Question

the equilibrium constant kp1 and kp2 for the reaction X = 2Y and Z = P+ Q respectively are in the ratio of 1:9 if the degree of dissociation of X And Z be equal then the ratio of total pressure at equilibrium was

Answer 1

I think this may be the right answer.....!

Answer 2

Answer: 36:1

Explanation: $X\rightleftharpoons 2Y$

initially           c                   0

at eq'm       $c- c\alpha$     $2c\alpha$

Total moles = $c- c\alpha+2c\alpha= c+c\alpha$

So eq'm constant will be:

$K_{p}=\frac{(\frac{2c\alpha}{c+c\alpha})^2\times P^2}{\frac{c-c\alpha}{c+c\alpha}\times P}$

$K_{p}=\frac{4c^2(\alpha)^2}{c+c\alpha\times c-c\alpha}\times P^2$

2) $Z\rightleftharpoons P+Q$

initially           c                   0     0

at eq'm   $c- c\alpha$   $c\alpha$  $c\alpha$

Total moles = $c- c\alpha+c\alpha+c\alpha=c+c\alpha$

So eq'm constant will be:

$K'_{p}=\frac{\frac{c\alpha}{c+c\alpha}\times P'\times \frac{c\alpha}{c+c\alpha}\times P'}{\frac{c-c\alpha}{c+c\alpha}\times P'}$

$K'_{p}=\frac{c^2(\alpha)^2}{c+c\alpha\times (c-c\alpha)}\times P'}$

$\frac{k_p}{K'_p}=\frac{1}{9}$

Thus $\frac{1}{9}=\frac{4P}{P'}$

$\frac{P}{P'}=\frac{36}{1}$