Question

The value of $K{p}_1$ and $K{p}_2$ for the reactions
$X\rightleftharpoons y+z\to \left(1\right)$
And $A\rightleftharpoons 2B\to \left(2\right)$
Are in the 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. 3 : 1
• B. 1 : 9
• C. 36 : 1
• D. 1 : 1

Answer 1

The correct option is C 36 : 1

Suppose total pressure at equilibrium for reactions (1) and (2) are $P_1$ and $P_2$ respectively
Then $X\rightleftharpoons Y+.Z$
Initial 1 mole 0 0
At Equilibrium $1-\alpha$ $\alpha$ $\alpha$
$\therefore$ Total $=1+\alpha$
$P_x=\frac{1-\alpha }{1+\alpha }{P}_1,P_y=\frac{\alpha }{1+\alpha }{P}_1$
And $P_z=\frac{\alpha }{1+\alpha }{P}_1$
$K{p}_1=\frac{{\left(\frac{\alpha }{1+\alpha }{P}_1\right)}^2}{\frac{1-\alpha }{1+\alpha }{P}_1}=\frac{{\alpha }^2{P}_1}{1-{\alpha }^2}={\alpha }^2{P}_1$
$A\rightleftharpoons 2B$
Initial 1 mole 0
At Equilibrium $1-\alpha$ $2\alpha$
$\therefore$ Total $=1+\alpha$
$K{p}_2=\frac{{\left(\frac{2\alpha }{1+\alpha }{P}_2\right)}^2}{\frac{1-\alpha }{1+\alpha }{P}_2}=\frac{4{\alpha }^2}{1-{\alpha }^2}{P}_2$
$=4{\alpha }^2{p}_2$
$\frac{K{p}_1}{K{p}_2}=\frac{{\alpha }^2{P}_1}{4{\alpha }^2{P}_2}=\frac{P_1}{4P_2}=\frac{9}{1}$ (given)
Or, $\frac{P_1}{P_2}=\frac{36}{1}=36:1$