Question

The equilibrium constants $K_{p_1}$ and $K_{p_2}$ for the reactions $A\rightleftharpoons 2B$ and $P\rightleftharpoons Q+R$, respectively, are in the ratio of $2:3$. If the degree of dissociation of $A$ and $P$ are equal, the ratio of the total pressure at equilibrium is,

• A. $1:36$
• B. $1:9$
• C. $1:6$
• D. 1:4

Answer 1

Answer: (C)

At Equilibrium $1-\alpha 2\alpha$
Total moles $=1-\alpha +2\alpha =1+\alpha$
$\therefore P_A=($ mole fraction of $A)\times$ Initial pressure $=$ $\left(\frac{1-\alpha }{1+\alpha }\right)P_1$
Similarly $P_B=\left(\frac{2\alpha }{1+\alpha }\right)P$
$\therefore K_{p_1}=\frac{{\left(P_B\right)}^2}{\left(P_A\right)}=\frac{{\left[\left(\frac{2\alpha }{1+\alpha }\right)P_1\right]}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)P_1}=\frac{4{\alpha }^2{P}_1}{\left(1-{\alpha }^2\right)}$

$\therefore P_p=\left(\frac{1-\alpha }{1+\alpha }\right)P_2;P_Q=\left(\frac{\alpha }{1+\alpha }\right)P_2;$
$P_R=\left(\frac{\alpha }{1+\alpha }\right)P_2$
$K_{p_2}=\frac{P_R\times P_Q}{P_P}=\frac{\left(\frac{\alpha }{1+\alpha }\right)P_2\times \left(\frac{\alpha }{1+\alpha }\right)P_2}{\left(\frac{1-\alpha }{1+\alpha }\right)P_2}$
$=\frac{{\alpha }^2{P}_2}{\left(1-{\alpha }^2\right)}$
$\therefore \frac{K_{p_1}}{K_{p_2}}=\frac{2}{3}$ (given) $=\frac{4/{\alpha }^2{P}_1}{/\left(1/{\alpha }^2\right)}\times \left(\frac{/1/{\alpha }^2}{/{\alpha }^2{P}_2}\right)=\frac{4P_1}{P_2}$
$\therefore \frac{P_1}{P_2}=\frac{1}{6}$