第1章 化学反应中的能量关系
1) 某理想气体对恒定外压(93.1kPa)膨胀,其体积从50L变化至150L,同时吸收6.48kJ的热量,试计算内能的变化。
解:ΔU = Q + W , Q = 6.48 kJ
W = -p外(V2-V1) = -93.31×103Pa×(150-50)×10-3m3 = -9.331kJ
ΔU = (6.48-9.331)kJ =-2.851kJ
∴ 内能的变化为-2.851kJ
2) 已知下列热化学方程式:
Fe2O3(s) + 3CO(g) = 2Fe(s) + 3CO2(g), Qp=-27.6kJ•mol-1
Fe2O3(s) + CO(g) = 2Fe3O4(s) + CO2(g), Qp=-58.6kJ•mol-1
Fe3O4(s) + CO(g) = 3FeO(s) + CO2(g), Qp=38.1kJ•mol-1
不用查表,计算下列反应的Qp:FeO(s) + CO(g) = Fe(s) + CO2(g)
(提示:根据盖斯定律,利用已知反应方程式,设计一循环,消去Fe2O3和Fe3O4,而得到所需反应方程式。)
解:若以①、②、③分别表示所给的化学方程式,
根据盖斯定律,设计一循环:6c2e3e2e98abd1fd9a66519db9da8d90.png[3×①-②-2×③]
可得下列方程:FeO(s) + CO(g) =Fe(s) + CO2(g)
∴ 该反应的Qp = 6c2e3e2e98abd1fd9a66519db9da8d90.png×[3×(-27.6)-(-58.6)-2×38.1]kJ•mol-1 =-16.7 kJ•mol-1
3) 试用书末附录1提供的标准摩尔生成焓6c39f3cc18ad1ff8502118d1a2e03ba9.png数据,计算下列反应的2f95b825d2f5daab93c2c1100da7c7e4.png。
① CaO(s) + SO3(g) + 2H2O(l) = CaSO4•2H2O(s)
② C6H6(l) + 793b05c90d14a117ba52da1d743a43ab1.pngO2(g) = 6 CO2(g) + 3H2O(l)
③ 2Al(s) + Fe2O3(s) = 2Fe(s) + Al2O3(s)
解: ① CaO(s) + SO3(g) + 2H2O(l) = CaSO4•2H2O(s)
6c39f3cc18ad1ff8502118d1a2e03ba9.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -635.5 -395.26 -285.85 -2021.12
word/media/image6_1.png= 6c39f3cc18ad1ff8502118d1a2e03ba9.png( CaSO4•2H2O,s) -6c39f3cc18ad1ff8502118d1a2e03ba9.png(CaO,s)-6c39f3cc18ad1ff8502118d1a2e03ba9.png(SO3,g) -26c39f3cc18ad1ff8502118d1a2e03ba9.png(H2O,l)
= [(-2021.12) -(-635.5)-(-395.26) -2×(-285.85)] kJ mol56cb2620366f4925e40545641d7b6f65.png=-418.66 kJ mol56cb2620366f4925e40545641d7b6f65.png
② C6H6(l) + 793b05c90d14a117ba52da1d743a43ab1.pngO2(g) = 6 CO2(g) + 3H2O(l)
6c39f3cc18ad1ff8502118d1a2e03ba9.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png 49.04 0 -393.51 -285.85
word/media/image6_1.png= 66c39f3cc18ad1ff8502118d1a2e03ba9.png(CO2,g) +36c39f3cc18ad1ff8502118d1a2e03ba9.png(H2O,l) -6c39f3cc18ad1ff8502118d1a2e03ba9.png(C6H6,l)
= [6×(-393.51) +3×(-285.85)-49.04] kJ mol56cb2620366f4925e40545641d7b6f65.png=-3267.65 kJ mol56cb2620366f4925e40545641d7b6f65.png
③ 2Al(s) + Fe2O3(s) = 2Fe(s) + Al2O3(s)
6c39f3cc18ad1ff8502118d1a2e03ba9.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png 0 -822.2 0 -1669.79
word/media/image6_1.png=6c39f3cc18ad1ff8502118d1a2e03ba9.png( Al2O3,s) -6c39f3cc18ad1ff8502118d1a2e03ba9.png( Fe2O3,s)
= [(-1669.79)-(-822.2)] kJ mol56cb2620366f4925e40545641d7b6f65.png=-847.59 kJ mol56cb2620366f4925e40545641d7b6f65.png
4) 判断下列反应或过程中熵变的数值是正值还是负值。
① 2C(g) + O2(g) = 2CO(g)
② 2NO2(g) = 2NO(g) + O2(g)
③ Br2(l) = Br2(g)
解: ① 气体分子数增大的反应,熵变增大
② 气体分子数增大的反应,熵变增大
③ 同一物质气态的熵值大于液态,该反应熵变增大
5)利用书末附录1所提供的bad052accc19031e4375c058fe4cc821.png数据计算下列反应的d1b7b58feacb147d686a98f6b5703da2.png,并判断这些反应能否自发进行。
① SiO2(s,石英) + 4 HCl(g) = SiCl4 (g) + 2H2O(g)
② CO(g) + H2O(g) = CO2(g) + H2(g)
③ Fe2O3(s) + 3CO(g) = 2Fe(s) + 3CO2(g)
解: ① SiO2(s,石英) + 4 HCl(g) = SiCl4 (g) + 2H2O(g)
bad052accc19031e4375c058fe4cc821.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -805.0 -95.27 -569.9 -228.59
f3fba57dee020a8c598dbf934a48a00b.png=bad052accc19031e4375c058fe4cc821.png( SiCl4,g) +2×bad052accc19031e4375c058fe4cc821.png(H2O,g)-bad052accc19031e4375c058fe4cc821.png(SiO2,石英)-4×bad052accc19031e4375c058fe4cc821.png(HCl,g)
= [(-569.9)+2×(-228.59)-(-805.0)-4×(-95.27)] kJ mol56cb2620366f4925e40545641d7b6f65.png= 159.0 kJ mol56cb2620366f4925e40545641d7b6f65.png> 0
∴ 各物质在标准状态下,反应不能正向进行。
② CO(g) + H2O(g) = CO2(g) + H2(g)
bad052accc19031e4375c058fe4cc821.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -137.30 -228.59 -394.38 0
f3fba57dee020a8c598dbf934a48a00b.png=bad052accc19031e4375c058fe4cc821.png(CO2,g)-bad052accc19031e4375c058fe4cc821.png(CO,g)-bad052accc19031e4375c058fe4cc821.png(H2O,g)
= [(-394.38)-(-137.30)-(-228.59)] kJ mol56cb2620366f4925e40545641d7b6f65.png=-28.49 kJ mol56cb2620366f4925e40545641d7b6f65.png< 0
∴ 各物质在标准状态下,反应可以正向进行。
③ Fe2O3(s) + 3CO(g) = 2Fe(s) + 3CO2(g)
bad052accc19031e4375c058fe4cc821.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -741.0 -137.30 0 -394.38
f3fba57dee020a8c598dbf934a48a00b.png= 3×bad052accc19031e4375c058fe4cc821.png(CO2,g)-bad052accc19031e4375c058fe4cc821.png(Fe2O3,s)-3×bad052accc19031e4375c058fe4cc821.png(CO,g)
= [3×(-394.4)-(-741.0)-3×(-137.30)] kJ mol56cb2620366f4925e40545641d7b6f65.png=-30.24 kJ mol56cb2620366f4925e40545641d7b6f65.png< 0
∴ 各物质在标准状态下,反应可以正向进行。
6) 利用书末附录1所提供的6c39f3cc18ad1ff8502118d1a2e03ba9.png和0949ebe07118227405eb6bf10479d7c6.png数据,计算下列反应在298K时的d1b7b58feacb147d686a98f6b5703da2.png。
① N2(g) + 3H2(g) = 2NH3(g)
② 2HgO(s) = 2Hg(l) + O2(g)
③ CH4(g) + 2O2(g) = CO2(g) + 2H2O(l)
解:① word/media/image6_1.png=2×6c39f3cc18ad1ff8502118d1a2e03ba9.png(NH3,g) = [2×(-45.96)] kJ mol56cb2620366f4925e40545641d7b6f65.png=-91.92 kJ mol56cb2620366f4925e40545641d7b6f65.png
3867c7ca5a73e92a87ef0260068b66ec.png= (2×192.70-3×130.70-191.60) J K56cb2620366f4925e40545641d7b6f65.png mol56cb2620366f4925e40545641d7b6f65.png=-198.30 J K56cb2620366f4925e40545641d7b6f65.png mol56cb2620366f4925e40545641d7b6f65.png
word/media/image13_1.png=2f95b825d2f5daab93c2c1100da7c7e4.png-T 3867c7ca5a73e92a87ef0260068b66ec.png= [-91.92-298×(-198.30)×10word/media/image14_1.png] kJ mol56cb2620366f4925e40545641d7b6f65.png
=-32.83 kJ mol56cb2620366f4925e40545641d7b6f65.png
② word/media/image6_1.png=-26c39f3cc18ad1ff8502118d1a2e03ba9.png(HgO,s) = -2×(-90.71) kJ mol56cb2620366f4925e40545641d7b6f65.png=181.42 kJ mol56cb2620366f4925e40545641d7b6f65.png
3867c7ca5a73e92a87ef0260068b66ec.png= [(205.14 + 2×77.4-2×72.0)×10word/media/image14_1.png] kJ mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png= 0.216 kJ mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png
word/media/image13_1.png=2f95b825d2f5daab93c2c1100da7c7e4.png-T 3867c7ca5a73e92a87ef0260068b66ec.png= (181.42-298× 0.216) kJ mol56cb2620366f4925e40545641d7b6f65.png= 117.05 kJ mol56cb2620366f4925e40545641d7b6f65.png
③ word/media/image6_1.png= 26c39f3cc18ad1ff8502118d1a2e03ba9.png(H2O,l) +6c39f3cc18ad1ff8502118d1a2e03ba9.png(CO2,g)-6c39f3cc18ad1ff8502118d1a2e03ba9.png(CH4,g)
= [2×(-285.85)+(-393.51)-(-74.85)] kJ mol56cb2620366f4925e40545641d7b6f65.png=-890.36 kJ mol56cb2620366f4925e40545641d7b6f65.png
3867c7ca5a73e92a87ef0260068b66ec.png= [2× 69.96+213.79-2×205.14-186.38]×10word/media/image14_1.pngkJ mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png
=-0.243kJ mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png
word/media/image13_1.png=2f95b825d2f5daab93c2c1100da7c7e4.png-T×3867c7ca5a73e92a87ef0260068b66ec.png= [-890.36-298×(-0.243)] kJ mol56cb2620366f4925e40545641d7b6f65.png=-817.95 kJ mol56cb2620366f4925e40545641d7b6f65.png
7) 用二氧化锰制取金属锰可采取下列两种方法:
① MnO2(s) + 2H2(g) = Mn(s) + 2H2O(g)
② MnO2(s) + 2C(s) = Mn(s) + 2CO(g)
上述两个反应在25℃、100kPa下能否自发进行?如果希望反应温度尽可能低一些,试通过计算,说明采用何种方法比较好?
解: ① MnO2(s) + 2H2(g) = Mn(s) + 2H2O(g)
6c39f3cc18ad1ff8502118d1a2e03ba9.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -520.9 0 0 -241.84
0949ebe07118227405eb6bf10479d7c6.png/ J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png 53.1 130.70 31.76 188.85
bad052accc19031e4375c058fe4cc821.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -466.1 0 0 -228.59
47f9cc8690477a16267f17368fc20c1a.png= 2×6c39f3cc18ad1ff8502118d1a2e03ba9.png(H2O, g) + 6c39f3cc18ad1ff8502118d1a2e03ba9.png(Mn,s) -6c39f3cc18ad1ff8502118d1a2e03ba9.png(MnO2,s) -2×6c39f3cc18ad1ff8502118d1a2e03ba9.png(H2,g)
=[2×(-241.84) -(-520.9) ] kJ mol56cb2620366f4925e40545641d7b6f65.png= 37.22 kJ mol56cb2620366f4925e40545641d7b6f65.png
2195fad2ada99a9cc6a0d27fca4b0d06.png= 2×0949ebe07118227405eb6bf10479d7c6.png(H2O, g) + 0949ebe07118227405eb6bf10479d7c6.png(Mn,s) -0949ebe07118227405eb6bf10479d7c6.png(MnO2,s) -2×0949ebe07118227405eb6bf10479d7c6.png(H2,g)
=[2×188.85+31.76-53.1-2 ×130.70] J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png =94.96 J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png
4fd03b1bbee11fc51b29ba80306464a3.png=47f9cc8690477a16267f17368fc20c1a.png-T 2195fad2ada99a9cc6a0d27fca4b0d06.png= (37.22-298× 94.96×10fe931e0335f1fb013033039e2b5337c7.png)kJ mol56cb2620366f4925e40545641d7b6f65.png = 8.92 kJ mol56cb2620366f4925e40545641d7b6f65.png> 0
或者4fd03b1bbee11fc51b29ba80306464a3.png= 2×bad052accc19031e4375c058fe4cc821.png(H2O, g) + bad052accc19031e4375c058fe4cc821.png(Mn,s) -bad052accc19031e4375c058fe4cc821.png(MnO2,s) -2bad052accc19031e4375c058fe4cc821.png(H2,g)
=[2×(-228.59) -(-466.1)] kJ mol56cb2620366f4925e40545641d7b6f65.png= 8.92 kJ mol56cb2620366f4925e40545641d7b6f65.png> 0
② MnO2(s) + 2C(s) = Mn(s) + 2CO(g)
6c39f3cc18ad1ff8502118d1a2e03ba9.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -520.9 0 0 -110.54
0949ebe07118227405eb6bf10479d7c6.png/ J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png 53.1 5.69 31.76 198.01
bad052accc19031e4375c058fe4cc821.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -466.1 0 0 -137.30
0f9e268f86ca19e0012853f492705673.png = 2×6c39f3cc18ad1ff8502118d1a2e03ba9.png(CO, g) + 6c39f3cc18ad1ff8502118d1a2e03ba9.png(Mn,s) -6c39f3cc18ad1ff8502118d1a2e03ba9.png(MnO2,s) -2×6c39f3cc18ad1ff8502118d1a2e03ba9.png(C,g)
= [2×(-110.54) -(-520.9)] kJ mol56cb2620366f4925e40545641d7b6f65.png = 299.82 kJ mol56cb2620366f4925e40545641d7b6f65.png
faa83e607e00b77244ba1eaefa21dc97.png= 2×0949ebe07118227405eb6bf10479d7c6.png( CO, g) + 0949ebe07118227405eb6bf10479d7c6.png(Mn,s) -0949ebe07118227405eb6bf10479d7c6.png(MnO2,s) -2×0949ebe07118227405eb6bf10479d7c6.png( C,g)
=[2×198.01 + 31.76-53.1-2 ×5.69] J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png=363.3 J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png
8d569baf6fccb1205ba248541620f44b.png=0f9e268f86ca19e0012853f492705673.png-T faa83e607e00b77244ba1eaefa21dc97.png= (299.82-298×363.3×10fe931e0335f1fb013033039e2b5337c7.png) kJ mol56cb2620366f4925e40545641d7b6f65.png= 191.56 kJ mol56cb2620366f4925e40545641d7b6f65.png> 0
或者 8d569baf6fccb1205ba248541620f44b.png= 2×bad052accc19031e4375c058fe4cc821.png(CO, g) + bad052accc19031e4375c058fe4cc821.png(Mn,s) -bad052accc19031e4375c058fe4cc821.png(MnO2,s) -2bad052accc19031e4375c058fe4cc821.png(C,g)
=[2×(-137.30) -(-466.1)] kJ mol56cb2620366f4925e40545641d7b6f65.png= 191.5 kJ mol56cb2620366f4925e40545641d7b6f65.png> 0
∴ 以上两反应在25℃和100kPa下均不能正向进行。
要使反应能够正向进行,则必须2f95b825d2f5daab93c2c1100da7c7e4.png-T 3867c7ca5a73e92a87ef0260068b66ec.png< 0,即T >af830b96faec4dd8879c33062e7fd1ce.png
∴ T1 >435120b533ed77087c54930b2bc0e520.pngK= 391.95 K
T2 >28ffd0103fa90cc4bacd1ecfb12a19f7.pngK= 825.3 K
故从温度考虑,第一种方法较好。
8) 反应 C2H4(g)+ H2(g)= C2H6(g) 2f95b825d2f5daab93c2c1100da7c7e4.png= —136.98kJ•mol-1,3867c7ca5a73e92a87ef0260068b66ec.png= —120.66J•mol-1•K-1, 试求该反应在25℃时的d1b7b58feacb147d686a98f6b5703da2.png,并指出该反应在25℃、100kPa下能否自发进行。
解:word/media/image13_1.png=2f95b825d2f5daab93c2c1100da7c7e4.png-T×3867c7ca5a73e92a87ef0260068b66ec.png= [-136.98-298×(-120.66) ×10fe931e0335f1fb013033039e2b5337c7.png] kJ mol56cb2620366f4925e40545641d7b6f65.png
=-101.02 kJ mol56cb2620366f4925e40545641d7b6f65.png<0
∴ 该反应在25℃、100kPa下能自发进行。
9) 25℃、100kPa下,CaSO4(s) CaO(s) + SO3(g),已知:2f95b825d2f5daab93c2c1100da7c7e4.png=401.92kJ•mol-1,3867c7ca5a73e92a87ef0260068b66ec.png=189.13J •mol-1•K-1,问:
① 上述反应能否自发进行?
② 对上述反应,是升高温度有利,还是降低温度有利?
③ 若使上述反应正向进行,其所需的最低温度是多少?
解:① word/media/image13_1.png=2f95b825d2f5daab93c2c1100da7c7e4.png-T×3867c7ca5a73e92a87ef0260068b66ec.png= [401.92-298×189.13 ×10fe931e0335f1fb013033039e2b5337c7.png] kJ mol56cb2620366f4925e40545641d7b6f65.png
=345.56 kJ mol56cb2620366f4925e40545641d7b6f65.png>0 ∴ 该反应在25℃、100kPa下不能自发进行。
② ∵ 2f95b825d2f5daab93c2c1100da7c7e4.png > 0, 3867c7ca5a73e92a87ef0260068b66ec.png>0 , 若使上述反应正向进行,即word/media/image13_1.png=2f95b825d2f5daab93c2c1100da7c7e4.png-T×3867c7ca5a73e92a87ef0260068b66ec.png<0
∴ 升高温度对反应有利。
③ 若使上述反应正向进行,则word/media/image13_1.png<0,即2f95b825d2f5daab93c2c1100da7c7e4.png-T×3867c7ca5a73e92a87ef0260068b66ec.png<0
∴ T >af830b96faec4dd8879c33062e7fd1ce.png= 5a79ecd220f0e0dbe273963ce8beb877.pngK=2125 K
10) ① 利用附录1所提供的数据,计算下列反应在298K时的2f95b825d2f5daab93c2c1100da7c7e4.png和d1b7b58feacb147d686a98f6b5703da2.png。
CuS(s) + H2(g) Cu(s) + H2S(g)
② 求该反应在1000K时的d1b7b58feacb147d686a98f6b5703da2.png。
解:① CuS(s) + H2(g) Cu(s) + H2S(g)
6c39f3cc18ad1ff8502118d1a2e03ba9.png/ kJ mol -48.5 0 0 -20.17
bad052accc19031e4375c058fe4cc821.png/ kJ mol56cb2620366f4925e40545641d7b6f65.png -48.9 0 0 -33.05
0949ebe07118227405eb6bf10479d7c6.png/ J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png 66.5 130.70 33.30 205.88
2f95b825d2f5daab93c2c1100da7c7e4.png=6c39f3cc18ad1ff8502118d1a2e03ba9.png(H2S,g) +6c39f3cc18ad1ff8502118d1a2e03ba9.png(Cu,s)-6c39f3cc18ad1ff8502118d1a2e03ba9.png(CuS,s)-6c39f3cc18ad1ff8502118d1a2e03ba9.png(H2,g)
= [(-20.17)-(-48.5)] kJ mol56cb2620366f4925e40545641d7b6f65.png = 28.33 kJ mol56cb2620366f4925e40545641d7b6f65.png
d1b7b58feacb147d686a98f6b5703da2.png=bad052accc19031e4375c058fe4cc821.png(H2S,g) +bad052accc19031e4375c058fe4cc821.png(Cu,s)-bad052accc19031e4375c058fe4cc821.png(CuS,s)-bad052accc19031e4375c058fe4cc821.png(H2,g)
= [(-33.05)-(-48.9)] kJ mol56cb2620366f4925e40545641d7b6f65.png = 15.85 kJ mol56cb2620366f4925e40545641d7b6f65.png
② 3867c7ca5a73e92a87ef0260068b66ec.png= 0949ebe07118227405eb6bf10479d7c6.png(H2S,g) +0949ebe07118227405eb6bf10479d7c6.png(Cu,s)-0949ebe07118227405eb6bf10479d7c6.png(CuS,s)-0949ebe07118227405eb6bf10479d7c6.png(H2,g)
= [205.88+33.30-66.5-130.70]J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png=41.98 J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png
或者 3867c7ca5a73e92a87ef0260068b66ec.png(298K) = 30be68584bfebaa809f173a303ab98d5.png=1ab89355fcba004a75185b22a194e680.png = 41.88 J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png
1000 K时 d1b7b58feacb147d686a98f6b5703da2.png(1000K) =2f95b825d2f5daab93c2c1100da7c7e4.png(298) -T3867c7ca5a73e92a87ef0260068b66ec.png
= [28.33-1000×41.98×10fe931e0335f1fb013033039e2b5337c7.png] kJ mol56cb2620366f4925e40545641d7b6f65.png=-13.65 kJ mol56cb2620366f4925e40545641d7b6f65.png
11) 金属镍在一定条件下可以与CO生成Ni(CO)4 (四羰基镍) Ni(s) + 4CO(g) Ni(CO)4(g)。据此可进行镍的提纯。过程是:在一定温度下以CO通过粗镍,生成Ni(CO)4,在另一温度下Ni(CO)4分解为纯镍及CO。试计算在100kPa,Ni(CO)4的生成和分解温度。
(已知:Ni(CO)4(g)的6c39f3cc18ad1ff8502118d1a2e03ba9.png=-602.3 kJ•mol-1,0949ebe07118227405eb6bf10479d7c6.png= 401.7 J•mol-1•K-1。Ni(s)的0949ebe07118227405eb6bf10479d7c6.png= 29.9 J•mol-1•K-1,其他有关数据可查表)
解: Ni(s) + 4CO(g) Ni(CO)4(g)
6c39f3cc18ad1ff8502118d1a2e03ba9.png/ kJ mol 0 -110.54 -602.3
0949ebe07118227405eb6bf10479d7c6.png/ J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png 29.9 198.01 401.7
2f95b825d2f5daab93c2c1100da7c7e4.png=6c39f3cc18ad1ff8502118d1a2e03ba9.png(Ni(CO)4,g)-6c39f3cc18ad1ff8502118d1a2e03ba9.png(Ni,s)-4×6c39f3cc18ad1ff8502118d1a2e03ba9.png(CO,g)
=[-602.3-4×(-110.54)] kJ mol56cb2620366f4925e40545641d7b6f65.png=-160.14 kJ mol56cb2620366f4925e40545641d7b6f65.png
3867c7ca5a73e92a87ef0260068b66ec.png=0949ebe07118227405eb6bf10479d7c6.png(Ni(CO)4,g)-0949ebe07118227405eb6bf10479d7c6.png(Ni,s)-0949ebe07118227405eb6bf10479d7c6.png(CO,g)
= [401.7-29.9-4×198.01] J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png=-420.24 J mol56cb2620366f4925e40545641d7b6f65.png K56cb2620366f4925e40545641d7b6f65.png
计算在100kPa,Ni(CO)4的生成温度,即上述反应正向进行:word/media/image13_1.png<0, T >af830b96faec4dd8879c33062e7fd1ce.png
T > 2af017e1b51c72b64eec5571687a3995.png ∴ T <381 K
计算在100kPa,Ni(CO)4的分解温度,即上述反应正向进行:word/media/image13_1.png> 0, T <af830b96faec4dd8879c33062e7fd1ce.png
T < 2af017e1b51c72b64eec5571687a3995.png ∴ T > 381 K
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