If the equilibrium constant for a reaction is $10$,then the value of $\Delta G^o$ is ....... $(R = 8 \, J \, K^{-1} \, mol^{-1}, T = 300 \, K)$

The equilibrium constant $(K)$ of a reaction may be written as :

For the equilibrium $H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$ at $1 \ atm$ and $298 \ K$,which of the following statements is correct?

For a chemical reaction, the standard Gibbs energy change, $\Delta G^{\circ}$ is $-7.64 \times 10^4 \ J \ mol^{-1}$. What is the value of equilibrium constant $(K)$?

For the reaction $A_{(g)} \rightarrow B_{(g)},$ the value of the equilibrium constant at $300 \ K$ and $1 \ atm$ is equal to $100.0.$ The value of $\Delta_{r}G^{\circ}$ for the reaction at $300 \ K$ and $1 \ atm$ in $J \ mol^{-1}$ is $-xR,$ where $x$ is ........... (Rounded off to the nearest integer) ($R = 8.31 \ J \ mol^{-1} K^{-1}$ and $\ln 10 = 2.3$)

Calculate $\Delta G^\circ$ for the reaction, $CH_4(g) + H_2(g) \rightarrow C_2H_6(g)$ at $298 \text{ K}$, given $K_p = 2 \times 10^{17}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.