At 200°C, dissociates as follows: (g) ⇌ (g) + (g). It was found that the equilibrium vapours are 62 — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Reaction: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ Molar mass of PCl$_5$ = 31 + 5(35.5) = 208.5 g/mol. \textbf{Theoretical (undissociated) vapour density:} \[ D = \frac{M_{\text{PCl}_5}}{2} = \frac{208.5}{2} = 104.25 \] \textbf{Observed (equilibrium) vapour density:} $d = 62$ (given) \textbf{Degree of dissociation} for a reaction where 1 mole gives 2 moles ($y = 2$ product moles per mole of PCl$_5$): \[ \alpha = \frac{D - d}{d(y - 1)} = \frac{104.25 - 62}{62 \times (2 - 1)} = \frac{42.25}{62} \approx 0.681 \] \textbf{Answer: D} — $\alpha \approx 68.1\%$