Calculate a) $\triangle G^{\circ}$ and b) the equilibrium constant for the formation of $\mathrm{NO}_{2}$ from $\mathrm{NO}$ and $\mathrm{O}_{2}$ at $298 \mathrm{~K}$
$\mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longleftrightarrow \mathrm{NO}_{2}(\mathrm{~g})$
where $\triangle_{f} G^{\circ}\left(\mathrm{NO}_{2}\right)=52.0 \mathrm{~kJ} / \mathrm{mol}$
$\Delta_{f} G^{\circ}(\mathrm{NO})=87.0 \mathrm{~kJ} / \mathrm{mol}$
$\Delta_{f} G^{\circ}\left(\mathrm{O}_{2}\right)=0 \mathrm{~kJ} / \mathrm{mol}$
(a) For the given reaction,
$\Delta G^{\circ}=\Delta G^{\circ}$ ( Products) $-\Delta G^{\circ}$ ( Reactants)
$\Delta G^{\circ}=52.0-\{87.0+0\}$
$=-35.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(b) We know that,
$\Delta G^{\circ}=R T \log K_{c}$
$\Delta G^{\circ}=2.303 \mathrm{RT} \log K_{c}$
$K_{c}=\frac{-35.0 \times 10^{-3}}{-2.303 \times 8.314 \times 298}$
$=6.134$
$\therefore K_{c}=$ antilog $(6.134)$
$=1.36 \times 10^{6}$
Hence, the equilibrium constant for the given reaction $K_{c}$ is $1.36 \times 10^{6}$
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