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1、 39 七、化学热力学 ( A) 1. U = 87.2 kJmol1 H = U + pV = U + nRT H = 87.2 + (2 4) 8.314 298 10 3 = 92.2 kJmol1 2. H 298 = 393.13 (395.0) = 1.87 kJmol1 3. rH m =12 1H m 2H m 3H m = (27.59)/2 (58.52)/6 (+38.04)/3 = 16.72 kJmol1 4. rH m = 32H m 21H m =3 (483.7) 2 (92.4) = 1266.3 kJmol1 在此温度下,正反应是放热的。 5.( 1)熵增
2、( 2)熵减 ( 3)熵增( 4)熵减( 5)熵增( 6)熵增 6. rG m =12 1G m +52 2G m =12 (142.0)+52 (113.6) = 355 kJmol1 7. rH m = fHm,CO2 + 2fHm,H2O(l) fHm,CH4 = (392.9) + 2(285.5) (74.82) = 889.08 kJmol1 rS m = Sm,CO2 + 2Sm,H2O(l) Sm,CH4 2Sm,O2 = 213.6 + 2 16.75 186.01 2 205.0 = 348.91 Jmol1K1 rG m =rH m TrS m = 889.08 298.
3、15 (348.91)103 = 785.05 kJmol1 8. Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g) fHm,CO = 110.5 kJmol1 fHm,Fe2O3 = 820 kJmol1 rH m = 3fHm,CO fHm,Fe2O3 = 3 (110.5) (820) = 488.5 kJmol1 rS m = 3Sm,CO + 2Sm,Fe 3Sm,C Sm, Fe2O3 = 3 198 + 2 27 3 5.5 180 = 451.5 Jmol1K1 假设在 298KT 区间, rH m 、 rS m 基本 保持不变,令 rG m,T =0 则 :
4、T = rH m/rS m = 488.5 103/451.5 = 1082 K 故在 1 atm 下,当 T 1082 K 时, Fe2O3 能用碳来还原。 9. TiO2 + 2C Ti + 2CO rH m = 2fHm,CO fHm,TiO2= 2 (110.5)(912) = 691 kJmol1 rS m = 2Sm,CO + Sm,Ti 2Sm,C Sm,TiO2 = 2198 + 30 2 5.5 50.5 = 364.5 Jmol1K1 令在 298K T 温度区间内, rH m、 rS m 基本不变, rG m,T = 0 则 : T = rH m /rS m = 691/
5、0.3645 = 1896 K 16 13 40 10. 查表得 : Al2O3 (s) Al (s) CO2 (g) CO (g) C (石墨 ) rH m (kJmol1) 1676 393.51 110.52 S m (Jmol1K1) 50.92 28.33 213.6 197.56 5.74 对于反应 : 2Al2O3(s) + 3C(s) 4Al(s) + 3CO2(g) (1) 1H m = 3fHm,CO22fHm,Al2O3 = 3(393.51)2(1676) = 2171.47 kJmol1 1S m = 3(213.6)+4(28.33)2(50.92)3(5.74)
6、= 635.06 Jmol1K1 1G m = 2171.47298.15635.06103 = 1982.13 kJmol1 对于反应 : Al2O3(s) + 3CO(g) 2Al(s) + 3CO2(g) (2) 2H m = 3fHm,CO2 3fHm,CO fHm,Al2O3 = 3(393.51)3(110.52)(1676) = 827.03 kJmol1 2S m = 3(213.6) + 2(28.33)3(197.56) 50.92 = 53.86 Jmol1K1 2G m = 827.03298.1553.86103 = 810.97 kJmol1 对于焦炭还原 Al2O
7、3 来炼金属铝 : T = 2171.47/0.63506 = 3419 K 需要高出 3419 K 温度,很难实现这样的条件。 ( B) 1. (1) 电极反应 负极: H2(g) + 2OH 2e 2H2O 正极: HgO(s) + H2O + 2e Hg(l) + 2OH 电池反应 : HgO(s) + H2(g) Hg(l) + H2O(l) (2) rG m = nFo = 2965000.9260 = 178.7 kJmol1 H2(g) + 1/2O2(g) H2O(l) rG m,(1) = 237.2 kJmol1 (1) HgO(s) + H2(g) Hg(l) + 2H2
8、O(l) rG m,(2) = 178.7 kJmol1 (2) (2) (1) 得 HgO(s) Hg(l) + 1/2O2(g) (3) rG m,(3) = rG m,(2) rG m,(1) = 178.7(237.2) = 58.5 kJmol1 由 lnKp = (G /RT)得 : lnKp = (58.5103)/(8.314298.15)= 23.5999 Kp = 5.6321011 而 Kp = pO21/2 或 Kp = (pO2/p0)1/2 41 pO2= (Kp )2 = 3.17 1021 (atm) 42 pO2= (Kp )2 p0 = 3.17210211
9、01.3 kPa = 3.211019 kPa 由 rG m = rH mTrS m 得 : rH m = rG m + TrS m =58.5 + 298.1577.1+ (1/2)205.073.2103298.15 = 90.2 kJmol1 (3) HgO(s) 在空气中分解 , pO2 = 0.21101.3 kPa , 由 rG m = RTlnK 得 : lnK1 = (1)1rmHRT + rmSR (1) lnK2 = (2)2rmHRT + rmSR (2) ln(K2 /K1 ) = ln(p2/p1)1/2 = (1/2)ln(p2/p1) = (rH m /R)(1/
10、T11/T2) rH m(1) = rH m(2), rS m(1) =rS m(2) 1/T2 = 1/T1 Rln(p2/p1)/2rH m = 1/298.15 8.314ln(0.21/3.211021)/(290.2103) = 1/298.15 2.013103 = 1.251103, T2 = 799 K 2. (1) 将 8(2)式 + 9(3)式 (1)式得 : 8C(石墨 ) + 9H2(g) C8H18(g) rH m = 82H m + 93H m 1H m = 8(393.5)+9(285.8) (5512.4) = 207.8 kJmol1 rS m = 463.7
11、 85.694 9130.6 = 757.25 Jmol1K1 rG m = rH m TrS m = 207.8298(757.25)103 = 17.861 kJmol1 rG m = RTlnKp Kp = exp(rG m /RT) = 7.40104 (2) Kx = Kp (p/po)n n = 8 Kx = Kp (p/ po) 8 当压力不高时, Kp 为常数, Kx 随总压的增加而增大, 有利于正辛烷生成。或者直接从 n = 8, 增大压力,平衡向缩小体积的方向移动,即有利于正辛烷的生成。 (3) 2ln p rmdK HdT RT, 而 rH m Kp = 1.43 103
12、8 rGm0,即 Ni(OH)2 在干燥的空气中不能自分解为 Ni(s)、 O2(g)和 H2O(l)。 4电池反应为 Ag2O(s) + H2(g) 2Ag(s) + H2O(l) (1) 1 = 1.172 V 818CHx818CHx 818CHx 2Hx 正 负负2O 0.21 atmp 2 12O( ) 0 .2 1 0 .4 6pp 44 化学反应为 H2(g) + (1/2)O2(g) H2O(l) (2) 2G m = 228.6 kJmol1 为标准大气压, 1G m = n1F1 = 2965001.172 = 226.2 kJmol1 为求 rG , 必须根据热力学原理设
13、计如下途径的反应: H2 (g) + 1/2O2 (g) 2 mG H2O (g) (101.3 kPa) G1 rG H2O (g) (3.168 kPa) G2 平衡态 G2 = 0 H2O (l) (101.3 kPa) H2O (l) (3.168 kPa) 由以上反应得: rG = G1 + G2 + G2 + G3 + 2G m = G1 + 2G m = 2G m + nRTln(p2/p1) = 228.6 + 18.314298.2ln(3.168/101.3) = 2.372105 (Jmol1) = 237.2 (kJmol1) 反应 (1)反应 (2),得反应 (3) Ag2O(s) 2Ag(s) + (1/2)O2(g) 3G m = 1G m2G m = 226.2 + 237.2 = 11.0 kJmol1 固体 Ag2O 的 rG m 即为 3G m, 等于 Ag2O(s)的标准生成自由能, rG m = 11.0 kJmol1 Kp = ( /p )1/2 3G m = RTlnKp = RTln ( /p )1/2 ln( /p ) = 43 2 2 1 . 1 0 1 0 8 . 8 7 48 . 3 1 4 2 9 8 . 2mGRT /p