Given the following data:
2C(s) + H2(g) --> C2H2(g)
delta H° = 227.0 kJ
C2H2(g) + 5/2O2(g) --> 2CO2(g) + H2O(l)
delta H° = -1300.0 kJ
CO2(g) --> C(s) + O2(g)
delta H° = 393.5 kJ
Calculate delta H° for the reaction
H2O(l) --> H2(g) + 1/2O2(g)
Please help, I'm stuck :( thank you
Hess' Law?
2016-03-01 11:18 am
回答 (1)
2016-03-01 1:20 pm
✔ 最佳答案
C₂H₂(g) →2C(s) + H₂(g) .... ΔH° = -227.0 kJ2CO₂(g) + H₂O(l) → C₂H₂(g) + (5/2)O₂(g) .... ΔH° = +1300.0 kJ
2C(s) + 2O2(g) → 2CO₂ .... ΔH° = 2×(-393.5) kJ
Add the above three thermochemical equations, and cross out C₂H₂(g), 2C(s), 2CO₂(g) and 2O₂(g) on the both sides. Then,
H₂O(l) → H₂(g) + (1/2)O₂(g) .... ΔH° = -227 + 1300 + 2×(-393.5) = +286 kJ
收錄日期: 2021-04-18 14:33:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160301031825AADudov