MCQ
The correct relationship between free energy change in a reaction and the corresponding equilibrium constant ${K_c}$ is
- A$\Delta G = RT\ln {K_c}$
- B$ - \Delta G = RT\ln {K_c}$
- C$\Delta {G^o} = RT\ln {K_c}$
- ✓$ - \Delta {G^o} = RT\ln {K_c}$
✓
Answer
Correct option: D.
$ - \Delta {G^o} = RT\ln {K_c}$
d
$\because \quad \Delta G =\Delta G ^{\circ}+ RTlnQ$, where $Q$ is reaction quotient at equilibrium, $\Delta G =0$ and $Q = K _{ c }$
$\because \quad \Delta G =\Delta G ^{\circ}+ RTlnQ$, where $Q$ is reaction quotient at equilibrium, $\Delta G =0$ and $Q = K _{ c }$
$\therefore \quad-\Delta G = RTlnK _{ c }$
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