Paragraph: Thermal decomposition of gaseous $\mathrm{X}_{2}$ to gaseous $\mathrm{X}$ at $298 \mathrm{~K}$…
Paragraph:
Thermal decomposition of gaseous $\mathrm{X}_{2}$ to gaseous $\mathrm{X}$ at $298 \mathrm{~K}$ takes place according to the following equation:
$\mathrm{X}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{X}(\mathrm{g})$
The standard reaction Gibbs energy, $\Delta_{r} G^{\circ}$, of this reaction is positive. At the start of the reaction, there is one mole of $\mathrm{X}_{2}$ and no $\mathrm{X}$. As the reaction proceeds, the number of moles of $\mathrm{X}$ formed is given by $\beta$. Thus, $\beta_{\text {equilibrium }}$ is the number of moles of $\mathrm{X}$ formed at equilibrium. The reaction is carried out at a constant total pressure of $2$ bar. Consider the gases to behave ideally. (Given : $R=0.083 \mathrm{~L}$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )
Question:
The equilibrium constant $K_{P}$ for this reaction at $298 \mathrm{~K},$ in terms of $\beta_{equilibrium}$ is