If the change in standard Gibbs free energy for a reaction is less than $0$,then the value of the equilibrium constant $K_c$ is:

At $298 \ K$,if the standard Gibbs energy change $\Delta_r G^{\ominus}$ of a reaction is $-115 \ kJ \ mol^{-1}$,the value of $\log_{10} K_{p}$ will be $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.

For the reaction $CH_{4(g)} + H_{2(g)} \longrightarrow C_2H_{6(g)}$,$K_p = 3.356 \times 10^{17}$,calculate $\Delta G^{\circ}$ for the reaction at $298 \ K$.

The equilibrium constant of the following given reaction is $K_p = 6.022 \times 10^{-5}$ at $298 \ K$ temperature. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. Calculate the value of $\Delta_r G^o$.

$A$ reaction attains equilibrium when the free energy change accompanying it is

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