$$ \begin{aligned} &\text { Use the following data : }\\ &\begin{array}{|c|c|c|} \hline \text { Substance } & \frac{\Delta_f \mathrm{H}^{\ominus}(500 \mathrm{~K})}{\mathrm{kJ} \mathrm{~mol}^{-1}} & \frac{\mathrm{~S}^{\ominus}(500 \mathrm{~K})}{\mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\ \hline \mathrm{AB}(\mathrm{~g}) & 32 & 222 \\ \hline \mathrm{~A}_2(\mathrm{~g}) & 6 & 146 \\ \hline \mathrm{~B}_2(\mathrm{~g}) & x & 280 \\ \hline \end{array} \end{aligned} $$

One mole each of $\mathrm{A}_2(\mathrm{~g})$ and $\mathrm{B}_2(\mathrm{~g})$ are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K .

$$ \mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{~g}) $$

The value of $x\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\_\_\_\_$ . (Nearest integer)

(Given : $\log \mathrm{K}=2.2 \quad \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )