A) \[163\ kJ\ mo{{l}^{-1}}\]
B) \[2.4\times {{10}^{2}}\ kJ\ mo{{l}^{-1}}\]
C) \[1.63\ kJ\ mo{{l}^{-1}}\]
D) \[2.38\times {{10}^{6}}\ kJ\ mo{{l}^{-1}}\]
Correct Answer: A
Solution :
[a] As we know that, \[\Delta {{G}^{o}}=-2.303RT\,\log \,{{K}_{p}}\] |
Therefore, \[\Delta {{G}^{o}}=-2.303\times (8.314)\times (298)\,\] |
\[(\log \,2.47\times {{10}^{-29}})\] |
\[\Delta {{G}^{o}}=16,3000\,J\,mo{{l}^{-1}}\]\[=163\,KJ\,mo{{l}^{-1}}\] |
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