At what temperature, if any, will a reaction whose ∆Hº = 321.8 kJ/mol and ∆Sº = -296.5 J/mol•K become spontaneous?
回答 (1)
2016-06-20 9:05 pm
For spontaneous reaction : ΔG° < 0
Since ΔG° = ΔH° - TΔS°, then
ΔH° - TΔS° < 0
321.8 - T(-296.5) < 0
296.5T + 321.8 < 0
296.5T < 321.8
T = 321.8/296.5 K
T < 1.09 K
But T must NOT be lower than 0 K.
Hence, the reaction is NOT spontaneous at any temperature.
Since ΔG° = ΔH° - TΔS°, then
ΔH° - TΔS° < 0
321.8 - T(-296.5) < 0
296.5T + 321.8 < 0
296.5T < 321.8
T = 321.8/296.5 K
T < 1.09 K
But T must NOT be lower than 0 K.
Hence, the reaction is NOT spontaneous at any temperature.
收錄日期: 2021-04-18 15:07:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160620123400AAXok0l