Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\mathrm{X}\left(\mathrm{g}\right) \rightleftharpoons \mathrm{Y}\left(\mathrm{g}\right) + \mathrm{Z}\left(\mathrm{g}\right) {\mathrm{K}}_{{\mathrm{p}}_1}=3$

 $\mathrm{n}$                                   Initial concentration

$\mathrm{n}-\alpha \mathrm{n} \alpha \mathrm{n} \alpha \mathrm{n}$     at equilibrium
Let total pressure be ${\mathrm{p}}_1$

Using dalton's law of partial pressures and equation for equilibrium constant:
$\therefore {\mathrm{K}}_{{\mathrm{p}}_1}=\frac{{\left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\times {\mathrm{p}}_1\right)}^2}{\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}{\mathrm{p}}_1}$

$\Rightarrow 3=\frac{{\mathrm{\alpha }}^2\times {\mathrm{p}}_1}{1-{\mathrm{\alpha }}^2}$       ....(X)

$\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons 2 \mathrm{B}( \mathrm{g}) {\mathrm{k}}_{{\mathrm{p}}_2}=1$

 $\mathrm{n}$                         Initial mole

 $\mathrm{x}-\mathrm{\alpha n} 2\mathrm{\alpha n}$   at equilibrium
Let total pressure be ${\mathrm{p}}_2$
Using dalton's law of partial pressures and equation for equilibrium constant

$\therefore {\mathrm{k}}_{{\mathrm{p}}_2}=\frac{{\left(\frac{2\alpha }{1+\alpha }\times {\mathrm{p}}_2\right)}^2}{\frac{1-\alpha }{1+\alpha }\times {\mathrm{p}}_2}$

$\Rightarrow 1=\frac{4{\mathrm{\alpha }}^2\times {\mathrm{p}}_2}{1-{\mathrm{\alpha }}^2}$   ....(Y)
Thus, dividing equation (X) by equation(Y)

$\Rightarrow \frac{{\mathrm{k}}_{{\mathrm{p}}_1}}{{\mathrm{k}}_{{\mathrm{p}}_2}}=\frac{{\mathrm{p}}_1}{4{\mathrm{p}}_2}$

$\Rightarrow \frac{3}{1}=\frac{{\mathrm{p}}_1}{4{\mathrm{p}}_2}$

$\therefore \frac{{\mathrm{p}}_1}{{\mathrm{p}}_2}=12$