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热力学理论体系中冰(H2O,s)的热力学性质也较重要,本文拟结合热力学基本原理,探讨冰的热力学性质(ΔfGθm、Sθm)的获取.
研究所需热力学数据参见如下表1及表2[1,2].
表1. 25℃、100kPa下H2O的热力学性质
物质 | ΔfHθm(/kJ·mol-1) | ΔfGθm(/kJ·mol-1) | Sθm(/J·mol-1·K-1) | Cp,m(/J·mol-1·K-1) |
H2O(l) | -285.830 | -237.129 | 69.91 | 75.291 |
H2O(g) | -241.818 | -228.572 | 188.825 | 33.577 |
H2O(s) | -292.72 | ? | ? | 37.11 |
表2. 100kPa下H2O(s)的饱和蒸汽压
温度(/℃) | -20 | -15 | -10 | -5 | 0.01 |
冰的饱和蒸汽压(/kPa) | 0.103 | 0.165 | 0.260 | 0.414 | 0.610 |
1. 冰的饱和蒸汽压与ΔfGθm(H2O,S,T)
冰的升华反应参见如下式(1)所示:
H2O(s)=H2O(g) (1)
由基希霍夫公式可得:
ΔrHθm(T)=ΔrHθm(298.15K)+ (2)
且:ΔrHθm(298.15K)=∑(νi·ΔfHθm,i)
=ΔfHθm( H2O,g)- ΔfHθm( H2O,s)
=-241.818kJ·mol-1-(-292.72)kJ·mol-1
=50.902kJ·mol-1
ΔrCp,m=∑(νi·Cp,m,i)
=Cp,m(H2O,g)- Cp,m(H2O,s)
=33.577J·mol-1·K-1-37.11 J·mol-1·K-1
= -3.533J·mol-1·K-1
将 ΔrHθm(298.15K)及ΔrCp,m数值分别代入式(2)可得:
ΔrHθm(T)=50.902kJ·mol-1-3.533J·mol-1·K-1 ×10-3×(T-298.15K) (3)
同理可得:
ΔrSθm(T)=ΔrSθm(298.15K)+
=[188.825J·mol-1·K-1-Sθm(H2O,s,298.15K)]-3.533J·mol-1·K-1 ×ln(T/298.15K) (4)
结合式(3)、(4)可得:
ΔrGθm(T)=ΔrHθm(T)-T·ΔrSθm(T)
=50.902kJ·mol-1-3.533J·mol-1·K-1 ×10-3×(T-298.15K)-T×{[188.825J·mol-1·K-1-Sθm(H2O,s,298.15K)]-3.533J·mol-1·K-1 ×ln(T/298.15K) }×10-3 (5)
2. 冰的Sθm(H2O,S,298.15K)计算
由热力学等温方程可得:ΔrGθm(T)=-RT·ln(p(H2O,g)/pθ) (6)
结合式(6)及表2可得表3:
表3. 100kPa不同温度下H2O(s)蒸发反应的ΔrGθm
温度(/℃) | -20 | -15 | -10 | -5 | 0.01 |
ΔrGθm (/kJ·mol-1) | 14.476 | 13.751 | 13.022 | 12.233 | 11.581 |
将表3结果与式(5)结合可得Sθm(H2O,s,298.15K)的值,参见如下表4.
表4. Sθm(H2O,s,298.15K)数据
温度(/℃) | -20 | -15 | -10 | -5 | 0.01 |
Sθm(298.15K) (/J·mol-1·K-1) | 44.886 | 44.874 | 44.850 | 44.597 | 44.863 |
Sθm(298.15K,平均值) (/J·mol-1·K-1) | 44.814 |
即:Sθm(H2O,S,298.15K)=44.814J·mol-1·K-1
标准偏差 (7)
将相关数据代入式(7)可得[3]:S=0.012J·mol-1·K-1
所以Sθm(H2O,S,298.15K)=44.814±0.012J·mol-1·K-1
3. 冰的ΔfGθm(H2O,S,298.15K)计算
由式(1)可得:
ΔrSθm(298.15K)=Sθm(H2O, g, 298.15K)- Sθm(H2O, s, 298.15K)
=188.825J·mol-1·K-1 - 44.814J·mol-1·K-1
=144.011J·mol-1·K-1
ΔrGθm(298.15K)=ΔrHθm(298.15K)-T·ΔrSθm(298.15K)
=50.902kJ·mol-1-298.15K×144.011J·mol-1·K-1
=7.965kJ·mol-1 (8)
又因为:ΔrGθm(298.15K)=ΔfGθm(H2O, g, 298.15K)-ΔfGθm(H2O, s, 298.15K)
=-228.572kJ·mol-1-ΔfGθm(H2O, s, 298.15K) (9)
结合式(8)、(9)可得:
ΔfGθm(H2O, s, 298.15K)=-228.572kJ·mol-1-7.965kJ·mol-1=-236.537kJ·mol-1
4. 结论
25℃、100kPa下H2O(s)的热力学性质参见如下表5.
表5. 25℃、100kPa下H2O(s)的热力学性质
物质 | ΔfHθm(/kJ·mol-1) | ΔfGθm(/kJ·mol-1) | Sθm(/J·mol-1·K-1) | Cp,m(/J·mol-1·K-1) |
H2O(s) | -292.72 | -236.537 | 44.814 | 37.11 |
参考文献
[1] Lide D R. CRC Handbook of Chemistry and Physics. 89th ed, Chemical Co, 2008,17:2688.
[2]天津大学物理化学教研室编. 物理化学(上册,第四版).北京:高等教育出版社, 2001,12:259.
[3]余高奇,陈阳,李凤莲. 还原半反应的相关热力学计算. 大学化学,2013,28(3):61-67.
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