Write the formula relating the equilibrium constant $K$ and the standard Gibbs free energy change $\Delta G^{\circ}$.

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:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$

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 $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)

For the reaction at $25\,^oC$,${N_2O_4}_{(g)} \rightleftharpoons 2NO_{2_{(g)}}$,if $\Delta G_f^o$ for $N_2O_4$ and $NO_2$ are $23.49 \, KCal$ and $12.39 \, KCal$ respectively,then $K_p$ for the reaction is: (in $, atm$)

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.)

In the glycolysis process,during the phosphorylation of glucose,the equilibrium constant at $298 \ K$ is $3.6 \times 10^{-3}$. Find the value of $\Delta G^{\Theta}$. What does this indicate? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$