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)

1:6

$\underset{\underset{1-\alpha }{1}}{A}\rightleftharpoons \underset{\underset{2\alpha }{0}}{2B}$
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 $P_B=\left(\frac{2\alpha }{1+\alpha }\right)P$
$\therefore K_{p_1}=\frac{(P_B{)}^2}{\left(P_A\right)}=\frac{{\left[\left(\frac{2\alpha }{1+\alpha }\right)P_1\right]}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)P_1}=\frac{4{\alpha }^2{P}_1}{(1-{\alpha }^2)}$
$\underset{\underset{1-\alpha }{1}}{P}\rightleftharpoons \underset{\underset{\alpha }{0}}{Q}+\underset{\underset{\alpha }{0}}{R}$
Total moles $=1-\alpha +\alpha +\alpha =1+\alpha$
$\therefore P_P=\left(\frac{1-\alpha }{1+\alpha }\right)P_2;P_Q=\left(\frac{\alpha }{1+\alpha }\right)P_2;P_R=\left(\frac{\alpha }{1+\alpha }\right)P_2$
$K_{P_2}=\frac{P_R\times P_Q}{P_P}=\frac{\left(\frac{\alpha }{1+\alpha }\right)P_2\times \left(\frac{\alpha }{1+\alpha }\right)P_2}{\left(\frac{1-\alpha }{1+\alpha }\right)P_2}=\frac{{\alpha }^2{P}_2}{(1-{\alpha }^2)}$
$\therefore \frac{K_{P_1}}{K_{P_2}}=\frac{2}{3}\left(given\right)=\frac{4{\alpha }^2{P}_1}{(1-{\alpha }^2)}\times \left(\frac{1-a^2}{{\alpha }^2{P}_2}\right)=\frac{4P_1}{P_2}$
$\therefore \frac{P_1}{P_2}=\frac{1}{6}$