The correct relationship between standard free energy change $(\Delta G^o)$ and equilibrium constant $(K)$ is:

For a reaction,$\Delta G^{\circ} = -115 \, kJ$. What is the value of $\log \, K_p$ at $298 \, K$?

At $300 \ K$,for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the equilibrium constant $K_p = 1.8 \times 10^{-7}$. Calculate its standard Gibbs free energy change $\Delta G^0$.

The equilibrium constant for a reaction is $10$. What will be the value of $\Delta G^{\theta}$? $R = 8.314 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K$

Consider the reaction $X \rightleftharpoons Y$ at $300 \text{ K}$. If $\Delta H^\circ$ and $K$ are $28.40 \text{ kJ mol}^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta S^\circ$ for the reaction in $\text{J K}^{-1} \text{ mol}^{-1}$ is . . . . . . . (Nearest integer) (Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$, $\ln 10 = 2.3$, $\log 3 = 0.48$, $\log 2 = 0.30$)

Assertion: For every chemical reaction at equilibrium,the standard Gibbs energy change is zero.
Reason: At constant temperature and pressure,a chemical reaction is spontaneous in the direction of decreasing Gibbs energy.