In an equilibrium reaction for which $\Delta G^o = 0$,the equilibrium constant $K = $

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

Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$

Which of the following correctly represents the relationship between $\Delta G$ and $\Delta G^{\circ}$? $[P = \text{products}, R = \text{reactants}]$

Solid $KClO_4$ is taken in a container maintained at a constant pressure of $1 \, atm$. Upon heating,the following equilibrium is obtained:
$2KClO_{4(s)} \rightleftharpoons 2KCl_{(s)} + 3O_{2(g)}$
If $\Delta H^o = 25 \, kcal/mol$ and $\Delta S^o = 50 \, cal/K \cdot mol$,at what temperature will equilibrium be established in the container (in $, K$)? (Ignore variation of $\Delta H^o$ and $\Delta S^o$ with temperature.)