A) \[\Delta G{}^\circ =RT\,\ln K\]
B) \[K={{e}^{-\Delta G{}^\circ /2303RT}}\]
C) \[\Delta G{}^\circ =-RT{{\log }_{10}}K\]
D) \[K={{10}^{-\Delta G{}^\circ /2.303\,RT}}\]
E) \[\Delta G{}^\circ =R\,\ln \,K\]
Correct Answer: D
Solution :
The relationship between equilibrium constant (K), standard free energy\[(\Delta G{}^\circ )\]and temperature \[(T)\]is given as \[\Delta G{}^\circ =-\,2.303\text{ }RT\,\,\log \,K\] \[-\frac{\Delta G{}^\circ }{2.303RT}=\log K\] \[K={{10}^{-\Delta G{}^\circ /2.303\,RT}}\]You need to login to perform this action.
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