At $298 \ K$,$\Delta_r G^{\ominus}$ for the following reaction is $165.469 \ kJ \ mol^{-1}$. What is the equilibrium constant for this reaction? $(R = 8.3 \ J \ mol^{-1} \ K^{-1})$
$\frac{3}{2} O_{2(g)} \longrightarrow O_{3(g)}$

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 $\Delta G^{\circ} \ (kJ/mol)$ at $127 \ ^{\circ}C$ for a reaction with $K_{equilibrium} = 10^5$.

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

For a homogeneous gaseous reaction,the equilibrium constant $K_p$ is $10^{-8}$. The standard Gibbs free energy change for the reaction is ........... $kcal$. $(R = 2.0 \, cal \, K^{-1} \, mol^{-1}, T = 298 \, K)$

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