For a hypothetical reversible reaction $\frac{1}{2} A_{2(g)} + \frac{3}{2} B_{2(g)} \rightarrow AB_{3(g)}$; $\Delta H = -20 \, kJ$. If the standard entropies of $A_2, B_2$ and $AB_3$ are $60, 40$ and $50 \, J K^{-1} mol^{-1}$ respectively,at what temperature $(K)$ will the reaction be in equilibrium?

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

The equilibrium constant for a reaction is $20$. What is the value of $\Delta G^{\circ}$ at $300 \ K$? (Given: $R = 8 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,$\ln(20) \approx 2.996$)

Assuming ideal behaviour,the magnitude of $\log \, K$ for the following reaction at $25^{\circ} C$ is $x \times 10^{-1}$. The value of $x$ is $......$. (Integer answer)
$3 HC \equiv CH_{(g)} \rightleftharpoons C_6H_{6(\ell)}$
[Given: $\Delta_f G^{\circ}(HC \equiv CH) = -2.04 \times 10^5 \, J \, mol^{-1}$
$\Delta_f G^{\circ}(C_6H_6) = -1.24 \times 10^5 \, J \, mol^{-1}; R = 8.314 \, J \, K^{-1} \, mol^{-1}$]

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 $\Delta G^0$.