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The values of $K_{p_1}$ and $K_{p_2}$ for the reactions
$X \rightleftharpoons Y + Z$ ....(1) and
$A \rightleftharpoons 2B$ ....(2)
are in ratio of 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio :
$X \rightleftharpoons Y + Z$ ....(1) and
$A \rightleftharpoons 2B$ ....(2)
are in ratio of 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio :
A
36 : 1
B
1 : 1
C
3 : 1
D
1 : 9
Detailed Solution
For $X \rightleftharpoons Y + Z$: initially 1, 0, 0; at equilibrium $1 - \alpha$, $\alpha$, $\alpha$.
Total number of moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha$
$K_{p_1} = \dfrac{p_Y \times p_Z}{p_X} = \dfrac{\left(\dfrac{\alpha}{1 + \alpha}P_1\right)\left(\dfrac{\alpha}{1 + \alpha}P_1\right)}{\left(\dfrac{1 - \alpha}{1 + \alpha}\right)P_1} = \dfrac{\alpha^2 P_1}{1 - \alpha^2}$
For $A \rightleftharpoons 2B$: initially 1, 0; at equilibrium $1 - \alpha$, $2\alpha$.
Total number of moles at equilibrium $= 1 - \alpha + 2\alpha = 1 + \alpha$
$K_{p_2} = \dfrac{(p_B)^2}{p_A} = \dfrac{\left(\dfrac{2\alpha}{1 + \alpha}P_2\right)^2}{\left(\dfrac{1 - \alpha}{1 + \alpha}\right)P_2} = \dfrac{4\alpha^2 P_2}{1 - \alpha^2}$
The degree of dissociation $\alpha$ is the same in both, so $\dfrac{K_{p_1}}{K_{p_2}} = \dfrac{P_1}{4P_2}$
$\dfrac{9}{1} = \dfrac{P_1}{4P_2}$
$\dfrac{P_1}{P_2} = \dfrac{36}{1}$, i.e. 36 : 1
Total number of moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha$
$K_{p_1} = \dfrac{p_Y \times p_Z}{p_X} = \dfrac{\left(\dfrac{\alpha}{1 + \alpha}P_1\right)\left(\dfrac{\alpha}{1 + \alpha}P_1\right)}{\left(\dfrac{1 - \alpha}{1 + \alpha}\right)P_1} = \dfrac{\alpha^2 P_1}{1 - \alpha^2}$
For $A \rightleftharpoons 2B$: initially 1, 0; at equilibrium $1 - \alpha$, $2\alpha$.
Total number of moles at equilibrium $= 1 - \alpha + 2\alpha = 1 + \alpha$
$K_{p_2} = \dfrac{(p_B)^2}{p_A} = \dfrac{\left(\dfrac{2\alpha}{1 + \alpha}P_2\right)^2}{\left(\dfrac{1 - \alpha}{1 + \alpha}\right)P_2} = \dfrac{4\alpha^2 P_2}{1 - \alpha^2}$
The degree of dissociation $\alpha$ is the same in both, so $\dfrac{K_{p_1}}{K_{p_2}} = \dfrac{P_1}{4P_2}$
$\dfrac{9}{1} = \dfrac{P_1}{4P_2}$
$\dfrac{P_1}{P_2} = \dfrac{36}{1}$, i.e. 36 : 1