.6 J ⋅mol -1⋅k -1= - 121.68 J ⋅mol -1⋅k -1(2) 3 Fe(s) + 4H 2O (l) == Fe 3O 4 (s ) + 4 H 2 (g)∆f H θm / kJ ⋅mol -10 -285.83 -1118.4 0 S θm / J ⋅mol -1⋅k -1 27.3 69.91 146.4 130.68 ∴∆r H θm (298k) = [-1118.4 - (-285.83 ⨯ 4 ) ] kJ ⋅mol -1 = 24.92 kJ ⋅mol -1 ∆r S θm (298k) = [(130.68 ⨯ 4 + 146.4 ) - (27.3 ⨯ 3 + 69.91 ⨯ 4 )] J ⋅mol -1⋅k -118. 解： 2Fe 2O 3 (s) + 3C (s ,石墨) == 4 Fe (s) + 3 CO 2 (g)∆f H θm(298k)/ kJ ⋅mol -1- 824.2 S θm (298k)/ J ⋅mol -1⋅k -1 87.4 5.74 27.3 ∆f G θm(298k)/ kJ ⋅mol -1 -742.2 ∵ ∆r G θm = ∆r H θm - T • ∆r S θm∴ 301.32 kJ ⋅mol -1 = 467.87 kJ ⋅mol -1 - 298 k •∆r S θm∴∆r S θm = 558.89 J ⋅mol -1⋅k-1 ∴ ∆r S θm = 3 S θm ( CO 2(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 ∴S θm ( CO 2(g) 298k) = 1/3 (558.89 +192.02 - 109.2 ) J ⋅mol -1⋅k -1 = 213.90 J ⋅mol -1⋅k -1∆f H θm (298k, C (s ,石墨))=0 ∆f G θm (298k, C (s ,石墨))=0∆f H θm （298k, Fe (s)）=0 ∆f G θm （298k, Fe (s)）=0∆r H θm =3∆f H θm (298k, CO 2(g) ) -2∆f H θm (298k, Fe 2O 3 (s) )⇒ 467.87 kJ ⋅mol -1 =3∆f H θm (298k, CO 2(g) ) -2 ⨯ (- 824.2 kJ ⋅mol -1) ∴ ∆f H θm (298k, CO 2(g) ) = 1/3 (467.87-1648.4) kJ ⋅mol -1 = -393.51 kJ ⋅mol-1 同理 ∆r G θm =3∆f G θm (298k, CO 2(g) ) -2∆f G θm (298k, Fe 2O 3 (s) )⇒ 301.32 kJ ⋅mol -1 = 3∆f G θm (298k, CO 2(g) ) -2 ⨯ (-742.2 kJ ⋅mol -1 ) ∴ ∆f G θm (298k, CO 2(g) ) = 1/3 (301.32 - 1484.4 ) kJ ⋅mol-1 = -394.36 kJ ⋅mol -119.解 6CO 2(g) + 6H 2O （l ） == C 6H 12O 6 (s) + 6O 2（g ）∆f G θm (298k)/ kJ ⋅mol -1 -394.36 -237.18 902.9 0∴ ∆r G θm (298k) = [ 902.9 - (-237.18 ⨯ 6 ) - (-394.36 ⨯ 6 ) ] kJ ⋅mol -1 = 4692.14 kJ ⋅mol-1 >020．解(1) 4NH 3(g ) + 5O 2(g ) == 4NO (g ) + 6H 2O (l )∆f G θm(298k ) /kJ ⋅mol -1-16.4 0 86.57 -237.18 ∴∆r G θm (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) 2SO 3(g ) == 2SO 2(g ) + O 2(g )∆f G θm (298k ) / kJ ⋅mol -1 -371.1 -300.19 0∴∆r G θm (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） MgCO 3(s ) == MgO (s ) + CO 2(g )∆f H θm (298k)/ kJ ⋅mol-1-1111.88 -601.6 -393.51 S θm (298k)/ J ⋅mol -1⋅k -165.6 27.0 213.8 ∆f G θm (298k ) / kJ ⋅mol -1 -1028.28 -569.3 -394.36 ∴ ∆r H θm (298k) = [ -601.6 + (-393.51) - (-1111.88)] = 116.77 kJ ⋅mol -1∆r S θm (298k) = [ 213.8+ 27.0 - 65.6] = 175.2 J ⋅mol -1⋅k -1∆r G θm (298K ) = [ (-394.36) +(-569.3)-(-1028.28)] = 64.62 kJ ⋅mol -1(2) ∆r G θm (1123K ) = ∆r H θm (298k)-T ⋅ ∆r S θm (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 K θ(1123k )= - ∆r G θm (1123k )∴ 8.315 J ⋅mol -1⋅k -1⨯1123 k ⋅ln K θ(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 -10 0 -45.9 S 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。