A) \[K={{e}^{-\Delta G/RT}}\]
B) \[K={{e}^{-\Delta {{G}^{o}}/RT}}\]
C) \[K={{e}^{-\Delta H/RT}}\]
D) \[K={{e}^{-\Delta {{H }^{o}}/RT}}\]
Correct Answer: B
Solution :
Gibbs free energy \[(\Delta {{G}^{o}})\] is related with equilibrium constant \[(K)\]as \[\Delta {{G}^{o}}=-RT\ln K\] \[-\frac{\Delta {{G}^{o}}}{RT}=\ln \,K\] \[\therefore \] \[K={{e}^{\frac{-\Delta {{G}^{o}}}{RT}}}\]You need to login to perform this action.
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