Using the data provided, find the value of equilibrium constant for the following reaction at $298 \mathrm{K}$ and $1$ atm pressure. ${\mathrm{NO}}_{\left(\mathrm{g}\right)}+\frac{1}{2}{\mathrm{O}}_2( \mathrm{g})\rightleftharpoons {\mathrm{NO}}_2( \mathrm{g})$
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0\left({\mathrm{NO}}_{\left(\mathrm{g}\right)}\right)=90.4\mathrm{kJ}.{\mathrm{mol}}^{-1}$

${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0\left({\mathrm{NO}}_2\left(\mathrm{g}\right)\right)=32.48\mathrm{kJ}\cdot {\mathrm{mol}}^{-1}$

${\mathrm{\Delta S}}^{\circ }@298\mathrm{K}=-70.8\mathrm{J}\cdot {\mathrm{K}}^{-1}\cdot {\mathrm{mol}}^{-1}$

$\mathrm{antilog}(0.50)=3162$

The given reaction is

${\mathrm{NO}}_{\left(\mathrm{g}\right)}+\frac{1}{2}{\mathrm{O}}_2( \mathrm{g})\rightleftharpoons {\mathrm{NO}}_2( \mathrm{g})$

The enthalpy of the above reaction given by the difference of enthalpies of reactant side and product side.

Therefore,

$$\begin{gathered} {∆}_{\mathrm{c}}\mathrm{H} = {∆}_{\mathrm{f}}\mathrm{H}\left({\mathrm{NO}}_2\right)-\left[{∆}_{\mathrm{f}}\mathrm{H}\left(\mathrm{NO}\right)+\frac{1}{2}{∆}_{\mathrm{f}}\mathrm{H}\left({\mathrm{O}}_2\right)\right] \\ {∆}_{\mathrm{c}}\mathrm{H} = 32.8-\left[90.4+\frac{1}{2}\left(0\right)\right] \\ {∆}_{\mathrm{c}}\mathrm{H} = -57.6 {\mathrm{KJmol}}^{-1} \end{gathered}$$

Also, we know that Gibb's free energy can be determined by

$$\begin{gathered} ∆\mathrm{G} = ∆\mathrm{H}-\mathrm{T}∆\mathrm{S} \\ ∆\mathrm{G} = 57600 {\mathrm{Jmol}}^{-1}-298\times \left(-70.8\right) {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \\ ∆\mathrm{G} =- 36501.5 {\mathrm{Jmol}}^{-1} \end{gathered}$$

Again, Using relation of Gibb's free energy with equilibrium constant i.e.

$$\begin{gathered} ∆\mathrm{G} = -2.303\times 8.31\times 298 \mathrm{logk} =-36501.5 {\mathrm{Jmol}}^{-1} \\ \mathrm{K}= 3.162 \times {10}^{6 } \end{gathered}$$