Starting with 2 moles and 1 mol of in 1 L flask, the equilibrium mixture required 0.4 moles of MnO4^ — Chemical Equilibrium Chemistry Question
Question
Starting with 2 moles $SO_2$ and 1 mol of $O_2$ in 1 L flask, the equilibrium mixture required 0.4 moles of MnO4^- for complete reaction in acidic medium. $K_c$ for the reaction: 2$SO_2$(g) + $O_2$(g) ⇌ 2$SO_3$(g) is:
💡 Solution & Explanation
Reaction: $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ \textbf{Step 1 — Find equilibrium $n_{\text{SO}_2}$ from permanganate data:} In acidic solution, $\text{MnO}_4^-$ oxidises $\text{SO}_2$ to $\text{SO}_4^{2-}$: $5\text{SO}_2 + 2\text{MnO}_4^- + 2\text{H}_2\text{O} \to 5\text{SO}_4^{2-} + 2\text{Mn}^{2+} + 4\text{H}^+$ Mole ratio: $n_{\text{SO}_2} = \frac{5}{2} \times n_{\text{MnO}_4^-} = \frac{5}{2} \times 0.4 = 1.0\ \text{mol}$ \textbf{Step 2 — ICE table} ($2-2x = 1 \Rightarrow x = 0.5$): \begin{center} \begin{tabular}{lccc} & $2\text{SO}_2$ & $\text{O}_2$ & $2\text{SO}_3$ \\ Initial & 2 & 1 & 0 \\ Change & $-2x$ & $-x$ & $+2x$ \\ Equil. & $2-2x=1$ & $1-x=0.5$ & $2x=1$ \\ \end{tabular} \end{center} \textbf{Step 3 — $K_c$} (V = 1 L): \[ K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 \cdot [\text{O}_2]} = \frac{1^2}{1^2 \times 0.5} = 2 \] \textbf{Answer: A} — $K_c = 2\ \text{M}^{-1}$