Chemical Equilibrium Question 4
Question 4 - 2024 (31 Jan Shift 2)
$A _{(g)} \rightleftharpoons B _{(g)}+\frac{C}{2}(g)$ 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} & A _{(g)} \rightleftharpoons B _{(g)}+\frac{C}{2}(g) \\ & t=t _{eq}(1-\alpha) \quad \alpha \quad \frac{\alpha}{2} \\ & P _B=\frac{\alpha}{\left(1+\frac{\alpha}{2}\right)} \cdot P, P _A=\frac{(1-\alpha)}{\left(1+\frac{\alpha}{2}\right)} \cdot P, P _C=\frac{\frac{\alpha}{2}}{\left(1+\frac{\alpha}{2}\right)} \cdot P \\ & K _P=\frac{P _B \cdot P _C^{\frac{1}{2}}}{P _A} \\ & =\frac{(\alpha)^{\frac{3}{2}}(P)^{\frac{1}{2}}}{(1-\alpha)(2+\alpha)^{\frac{1}{2}}} \end{aligned} $
