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热力学第二定律的宗旨是解决热力学过程的自发性问题,其中熵变计算是热力学第二定律应用的关键环节.
本文拟结合平衡态热力学及准静态过程假说基本原理,分别计算指定热力学不可逆过程的Q、W、ΔU、
ΔH、ΔSClo、ΔSAmb及ΔSIso,旨在探寻不可逆过程熵变计算的一般性规律.
本文计算体系分别选取①298.15K、标态下的不可逆化学反应N2(g)+3H2(g)=2NH3(g)、②298.15K、标
态下的不可逆相变H2O(l)=H2O(g)及③298.15K、100kPa下1molN2,反抗50kPa恒外压,恒温不可逆膨胀至
60kPa的热力学过程.
298.15K、100kPa下有关物质的热力学性质参见如下表1.
表1. 298.15K,标态下相关物质的热力学性质[1].
N2(g)+3H2(g)=2NH3(g)的热力学计算
依热力学基本原理可得:
ΔrHθm(298.15K)=∑(νi·ΔfHθm,i(298.15K))
=2ΔfHθm(NH3,g, 298.15K)-3ΔfHθm(H2,g, 298.15K)-ΔfHθm(N2,g, 298.15K)
=2×(-46.11kJ·mol-1)
=-92.22kJ·mol-1 (1)
ΔrGθm(298.15K)=∑(νi·ΔfGθm,i(298.15K))
=2ΔfGθm(NH3,g, 298.15K)-3ΔfGθm(H2,g, 298.15K)-ΔfGθm(N2,g, 298.15K)
=2×(-16.45kJ·mol-1)
=-32.90kJ·mol-1 (2)
ΔrSθm(298.15K)=∑(νi·Sθm,i(298.15K))
=2Sθm(NH3,g, 298.15K)-3Sθm(H2,g, 298.15K)-Sθm(N2,g, 298.15K)
=2×192.45J·mol-1·K-1-3×130.684J·mol-1·K-1-191.61J·mol-1·K-1
=-198.762J·mol-1·K-1 (3)
1.1 平衡态热力学的计算
由式(3)可得:ΔSClo=-198.762J·mol-1·K-1
平衡态热力学认为恒温恒压下化学反应[2],ΔrHθm(298.15K)=Qp, W'(非体积功)=0.
结合式(1)可得:Qp=ΔrHθm(298.15K)=-92.22kJ·mol-1
则:ΔSAmb=-Qp/T=-ΔrHθm(298.15K)/T (4)
将式(1)结果代入式(4)可得:
ΔSAmb=92.22kJ·mol-1/298.15K=309.307J·mol-1·K-1
ΔSIso=ΔSClo+ΔSAmb=-198.762J·mol-1·K-1 +309.307J·mol-1·K-1
=110.545J·mol-1·K-1>0 (5)
式(5)结果证实298.15K及标态下,N2(g)+3H2(g)=2NH3(g)为自发反应.
另恒温条件下,ΔU(298.15K)=ΔrHθm(298.15K)-∑ν(B,g)·RT
=-92.22kJ·mol-1-(-2)×8.314J·mol-1·K-1×298.15K
=-87.262kJ·mol-1
WT(体积功)=ΔU(298.15K)-Qp
=-87.262kJ·mol-1-(-92.22kJ·mol-1)
=4.958kJ·mol-1
1.2 准静态过程假说的计算
依题ΔSClo=-198.762J·mol-1·K-1
准静态过程假说认为恒温恒压下化学反应[3,4],ΔrHθm(298.15K)=Qp+ W',
W'= ΔrGθm(298.15K)
ΔSAmb=(-Qp-W')/T
=[-Qp- ΔrGθm(298.15K)]/T
=-ΔrHθm(298.15K)/T (6)
将式(1)结果代入式(6)可得:
ΔSAmb=92.22kJ·mol-1/298.15K=309.307J·mol-1·K-1
ΔSIso=ΔSClo+ΔSAmb=-198.762J·mol-1·K-1 +309.307J·mol-1·K-1
=110.545J·mol-1·K-1>0 (7)
式(7)结果证实298.15K及标态下,N2(g)+3H2(g)=2NH3(g)为自发反应, 这与1.1结果一致.
另恒温恒压条件下,ΔU(298.15K)=ΔrHθm(298.15K)-∑ν(B,g)·RT
=-92.22kJ·mol-1-(-2)×8.314J·mol-1·K-1×298.15K
=-87.262kJ·mol-1
Qp=ΔrHθm(298.15K)- ΔrGθm(298.15K)
=-92.22kJ·mol-1-(-32.90kJ·mol-1)
=-59.32kJ·mol-1
WT(体积功)=WV(体势变)
=ΔU(298.15K)-Qp-W'
=ΔU(298.15K)-Qp-ΔrGθm(298.15K)
=-87.262kJ·mol-1-(-59.32kJ·mol-1)-(-32.90kJ·mol-1)
=4.958kJ·mol-1
N2(g)+3H2(g)=2NH3(g)的热力学计算结果参见如下表2.
表2. N2(g)+3H2(g)=2NH3(g)的热力学计算结果
热力学性质 | 平衡态热力学 | 准静态过程假说 |
Q(/kJ·mol-1) | -92.22 | -59.32 |
WT(/kJ·mol-1) | 4.958 | 4.958 |
W'(/kJ·mol-1) | 0 | -32.90 |
ΔU(/kJ·mol-1) | -87.262 | -87.262 |
ΔH(/kJ·mol-1) | -92.22 | -92.22 |
ΔSClo(J·mol-1·K-1) | -198.762 | -198.762 |
ΔSAmb(J·mol-1·K-1) | 309.307 | 309.307 |
ΔSIso(J·mol-1·K-1) | 110.545 | 110.545 |
2. H2O(l)=H2O(g)的热力学计算
依热力学基本原理可得:
ΔrHθm(298.15K)=∑(νi·ΔfHθm,i(298.15K))
=ΔfHθm(H2O,g, 298.15K)-ΔfHθm(H2O,l, 298.15K)
=-241.818kJ·mol-1-(-285.830)kJ·mol-1
=44.012kJ·mol-1 (8)
ΔrGθm(298.15K)=∑(νi·ΔfGθm,i(298.15K))
=ΔfGθm(H2O,g, 298.15K)-ΔfGθm(H2O,l, 298.15K)
=-228.572kJ·mol-1-(-237.129)kJ·mol-1
=8.557kJ·mol-1 (9)
ΔrSθm(298.15K)=∑(νi·Sθm,i(298.15K))
=Sθm(H2O,g, 298.15K)-Sθm(H2O,l, 298.15K)
=188.825J·mol-1·K-1-69.91J·mol-1·K-1
=118.915J·mol-1·K-1 (10)
2.1 平衡态热力学的计算
依题ΔSClo=118.915J·mol-1·K-1
同一,平衡态热力学认为恒温恒压下化学反应,ΔrHθm(298.15K)=Qp, W'=0.
则:ΔSAmb=-Qp/T=-ΔrHθm(298.15K)/T (11)
将式(8)结果代入式(11)可得:
ΔSAmb=-44.012kJ·mol-1/298.15K=-147.617J·mol-1·K-1
ΔSIso=ΔSClo+ΔSAmb=118.915J·mol-1·K-1 -147.617J·mol-1·K-1
=-28.699J·mol-1·K-1<0 (12)
式(12)结果证实298.15K及标态下,H2O(l)=H2O(g)为非自发反应.
另恒温恒压条件下,ΔU(298.15K)=ΔrHθm(298.15K)-∑ν(B,g)·RT
=44.012kJ·mol-1 -1×8.314J·mol-1·K-1×298.15K
=41.533kJ·mol-1
Qp=ΔrHθm(298.15K)=44.012kJ·mol-1
WT(体积功)=WV(体势变)
=ΔU(298.15K)-Qp
=41.533kJ·mol-1- 44.012kJ·mol-1
=-2.479kJ·mol-1
2.2 准静态过程假说的计算
依题ΔSClo=118.915J·mol-1·K-1
同一,恒温恒压下化学反应,ΔrHθm(298.15K)=Qp+ W',
W'= ΔrGθm(298.15K)
则:ΔSAmb=(-Qp-W')/T=-ΔrHθm(298.15K)/T (13)
将式(8)结果代入式(13)可得:
ΔSAmb=-44.012kJ·mol-1/298.15K=-147.617J·mol-1·K-1
ΔSIso=ΔSClo+ΔSAmb=118.915J·mol-1·K-1 -147.617J·mol-1·K-1
=-28.699J·mol-1·K-1<0 (14)
式(14)结果证实298.15K及标态下,H2O(l)=H2O(g)为非自发反应, 这与2.1结果一致.
另恒温恒压条件下,ΔU(298.15K)=ΔrHθm(298.15K)-∑ν(B,g)·RT
=44.012kJ·mol-1-1×8.314J·mol-1·K-1×298.15K
=41.533kJ·mol-1
Qp=ΔrHθm(298.15K)- ΔrGθm(298.15K)
=44.012kJ·mol-1-8.557kJ·mol-1
=35.455kJ·mol-1
WT(体积功)=WV(体势变)
=ΔU(298.15K)-Qp-W'
=ΔU(298.15K)-Qp-ΔrGθm(298.15K)
=41.533kJ·mol-1-35.455kJ·mol-1-8.557kJ·mol-1
=-2.479kJ·mol-1
H2O(l)=H2O(g)的热力学计算结果参见如下表3.
表3. H2O(l)=H2O(g)的热力学计算结果
热力学性质 | 平衡态热力学 | 准静态过程假说 |
Q(/kJ·mol-1) | 44.012 | 35.455 |
WT(/kJ·mol-1) | -2.479 | -2.479 |
W'(/kJ·mol-1) | 0 | 8.557 |
ΔU(/kJ·mol-1) | 41.533 | 41.533 |
ΔH(/kJ·mol-1) | 44.012 | 44.012 |
ΔSClo(J·mol-1·K-1) | 118.915 | 118.915 |
ΔSAmb(J·mol-1·K-1) | -147.617 | -147.617 |
ΔSIso(J·mol-1·K-1) | -28.699 | -28.699 |
由表2及3数据可知:平衡态热力学与准静态过程假说最大的分歧点是有效功(W'),平衡态热力学不承认系统自身可以产生有效功. 另恒压条件下体积功(WT)与体势变(WV)相等.
3. 理想气体pVT变化过程的热力学计算
以298.15K、100kPa下1molN2,反抗50kPa恒外压,恒温膨胀至60kPa为体系,参见如下图1.
依题恒温条件下[5]ΔSClo=nR·ln(V2/V1) (15)
将已知条件代入式(15)可得:ΔSClo=1mol×8.314J·mol-1·K-1×ln(41.314dm3/24.788dm3)
=4.247J·mol-1·K-1
3.1 平衡态热力学的计算
理想气体的pVT变化,W'=0.
另依热力学第一定律可得:dU=δQ-pedV (16)
恒温条件下,dU=0.
结合式(16)可得:δQ=pedV (17)
式(17)两边积分可得:Q=pe·ΔV=50kPa×(41.314dm3-24.788dm3)=0.826kJ
则:ΔSAmb=-Qp/T=-0.826kJ/298.15K=-2.771J·mol-1·K-1
ΔSIso=ΔSClo+ΔSAmb=4.247J·mol-1·K-1 -2.771J·mol-1·K-1
=1.476J·mol-1·K-1>0 (18)
式(18)结果证实298.15K及标态下,N2反抗50kPa恒外压进行的恒温膨胀为自发反应.
另对于恒温恒压条件下理想气体的pVT变化,
ΔU(298.15K)=ΔH(298.15K)=0
WT(体积功)=-pe·ΔV=-50kPa×(41.314dm3-24.788dm3)=-0.826kJ
3.2 准静态过程假说热力学的计算
理想气体的pVT变化,W'=0.
另依热力学第一定律可得:dU=δQ-pdV (19)
恒温条件下,dU=0.
结合式(19)可得:δQ=pdV (20)
式(20)两边积分可得:Q=nRT·ln(V2/V1)=nRT·ln(p1/p2)
=1mol×8.314J·mol-1·K-1×298.15K×ln(100kPa/60kPa)
=1.266kJ
则:ΔSAmb=-Qp/T+[∫(p-pe)·dV]/T
=-Qp/T+Qp/T-pe·ΔV/T
=-50kPa×(41.314dm3-24.788dm3)/298.15K
=-2.771J·mol-1·K-1
ΔSIso=ΔSClo+ΔSAmb=4.247J·mol-1·K-1 -2.771J·mol-1·K-1
=1.476J·mol-1·K-1>0 (21)
另对于恒温恒压条件下理想气体的pVT变化,
ΔU(298.15K)=ΔH(298.15K)=0
WT(体积功)=-pe·ΔV=-50kPa×(41.314dm3-24.788dm3)=-0.826kJ
WV(体势变)=∫(-p·dV)
=nRT·ln(V1/V2)=nRT·ln(p2/p1)
=1mol×8.314J·mol-1·K-1×298.15K×ln(60kPa/100kPa)
=-1.266kJ
热力学计算结果参见如下表4.
表4. 理想气体恒温膨胀的热力学计算结果
热力学性质 | 平衡态热力学 | 准静态过程假说 |
Q(/kJ·mol-1) | 0.826 | 1.266 |
WT(/kJ·mol-1) | -0.826 | -0.826 |
WV(/kJ·mol-1) | 不存在 | -1.226 |
W'(/kJ·mol-1) | 0 | 0 |
ΔU(/kJ·mol-1) | 0 | 0 |
ΔH(/kJ·mol-1) | 0 | 0 |
ΔSClo(J·mol-1·K-1) | 4.247 | 4.247 |
ΔSAmb(J·mol-1·K-1) | -2.771 | -2.771 |
ΔSIso(J·mol-1·K-1) | 1.476 | 1.476 |
由表4数据可知:对于理想气体pVT变化,平衡态热力学与准静态过程假说最大的分歧点是体势变(WV),平衡态热力学不承认体势变的存在. 准静态过程假说认为体积功(WT)仅是体势变(WV)的一部分.
4. 结论
⑴准静态过程假说认为所有的热力学过程均为准静态过程,准静态过程包括自发、平衡与非自发过程三大类;
⑵ 准静态过程假说认为系统自身可产生有效功,有效功普遍存在于化学反应与相变之中,仅当化学反应(或相变)建立平衡后,系统才失去提供有效功的能力;
⑶通常情况下,体势变(WV)与体积功(WT)总是成对出现,体积功仅为体势变的一部分;
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,2001,12:45.
[3]余高奇. 热力学第一定律研究.http://blog.sciencenet.cn/u/yugaoqi666. 科学网博客, 2021,8.
[4]余高奇. 热力学第二定律研究.http://blog.sciencenet.cn/u/yugaoqi666. 科学网博客, 2021,8
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