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 $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.

The standard Gibbs free energy change $\Delta G^o$ is related to the equilibrium constant $K_p$ as:

Calculate the standard Gibbs energy change $(\Delta G^\circ)$ for a gaseous reaction at $298 \text{ K}$ if the equilibrium constant $K_p$ is $3.5 \times 10^{17}$ and the gas constant $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

Find the value of the equilibrium constant $(K)$ of a reaction at $300 \ K$, when standard Gibbs free energy change is $-25 \ kJ \ mol^{-1}$? (Consider $R = 8.33 \ J \ mol^{-1} \ K^{-1}$)

The reaction,$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$ is in equilibrium in a closed vessel at $298 \, K$. The partial pressure (in $atm$) of $CO_{2(g)}$ in the reaction vessel is closest to $....$ [Given : The change in Gibbs energies of formation at $298 \, K$ and $1 \, bar$ for $CaO_{(s)} = -603.501 \, kJ \, mol^{-1}$,$CO_{2(g)} = -394.389 \, kJ \, mol^{-1}$,$CaCO_{3(s)} = -1128.79 \, kJ \, mol^{-1}$,Gas constant,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

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 :-