Question

For a reaction at equilibrium
\(\mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} + \frac{1}{2}\mathrm{C(g)}\)
the relation between dissociation constant (K), degree of dissociation (\(\alpha\)) and equilibrium pressure (p) is given by :

• A. \(K=\dfrac{\alpha^{\frac{1}{2}}p^{\frac{3}{2}}}{\left(1+\frac{3}{2}\alpha\right)^{\frac{1}{2}}(1-\alpha)}\)
• B. \(K=\dfrac{\alpha^{\frac{3}{2}}p^{\frac{1}{2}}}{(2+\alpha)^{\frac{1}{2}}(1-\alpha)}\)
• C. \(K=\dfrac{(\alpha p)^{\frac{3}{2}}}{\left(1+\frac{3}{2}\alpha\right)^{\frac{1}{2}}(1-\alpha)}\)
• D. \(K=\dfrac{(\alpha p)^{\frac{3}{2}}}{(1+\alpha)(1-\alpha)^{\frac{1}{2}}}\)

Hint:

Start with 1 mol of A, write moles of A, B, C at equilibrium, convert to partial pressures using mole fractions, and substitute in \(K_p = p_B\,p_C^{1/2}/p_A\).

Answer 1

The Correct Option is B

Initial: A = 1 mol. At equilibrium: A = \(1-\alpha\), B = \(\alpha\), C = \(\alpha/2\).
Total moles = \(1-\alpha+\alpha+\frac{\alpha}{2} = 1+\frac{\alpha}{2} = \frac{2+\alpha}{2}\).
Partial pressures: \(p_A = \frac{(1-\alpha)p}{(2+\alpha)/2}\), \(p_B = \frac{\alpha p}{(2+\alpha)/2}\), \(p_C = \frac{(\alpha/2)p}{(2+\alpha)/2} = \frac{\alpha p}{2+\alpha}\).
\(K = \frac{p_B\,p_C^{1/2}}{p_A} = \frac{\frac{2\alpha p}{2+\alpha}\left(\frac{\alpha p}{2+\alpha}\right)^{1/2}}{\frac{2(1-\alpha)p}{2+\alpha}} = \frac{\alpha}{1-\alpha}\left(\frac{\alpha p}{2+\alpha}\right)^{1/2}\)
\(K = \frac{\alpha^{3/2}p^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}\)
Answer: (2) \(K=\dfrac{\alpha^{3/2}p^{1/2}}{(2+\alpha)^{1/2}(1-\alpha)}\)