Thermal decomposition of gaseous $\mathrm{X}_2$ to gaseous $\mathrm{X}$ at $298 \mathrm{~K}$ takes place according to the following equation :
$\mathrm{X}_2(g) \rightleftharpoons 2 \mathrm{X}(g)$
The standard reaction Gibbs energy, $\Delta_r G^{\circ}$, of this reaction is positive. At the stiur of the reaction, there is one mole of $X_2$ and no $X$. As the reaction proceeds, the number of roles of $X$ formed is given by $\beta$. Thus, $\beta_{\text {equitibrium }}$ 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 : $R=0.083 \mathrm{~L} \mathrm{bar}^{-1} \mathrm{~mol}^{-1}$ )
($1$) The equilibrium constant $K_P$ for this reaction at $298 \mathrm{~K}$, in terms of $\beta_{\text {equilibs, um }}$, is
($A$) $\frac{8 \beta_{\text {equilibrium }}^2}{2-\beta_{\text {equilibrium }}}$ ($B$) $\frac{8 \beta_{\text {equititrium }}^2}{4-\beta_{\text {equilibrium }}^2}$ ($C$) $\frac{4 \beta_{\text {equilibrium }}^2}{2-\beta_{\text {equilibrium }}}$ ($D$) $\frac{4 \ell_{\text {equitibrium }}^2}{4-\beta_{\text {equilibrium }}^2}$
($2$) The $INCORRECT$ statement among the following, for this reaction, is
($A$) Decrease in the total pressure will result in formation of more moles of gaseous $\mathrm{X}$
($B$) At the start of the reaction, dissociation of gaseous $\mathrm{X}_2$ takes place spontaneously
($C$) $\beta_{\text {equilibrium }}=0.7$
($D$) $\quad K_c<1$
Given the answer question ($1$) and ($2$)