The Deacon reaction is the oxidation of by : (g) + 1/4 (g) ⇌ 1/2 (g) + 1/2 (g). At a pressure of 730 — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Reaction: $\text{HCl}(g) + \tfrac{1}{4}\text{O}_2(g) \rightleftharpoons \tfrac{1}{2}\text{Cl}_2(g) + \tfrac{1}{2}\text{H}_2\text{O}(g)$ Total pressure = 730 mm; initial mixture: 8\% HCl, 92\% O$_2$. \textbf{Basis: 100 mol total.} Degree of decomposition of HCl = 0.08. $n_{\text{HCl}}^0 = 8$, $n_{\text{O}_2}^0 = 92$. HCl reacted: $0.08 \times 8 = 0.64$ mol. \textbf{Moles consumed/produced:} \[ \Delta n_{\text{O}_2} = -\tfrac{1}{4}(0.64) = -0.16,\quad \Delta n_{\text{Cl}_2} = +\tfrac{1}{2}(0.64) = +0.32,\quad \Delta n_{\text{H}_2\text{O}} = +0.32 \] \textbf{Equilibrium moles:} \[ n_{\text{HCl}} = 7.36,\quad n_{\text{O}_2} = 91.84,\quad n_{\text{Cl}_2} = 0.32,\quad n_{\text{H}_2\text{O}} = 0.32 \] \[ n_{\text{total}} = 7.36 + 91.84 + 0.32 + 0.32 = 99.84 \] \textbf{Partial pressure of O$_2$:} \[ P_{\text{O}_2} = \frac{91.84}{99.84} \times 730 \approx 671.5\ \text{mm} \] \textbf{Note on answer options:} The textbook key gives D (670.43 mm), which arises from using the initial 100 mol as denominator: $(91.84/100) \times 730 = 670.43$ mm. This is technically incorrect since the total moles changed. The correct calculation using equilibrium total moles gives $\approx 671.5$ mm (option A, 671.6 mm). The official answer key is D. \textbf{Answer key: D (670.43 mm); correct value: $\approx$ 671.5 mm (option A)}