全国高中化学竞赛教练员习题答案 七 化学热力学高中化学.doc

七、化学热力学 （A） 1. (U = －87.2 kJ·mol1 (H = (U + (pV = (U + (nRT ∴(H = －87.2 + (2－4) × 8.314 × 298 × 10－3kJ·mol(1 2. (H 298 = –393.13 – (–395.0) = 1.87 kJ·mol(1 3. (rH m =(1H m – (2H m – (3H m = (–27.59)/2 – (–58.52)/6 – (+38.04)/3 = －16.72 kJ·mol(1 4. (rH m = 3(2H m – 2(1H m =3 × (–483.7) – 2 × (–92.4) = –1266.3 kJ·mol(1 在此温度下，正反应是放热的。 5.（1）熵增 （2）熵减 （3）熵增（4）熵减（5）熵增（6）熵增 6. (rG m =(1G m +(2G m =×(–142.0)+×(–113.6) = –355 kJ·mol(1 7. (rH m = (fHm,CO2 + 2(fHm,H2O(l) – (fHm,CH4 = (–392.9) + 2×(–285.5) – (–74.82) = –889.08 kJ·mol(1 (rS m = Sm,CO2 + 2Sm,H2O(l) – Sm,CH4 – 2Sm,O2= 213.6 + 2 × 16.75 – 186.01 – 2 × 205.0 = –348.91 J·mol–1·K–1 (rG m =(rH m– T×(rS m = – 889.08 – 298.15 × (–348.91)×10–3 = –785.05 kJ·mol(1 8. Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g) (fHm,CO = –110.5 kJ·mol(1 (fHm,Fe2O3 = –820 kJ·mol(1 ∴(rH m = 3(fHm,CO– (fHm,Fe2O3= 3 × (–110.5) – (–820) = –488.5 kJ·mol(1 (rS m = 3Sm,CO + 2Sm,Fe – 3Sm,C – Sm, Fe2O3 = 3 × 198 + 2 × 27 – 3 × 5.5 – 180 = 451.5 J·mol–1·K–1 假设在298K—T区间，(rH m (rS m基本保持不变，令(rG m,T =0 则: 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。 Ti + 2CO (rH m = 2(fHm,CO– (fHm,TiO2= 2 × (–110.5)–(–912) = 691 kJ·mol(1 (rS m = 2Sm,CO + Sm,Ti – 2Sm,C – Sm,TiO2 = 2×198 + 30 – 2 × 5.5 – 50.5 = 364.5 J·mol–1·K–1 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 = 691/0.3645 = 1896 K 10. 查表得: α–Al2O3 (s) Al (s) CO2 (g) CO (g) C (石墨) (rH m (kJ·mol(1) –1676 — –393.51 –110.52 — S m (J·mol–1·K–1) 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 = 3(fHm,CO2–2(fHm,Al2O3= 3×(–393.51)–2×(–1676) = 2171.47 kJ·mol(1 (1S m = 3×(213.6)+4×(28.33)–2×(50.92)–3×(5.74) = 635.06 J·mol–1·K–1 (1G m = 2171.47–298.15×635.06×10–3 = 1982.13 kJ·mol(1 对于反应: Al2O3(s) + 3CO(g) 2Al(s) + 3CO2(g) (2) (2H m = 3(fHm,CO2 – 3×(fHm,CO － (fHm,Al2O3 = 3×(–393.51)–3×(–110.52)–(–1676) = 827.03 kJ·mol(1 (2S m = 3×(213.6) + 2×(28.33)–3×(197.56)– 50.92 = 53.86 J·mol–1·K–