Chemistry - Biomolecules Question with Solution | TestHub

From equation. $\begin{array}{lccccc}& X& \rightleftharpoons & Y& +& Z\\ \text{Initial mole }& 10& & 0& & 0\\ \text{mole at equilibrium }& (1-a)& & a& & a\end{array}$ $\begin{array}{rl}& K_{p_1}=\frac{p_Y\times p_z}{p_X}\\ & =\frac{\left[\frac{\alpha \times p_1}{1+\alpha }\right]\left[\frac{\alpha \times p_1}{1+\alpha }\right]}{\left[\frac{1-\alpha }{1+\alpha }\right]p_1}\\ & K_{P_1}=\frac{{\alpha }^2{p}_1}{1-{\alpha }^2}\end{array}$ From equation $\begin{array}{lccc}& A& \rightleftharpoons & 2B\\ \text{Initial mole }& 10& & 0\\ \text{mole at equilibrium }& (1-a)& & 2a\end{array}$ $\begin{array}{rl}& K_{p_2}=\frac{{\left[\frac{2\alpha }{1+\alpha }\cdot p_2\right]}^2}{\left[\frac{1-\alpha }{1+\alpha }\right]p_2}=\frac{4{\alpha }^2{p}_2}{1-{\alpha }^2}\end{array}$ From Eqs (i) and (ii) $\frac{K_{P_1}}{K_{P_2}}=\frac{p_1}{4p_2}$ $\begin{array}{ll}\Rightarrow & \frac{9}{1}=\frac{p_1}{4p_2}\\ \Rightarrow & \therefore \frac{p_1}{p_2}=\frac{36}{1}\end{array}$