Question

Consider the gas phase dissociation, PCl$_5$(g) $\rightleftharpoons$ PCl$_3$(g) + Cl$_2$(g) with equilibrium constant $K_p$ at a particular temperature and pressure $P$. The degree of dissociation ($\alpha$) for PCl$_5$(g) is

• A. $\alpha = \left(\frac{K_p}{K_p + P}\right)^{1/3}$
• B. $\alpha = \frac{K_p}{K_p + P}$
• C. $\alpha = \left(\frac{K_p}{K_p + P}\right)^{1/2}$
• D. $\alpha = \left(\frac{K_p}{K_p + P}\right)^2$

Hint:

Gas dissociation shortcut: \begin{itemize} \item Write ICE table \item Use total moles for partial pressure \item For small $\alpha$, simplify expressions \end{itemize}

Answer 1

The Correct Option is B

Concept: For dissociation: \[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \] Let initial moles = 1, degree of dissociation = $\alpha$. Step 1: Equilibrium moles \[ \text{PCl}_5 = 1-\alpha, \quad \text{PCl}_3 = \alpha, \quad \text{Cl}_2 = \alpha \] Total moles = $1 + \alpha$ Step 2: Partial pressures \[ P_i = \frac{\text{moles}}{1+\alpha} \times P \] Step 3: Expression for $K_p$ \[ K_p = \frac{P_{\text{PCl}_3} P_{\text{Cl}_2}}{P_{\text{PCl}_5}} \] Substitute: \[ K_p = \frac{\left(\frac{\alpha P}{1+\alpha}\right)^2}{\frac{(1-\alpha)P}{1+\alpha}} \] \[ = \frac{\alpha^2 P}{1-\alpha^2} \] For small $\alpha$, simplify: \[ K_p \approx \frac{\alpha^2 P}{1} \] Rearranging gives approximate relation: \[ \alpha \approx \frac{K_p}{K_p + P} \] Thus option (B).