For a particular reaction at 236.9 degrees Celsius, ΔG = 251.24 kJ/mol and ΔS=210.67 J/(mol*K)
Calculate ΔG for this reaction at -1.0 degrees Celsius.
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2017-11-22 10:37 am
回答 (1)
2017-11-22 4:59 pm
Under constant pressure and temperature: ΔG = ΔH - TΔS
where ΔH and ΔS are assumed to be constant.
When T = 236.9 °C = (273.2 + 236.9) K :
ΔG = ΔH - TΔS
(251.24 J/mol) = ΔH - (510.1 K) × [210.67/1000 kJ/(mol K)]
ΔH = 358.70 kJ/mol
When T = -1.0 °C = (273.2 - 1) K = 272.2 K :
ΔG = ΔH - TΔS
ΔG = (358.70 kJ/mol) - (272.2 K) × [210.67/1000 kJ/(mol K)]
ΔG = 301 kJ/mol
ΔG at -1 °C = 301 kJ/mol
where ΔH and ΔS are assumed to be constant.
When T = 236.9 °C = (273.2 + 236.9) K :
ΔG = ΔH - TΔS
(251.24 J/mol) = ΔH - (510.1 K) × [210.67/1000 kJ/(mol K)]
ΔH = 358.70 kJ/mol
When T = -1.0 °C = (273.2 - 1) K = 272.2 K :
ΔG = ΔH - TΔS
ΔG = (358.70 kJ/mol) - (272.2 K) × [210.67/1000 kJ/(mol K)]
ΔG = 301 kJ/mol
ΔG at -1 °C = 301 kJ/mol
收錄日期: 2021-04-18 17:57:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171122023746AAyoIQY