JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Equilibrium

Let the initial amount of each reactant be $c$ and the degree of dissociation $\alpha$. For $\mathrm{X}\rightleftharpoons \mathrm{Y}+\mathrm{Z}$ the equilibrium amounts are $c(1-\alpha )$, $c\alpha$ and $c\alpha$, totalling $c(1+\alpha )$, and expressing each partial pressure through the mole fractions and the total pressure $p_1$ gives $K_{p1}=\frac{{\alpha }^2{p}_1}{1-{\alpha }^2}=3$, so $p_1=\frac{3(1-{\alpha }^2)}{{\alpha }^2}$. For $\mathrm{A}\rightleftharpoons 2\mathrm{B}$ the amounts are $c(1-\alpha )$ and $2c\alpha$, again totalling $c(1+\alpha )$, and $K_{p2}=\frac{4{\alpha }^2{p}_2}{1-{\alpha }^2}=1$, so $p_2=\frac{1-{\alpha }^2}{4{\alpha }^2}$. Dividing one by the other, $\frac{p_1}{p_2}=3\times 4=12$. The value of x is 12.