Using the data provided, find the value of equilibrium constant for the following reaction at 298   K…

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.

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