A) \[-80kJ\,mo{{l}^{-1}}\]
B) \[100kJ\,mo{{l}^{-1}}\]
C) \[-100kJ\,mo{{l}^{-1}}\]
D) \[80kJ\,mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[2{{H}_{2}}O\rightleftharpoons {{H}_{3}}{{O}^{+}}+O{{H}^{-}}\] \[{{K}_{w}}={{10}^{-14}}\] \[\Delta {{G}^{o}}=-2.303RT\log \,K\] \[=-2.303\,RT\log {{10}^{-14}}\] \[=+2.303\times 8.314\times 298\times 14=80kJ\,mo{{l}^{-1}}\]
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