Chemical Equilibrium Question 4
Question 4 - 2024 (31 Jan Shift 2)
$\mathrm{A}{(\mathrm{g})} \rightleftharpoons \mathrm{B}{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{~g})$ The correct relationship between $K_{P}, \alpha$ and equilibrium pressure $P$ is
(1) $K_{P}=\frac{\alpha^{1 / 2} P^{1 / 2}}{(2+\alpha)^{1 / 2}}$
(2) $K_{P}=\frac{\alpha^{3 / 2} P^{1 / 2}}{(2+\alpha)^{1 / 2}(1-\alpha)}$
(3) $K_{P}=\frac{\alpha^{1 / 2} P^{3 / 2}}{(2+\alpha)^{3 / 2}}$
(4) $K_{P}=\frac{\alpha^{1 / 2} P^{1 / 2}}{(2+\alpha)^{3 / 2}}$
Show Answer
Answer (2)
Solution
$$ \begin{aligned} & \mathrm{A}{(\mathrm{g})} \rightleftharpoons \mathrm{B}{(\mathrm{g})}+\frac{\mathrm{C}}{2}(\mathrm{~g}) \ & \mathrm{t}=\mathrm{t}{\mathrm{eq}}(1-\alpha) \quad \alpha \quad \frac{\alpha}{2} \ & \mathrm{P}{\mathrm{B}}=\frac{\alpha}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}{\mathrm{A}}=\frac{(1-\alpha)}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P}, \mathrm{P}{\mathrm{C}}=\frac{\frac{\alpha}{2}}{\left(1+\frac{\alpha}{2}\right)} \cdot \mathrm{P} \ & \mathrm{K}{\mathrm{P}}=\frac{\mathrm{P}{\mathrm{B}} \cdot \mathrm{P}{\mathrm{C}}^{\frac{1}{2}}}{\mathrm{P}{\mathrm{A}}} \ & =\frac{(\alpha)^{\frac{3}{2}}(\mathrm{P})^{\frac{1}{2}}}{(1-\alpha)(2+\alpha)^{\frac{1}{2}}} \end{aligned} $$
