1. Write the reaction and assume initial moles

Given: $A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2}C_{(g)}$

Let initially we take 1 mole of $A$ and no products. 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

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

3. Partial pressures in terms of total pressure $P$

Using $p_i = \frac{n_i}{n_{\text{total}}}P$

So,

$p_A=\frac{1-\alpha}{(2+\alpha)/2}P=\frac{2(1-\alpha)}{2+\alpha}P$

$p_B=\frac{\alpha}{(2+\alpha)/2}P=\frac{2\alpha}{2+\alpha}P$

$p_C=\frac{\alpha/2}{(2+\alpha)/2}P=\frac{\alpha}{2+\alpha}P$

4. Expression for $K_P$

For the reaction $A \rightleftharpoons B + \frac{1}{2}C$

$K_P=\frac{p_B\,(p_C)^{1/2}}{p_A}$

Substitute the partial pressures:

$K_P=\frac{\left(\frac{2\alpha P}{2+\alpha}\right)\left(\frac{\alpha P}{2+\alpha}\right)^{1/2}}{\left(\frac{2(1-\alpha)P}{2+\alpha}\right)}$

Cancel the factor of 2:

$K_P=\frac{\alpha P}{(1-\alpha)(2+\alpha)}\left(\frac{\alpha P}{2+\alpha}\right)^{1/2}$

Now combine terms:

$K_P=\frac{\alpha^{3/2}P^{3/2}}{(1-\alpha)(2+\alpha)^{3/2}}\cdot \frac{1}{P}$

More directly,

$K_P=\frac{\alpha}{1-\alpha}\left(\frac{\alpha P}{2+\alpha}\right)^{1/2}$

Hence,

$K_P=\frac{\alpha^{3/2}P^{1/2}}{(1-\alpha)(2+\alpha)^{1/2}}$

5. Compare with the options

This matches:

$\boxed{K_P=\frac{\alpha^{3/2}P^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}}$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.