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

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

The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)

$K_p$ for the conversion of oxygen to ozone at $400 \ K$ is $1.0 \times 10^{-30}$,its standard Gibbs energy change in $kJ \ mol^{-1}$ is approximately

The free energy change for a reversible reaction at equilibrium is

For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,if the equilibrium pressure is $X \ bar$,then $\Delta_r G^o$ would be :-