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

Solution

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More