Calculate:
$(a)$ $\Delta G^{\circ}$ and
$(b)$ the equilibrium constant for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \, K$
$NO_{(g)} + 1/2 O_{2(g)} \longleftrightarrow NO_{2(g)}$
Given:
$\Delta G^{\circ}_f(NO_2) = 52.0 \, kJ/mol$
$\Delta G^{\circ}_f(NO) = 87.0 \, kJ/mol$
$\Delta G^{\circ}_f(O_2) = 0 \, kJ/mol$

$(a)$ For the given reaction,
$\Delta G^{\circ} = \Delta G^{\circ}_f(\text{Products}) - \Delta G^{\circ}_f(\text{Reactants})$
$\Delta G^{\circ} = 52.0 - \{87.0 + 0\} = -35.0 \, kJ \, mol^{-1}$
$(b)$ We know that,
$\Delta G^{\circ} = -2.303 \, RT \, \log K_c$
$\log K_c = \frac{-\Delta G^{\circ}}{2.303 \, RT}$
$\log K_c = \frac{-(-35.0 \times 10^3 \, J \, mol^{-1})}{2.303 \times 8.314 \, J \, K^{-1} \, mol^{-1} \times 298 \, K}$
$\log K_c = 6.134$
$\therefore K_c = \text{antilog}(6.134) = 1.36 \times 10^6$
Hence,the equilibrium constant for the given reaction $K_c$ is $1.36 \times 10^6$.