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(C) The relationship between Gibbs free energy change and the reaction quotient is given by: $\Delta G = \Delta G^{\circ} + RT \ln Q$. Substituting $Q

Vedclass · Accepted Answer

$\Delta G^{\circ} = \Delta G + 2.303 RT \log \frac{[R]}{[P]}$

Vedclass · Answer

(C) The relationship between Gibbs free energy change and the reaction quotient is given by: $\Delta G = \Delta G^{\circ} + RT \ln Q$. Substituting $Q = \frac{[P]}{[R]}$ and converting natural logarithm to base $10$ $(\ln x = 2.303 \log x)$,we get: $\Delta G = \Delta G^{\circ} + 2.303 RT \log \frac{[P]}{[R]}$. Rearranging this equation to solve for $\Delta G^{\circ}$: $\Delta G^{\circ} = \Delta G - 2.303 RT \log \frac{[P]}{[R]}$. Since $-\log \frac{[P]}{[R]} = \log \frac{[R]}{[P]}$,the equation can be written as: $\Delta G^{\circ} = \Delta G + 2.303 RT \log \frac{[R]}{[P]}$. Thus,option $C$ is correct.

Which of the following correctly represents the relationship between $\Delta G$ and $\Delta G^{\circ}$? $[P = \text{products}, R = \text{reactants}]$

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