Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\begin{array}{lccccc}& {\mathrm{AB}}_2(\mathrm{g})& \rightleftharpoons & \mathrm{AB}(\mathrm{g})& +& \frac{1}{2}{\mathrm{B}}_2(\mathrm{g})\\ t=0& p_0& & & & \\ t=t_{\text{eq }}& p_0(1-x)& & p_{0}x& & \frac{p_{0}x}{2}\end{array}$ $\mathrm{p}={\mathrm{p}}_0\mathrm{x}-{\mathrm{p}}_0\mathrm{x}+{\mathrm{p}}_0\mathrm{x}+\frac{{\mathrm{p}}_0\mathrm{x}}{2}$ $\begin{array}{c}p=p_0\left(1+\frac{x}{2}\right)\\ p_0=\frac{p}{\left(1+\frac{x}{2}\right)}\\ K_p=\frac{\left(p_{AB}\right){\left(p_{B_2}\right)}^{1/2}}{\left(p_{{\mathrm{AB}}_2}\right)}\\ K_p=\frac{\left(p_{0}x\right){\left(\frac{p_{0}x}{2}\right)}^{1/2}}{p_0(1-x)}\end{array}$ ${\mathrm{K}}_{\mathrm{p}}=\frac{\frac{\mathrm{px}}{\left(1+\frac{x}{2}\right)}{\left(\frac{\mathrm{p}}{1+\frac{x}{2}}\times \frac{\mathrm{x}}{2}\right)}^{1/2}}{\frac{\mathrm{p}(1-\mathrm{x})}{\left(1+\frac{x}{2}\right)}}$ Since $x\ll 1$ $\begin{array}{rl}& K_p=\frac{{\mathrm{p}}^{1/2}{\mathrm{x}}^{3/2}}{2^{1/2}}\\ & \mathrm{x}=\sqrt[3]{\frac{2{\mathrm{K}}_{\mathrm{p}}^2}{\mathrm{p}}}\end{array}$