Question

The values of $K_{P1}$ and $K_{P2}$ for the reactions,
$X\rightleftharpoons Y+Z...\left(i\right)A\rightleftharpoons 2B...\left(ii\right)$
are in the ratio $9:1$. If the initial concentrations and the degrees of dissociation of $X$ and $A$ are equal, then the smallest ratio of total pressures at equilibrium $\left(i\right)$ and $\left(ii\right)$ is $x:y$. Find the value of $x+y$.

Answer 1

$X\rightleftharpoons Y+Z...\left(i\right)A\rightleftharpoons 2B...\left(ii\right)$
$$\begin{array}{ccccc}& X& \rightleftharpoons & Y+& Z\\ \text{Initially}& a& & 0& 0\\ \text{At equilibrium}& a(1-\alpha )& & a\alpha & a\alpha \end{array}$$
$\text{Total number of moles at equilibrium}=a(1-\alpha )+2a\alpha =a(1+\alpha )$
$\therefore K_{P1}=\frac{P_Y\times P_Z}{P_X}=\frac{(\frac{a\alpha }{a(1+\alpha )}\times P_1)\times (\frac{a\alpha }{a(1+\alpha )}\times P_1)}{\frac{a(1-\alpha )}{a(1+\alpha )}\times P_1}=\frac{a^2{\alpha }^2{P}_1}{a(1+\alpha )a(1-\alpha )}=\frac{{\alpha }^2{P}_1}{(1+\alpha )(1-\alpha )}$
$$\begin{array}{ccc}& A\rightleftharpoons & 2B\\ \text{Initially}& a& 0\\ \text{At Equilibrium}& a(1-\alpha )& 2a\alpha \end{array}$$
$\text{Total number of moles at equilibrium}=a(1-\alpha )+2a\alpha =a(1+\alpha )$

$K_{P2}=\frac{(P_B{)}^2}{P_A}=\frac{(\frac{\left(2a\alpha \right)}{a(1+\alpha )}\times P_2{)}^2}{\frac{a(1-\alpha )}{a(1+\alpha )}\times P_2}=\frac{(2a\alpha {)}^2{P}_2}{a(1-\alpha )a(1+\alpha )}=\frac{(2\alpha {)}^2{P}_2}{(1-\alpha )(1+\alpha )}$
Now,
$\frac{K_{P1}}{K_{P2}}=\frac{P_1}{4P_2}$
$\Rightarrow \frac{9}{1}=\frac{P_1}{4P_2}(\because \frac{K_{P1}}{K_{P2}}=\frac{9}{1})$
$\Rightarrow \frac{P_1}{P_2}=\frac{36}{1}=36:1$
$\therefore x:y=36:1\Rightarrow x+y=36+1=37$