The correct relation is:

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(C) The equation for Gibbs free energy is given by:$\Delta G = \Delta G^{\circ} + RT \ln Q \dots(1)$At equilibrium,$\Delta G = 0$ and $Q = K$,therefor

Vedclass · Accepted Answer

$\Delta G = - RT \ln (K / Q)$

Vedclass · Answer

(C) The equation for Gibbs free energy is given by:$\Delta G = \Delta G^{\circ} + RT \ln Q \dots(1)$At equilibrium,$\Delta G = 0$ and $Q = K$,therefore:$0 = \Delta G^{\circ} + RT \ln K \implies \Delta G^{\circ} = - RT \ln K \dots(2)$Substituting equation $(2)$ into equation $(1)$:$\Delta G = - RT \ln K + RT \ln Q$$\Delta G = RT \ln Q - RT \ln K$$\Delta G = RT \ln (Q / K)$Wait,checking the options provided: $\Delta G = - RT \ln (K / Q)$ is equivalent to $\Delta G = RT \ln (Q / K)$. Thus,option $C$ is correct.

Similar Questions

Write the formula relating the equilibrium constant $K$ and $\Delta G^{\circ}$.

For a system in equilibrium,$\Delta G = 0$ under conditions of constant:

$STATEMENT-1$: For every chemical reaction at equilibrium,standard Gibbs energy of reaction is zero. $STATEMENT-2$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.

The equilibrium constant of a reaction is related to

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