Chemistry - Chemical Equilibrium Question with Solution | TestHub

The reaction is given by: ${\mathrm{A}}_{(\mathrm{s})}\rightleftharpoons 2{\mathrm{B}}_{(\mathrm{g})}+3{\mathrm{C}}_{(\mathrm{g})}$. The equilibrium constant is $K_{\mathrm{P}}$.

$$\begin{array}{rl}& t=t_{eq}{P}_B{P}_C\\ & K_P=P_B^2{P}_C^3\\ & \text{ ATQ, }{P}_C^{\prime }=2P_C;P_B^{\prime }\\ & K_P={\left(P_B^{\prime }\right)}^2{\left(P_C^{\prime }\right)}^3={\left(P_B^{\prime }\right)}^2{\left(2P_C\right)}^3\end{array}$$

Now, equating the expressions for $K_{\mathrm{P}}$ from equations (1) and (2):

$\Rightarrow {\mathrm{P}}_{\mathrm{B}}^2{\mathrm{P}}_{\mathrm{C}}^3={\left({\mathrm{P}}_{\mathrm{B}}^{\prime }\right)}^{2}8{\left({\mathrm{P}}_{\mathrm{C}}\right)}^3$

$\Rightarrow \frac{P_B^2}{8}={\left(P_B^{\prime }\right)}^2$

$\Rightarrow P_B^{\prime }=\frac{1}{2\sqrt{2}}{P}_B$