A) \[\Delta G=RT\,\ln \,{{K}_{c}}\]
B) \[-\Delta G=RT\,\ln \,{{K}_{c}}\]
C) \[\Delta {{G}^{o}}=RT\,\ln \,{{K}_{c}}\]
D) \[-\Delta {{G}^{o}}=RT\,\ln \,{{K}_{c}}\]
Correct Answer: D
Solution :
\[\Delta G=\Delta {{G}^{\odot \_}}+RT\,\ln \,K\] In case of equilibrium state, \[K={{K}_{c}}\] and\[\Delta G=0.\] Hence, \[0=\Delta {{G}^{\odot -}}+RT\,\ln \,{{K}_{c}}\] \[-\Delta {{G}^{\Theta }}=RT\,\ln \,{{K}_{c}}\]
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