1. Assume initial moles

Let initially we have $1$ mole of $A$.

For the reaction $A(g) \rightleftharpoons B(g) + \frac12 C(g)$ if degree of dissociation is $\alpha$, then at equilibrium:

• moles of $A = 1-\alpha$
• moles of $B = \alpha$
• moles of $C = \frac{\alpha}{2}$

2. Total moles at equilibrium

$n_{\text{total}}=(1-\alpha)+\alpha+\frac{\alpha}{2}=1+\frac{\alpha}{2}$

3. Partial pressures

If total pressure at equilibrium is $p$, then $p_A=\frac{1-\alpha}{1+\alpha/2}p$ $p_B=\frac{\alpha}{1+\alpha/2}p$ $p_C=\frac{\alpha/2}{1+\alpha/2}p$

4. Write equilibrium constant in terms of partial pressures

For the reaction, $K=\frac{p_B\,(p_C)^{1/2}}{p_A}$

Substitute the partial pressures: $K=\frac{\left(\frac{\alpha p}{1+\alpha/2}\right)\left(\frac{(\alpha/2)p}{1+\alpha/2}\right)^{1/2}}{\left(\frac{(1-\alpha)p}{1+\alpha/2}\right)}$

5. Simplify

First cancel the common factor: $K=\frac{\alpha}{1-\alpha}\left(\frac{(\alpha/2)p}{1+\alpha/2}\right)^{1/2}$

So, $K=\frac{\alpha}{1-\alpha}\cdot \frac{(\alpha p/2)^{1/2}}{(1+\alpha/2)^{1/2}}$

$K=\frac{\alpha^{3/2}p^{1/2}}{(1-\alpha)\,[2(1+\alpha/2)]^{1/2}}$

Now, $2\left(1+\frac\alpha2\right)=2+\alpha$

Hence, $K=\frac{\alpha^{3/2}p^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}$

or $K=\frac{\alpha^{3/2}p^{1/2}}{\sqrt{2+\alpha}\,(1-\alpha)}$

6. Match with options

This matches Option B: $K = \frac{\alpha^{3/2}p^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}$

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.