Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\; \; \; \; \; \; \; \; \; 2{\mathrm{AB}}_2\left(\text{g}\right)\rightleftharpoons \mathrm{2AB}\left(\text{g}\right)+B_2\left(g\right)$
At equilibrium$\text{2(1}-\text{x)}$$\text{2x}$$\text{x}$
Total moles =$\text{(2}\text{+}\text{x)}$
Partial pressures
${\text{P}}_{{\text{AB}}_{\text{2}}}\text{=}\frac{\text{2(1}-\text{x)}}{\text{2}\text{+}\text{x}}\text{p,}{\text{P}}_{\text{AB}}\text{=}\left(\frac{\text{2x}}{\text{2}\text{+}\text{x}}\right)\text{p,}{\text{P}}_{{\text{B}}_{\text{2}}}\text{=}\left(\frac{\text{x}}{\text{2+x}}\right)\text{p}$
${\text{K}}_{\text{p}}=\frac{\text{p}{\text{AB}}^2{\text{pB}}_2}{({\text{pAB}}_2{)}^2}=\frac{\left[\right(\frac{2x}{2+x}\left)\text{p}{]}^2\right[\frac{x\text{p}}{2+x}]}{[\frac{2(1-x)}{2+x}\text{p}{]}^2}$
${\text{K}}_{\text{p}}=\frac{4{\text{x}}^{\text{3}}\text{P}}{4\left(2+\text{x}\right){(1-\text{x})}^2}$
$\left(2+\text{x}\right)\to 2$($\text{x}$is small)
$\left(1-\text{x}\right)\to 1$
${\text{K}}_{\text{P}}=\frac{4{\text{x}}^{\text{3}}\text{p}}{8}=\frac{{\text{x}}^{\text{3}}\text{p}}{2}$
$\text{x}=\sqrt[3]{\frac{2\text{kp}}{\text{p}}}$