Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\text{X}\left(\text{g}\right)\rightleftharpoons \text{2Y}\left(\text{g}\right)$

1                    0

At equilibrium$1-\text{x}$$\text{2x}$

Total no. of moles =$1+\text{x}$

${\text{K}}_{{\text{p}}_{\text{1}}}\text{=}\frac{{\left[\frac{\text{2x}}{\text{1+x}}{\text{.p}}_{\text{1}}\right]}^{\text{2}}}{\left(\frac{\text{1}-\text{x}}{\text{1+x}}\right){\text{.p}}_{\text{1}}}\text{=}\frac{{\text{4x}}^{\text{2}}}{\left(\text{1}-\text{x}\right)}\frac{{\text{p}}_{\text{1}}}{\left(\text{1+x}\right)}\text{=}\frac{{\text{4x}}^{\text{2}}{\text{p}}_{\text{1}}}{\left(\text{1}-{\text{x}}^{\text{2}}\right)}$

$\text{Z}\left(\text{g}\right)\rightleftharpoons \text{P}\left(\text{g}\right)\text{+Q}\left(\text{g}\right)$

1                             0            0

At equilibrium$1-\text{x}$$\text{x}$$\text{x}$

Similarly,${\text{K}}_{{\text{p}}_{\text{2}}}\text{=}\frac{{\text{x}}^{\text{2}}}{\left(\text{1}-\text{x}\right)}\frac{{\text{p}}_{\text{2}}}{\left(\text{1+x}\right)}\text{=}\frac{{\text{x}}^{\text{2}}{\text{p}}_{\text{2}}}{\left(\text{1}-{\text{x}}^{\text{2}}\right)}$

$\therefore \frac{{\text{K}}_{\text{p1}}}{{\text{K}}_{\text{p2}}}\text{=}\frac{\text{1}}{\text{9}}$

So,$\frac{\text{4}\text{×}{\text{p}}_{\text{1}}}{{\text{p}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{9}}$

$\Rightarrow \frac{{\text{p}}_{\text{1}}}{{\text{p}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{36}}$