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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
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B
1:9
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C
1:6
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D
1:4
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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

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