Chemistry Question from JEE Main 2024 · 27 January, Shift 2

For a certain thermochemical reaction M $\rightarrow$ N at $T = 400 \, \text{K}$, $\Delta H^\circ = 77.2 \, \text{kJ/mol}$, $\Delta S^\circ = 122 \, \text{J/K}$, the log of the equilibrium constant $(\log K)$ is $-x \times 10^{-1}$. The value of $x$ is: (Given $R = 8.314 \, \text{J K}^{-1}\text{mol}^{-1}$)