Q14 / 180 · COMEDK 2026 (9 May Shift 1) · Chemistry
Consider the gaseous equilibrium
$2\text{AB}_{2(g)} \rightleftharpoons 2\text{AB}_{(g)} + \text{B}_{2(g)}$
The expression relating the degree of dissociation ($\alpha$) and equilibrium constant ($K_p$) and total pressure P is:
$2\text{AB}_{2(g)} \rightleftharpoons 2\text{AB}_{(g)} + \text{B}_{2(g)}$
The expression relating the degree of dissociation ($\alpha$) and equilibrium constant ($K_p$) and total pressure P is:
- A$\left(\dfrac{K_p}{2P}\right)^{1/3}$
- Correct answerB$\left(\dfrac{2K_p}{P}\right)^{1/3}$
- C$\left(\dfrac{2K_p}{P}\right)^{2}$
- D$\left(\dfrac{K_p}{2P}\right)^{1/2}$
Solution
The given reaction is $2\text{AB}_{2(g)} \rightleftharpoons 2\text{AB}_{(g)} + \text{B}_{2(g)}$
Let the initial moles of $\text{AB}_2$ be $2$.
Moles at equilibrium:
$\text{AB}_2 = 2(1-\alpha)$
$\text{AB} = 2\alpha$
$\text{B}_2 = \alpha$
Total moles at equilibrium $= 2(1-\alpha) + 2\alpha + \alpha = 2 + \alpha$
Partial pressures at equilibrium:
$P_{\text{AB}_2} = \dfrac{2(1-\alpha)}{2+\alpha}P$
$P_{\text{AB}} = \dfrac{2\alpha}{2+\alpha}P$
$P_{\text{B}_2} = \dfrac{\alpha}{2+\alpha}P$
The equilibrium constant $K_p$ is given by:
$K_p = \dfrac{(P_{\text{AB}})^2 P_{\text{B}_2}}{(P_{\text{AB}_2})^2}$
Substituting the partial pressures:
$K_p = \dfrac{\left(\dfrac{2\alpha}{2+\alpha}P\right)^2 \left(\dfrac{\alpha}{2+\alpha}P\right)}{\left(\dfrac{2(1-\alpha)}{2+\alpha}P\right)^2}$
$K_p = \dfrac{4\alpha^2 P^2}{(2+\alpha)^2} \times \dfrac{\alpha P}{2+\alpha} \times \dfrac{(2+\alpha)^2}{4(1-\alpha)^2 P^2}$
$K_p = \dfrac{\alpha^3 P}{(2+\alpha)(1-\alpha)^2}$
Assuming $\alpha$ is very small ($\alpha \ll 1$), we can approximate $2+\alpha \approx 2$ and $1-\alpha \approx 1$.
$K_p = \dfrac{\alpha^3 P}{2}$
$\alpha^3 = \dfrac{2K_p}{P}$
$\alpha = \left(\dfrac{2K_p}{P}\right)^{1/3}$
Answer: $\left(\dfrac{2K_p}{P}\right)^{1/3}$
Let the initial moles of $\text{AB}_2$ be $2$.
Moles at equilibrium:
$\text{AB}_2 = 2(1-\alpha)$
$\text{AB} = 2\alpha$
$\text{B}_2 = \alpha$
Total moles at equilibrium $= 2(1-\alpha) + 2\alpha + \alpha = 2 + \alpha$
Partial pressures at equilibrium:
$P_{\text{AB}_2} = \dfrac{2(1-\alpha)}{2+\alpha}P$
$P_{\text{AB}} = \dfrac{2\alpha}{2+\alpha}P$
$P_{\text{B}_2} = \dfrac{\alpha}{2+\alpha}P$
The equilibrium constant $K_p$ is given by:
$K_p = \dfrac{(P_{\text{AB}})^2 P_{\text{B}_2}}{(P_{\text{AB}_2})^2}$
Substituting the partial pressures:
$K_p = \dfrac{\left(\dfrac{2\alpha}{2+\alpha}P\right)^2 \left(\dfrac{\alpha}{2+\alpha}P\right)}{\left(\dfrac{2(1-\alpha)}{2+\alpha}P\right)^2}$
$K_p = \dfrac{4\alpha^2 P^2}{(2+\alpha)^2} \times \dfrac{\alpha P}{2+\alpha} \times \dfrac{(2+\alpha)^2}{4(1-\alpha)^2 P^2}$
$K_p = \dfrac{\alpha^3 P}{(2+\alpha)(1-\alpha)^2}$
Assuming $\alpha$ is very small ($\alpha \ll 1$), we can approximate $2+\alpha \approx 2$ and $1-\alpha \approx 1$.
$K_p = \dfrac{\alpha^3 P}{2}$
$\alpha^3 = \dfrac{2K_p}{P}$
$\alpha = \left(\dfrac{2K_p}{P}\right)^{1/3}$
Answer: $\left(\dfrac{2K_p}{P}\right)^{1/3}$
Hence, the correct answer is option: 'B'.