# Calculate a) ΔG°and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K

Question:

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 $\triangle_{f} G^{\circ}\left(\mathrm{NO}_{2}\right)=52.0 \mathrm{~kJ} / \mathrm{mol}$

$\Delta_{f} G^{\circ}(\mathrm{NO})=87.0 \mathrm{~kJ} / \mathrm{mol}$

$\Delta_{f} G^{\circ}\left(\mathrm{O}_{2}\right)=0 \mathrm{~kJ} / \mathrm{mol}$

Solution:

(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 $1.36 \times 10^{6}$