Two moles of an equimolar mixture of two alcohols R1-OH and R2-OH are esterified with one mole of ac — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Two alcohols (R$_1$-OH and R$_2$-OH, 1 mol each) + 1 mol acetic acid. At equilibrium, 80\% acid consumed, ester ratio R$_1$OAc : R$_2$OAc = 3 : 2. \textbf{Step 1 — Equilibrium amounts:} Total ester = 0.80 mol (80\% of 1 mol acid consumed). Acid remaining = 0.20 mol. Ratio 3:2 $\Rightarrow$ $n_{\text{R}_1\text{OAc}} = 0.48$ mol, $n_{\text{R}_2\text{OAc}} = 0.32$ mol (sum = 0.80). Water produced = 0.80 mol. Alcohol remaining: $n_{\text{R}_1\text{OH}} = 1-0.48 = 0.52$ mol, $n_{\text{R}_2\text{OH}} = 1-0.32 = 0.68$ mol. \textbf{Step 2 — $K_1$ for R$_1$-OH + CH$_3$COOH $\rightleftharpoons$ R$_1$OAc + H$_2$O:} \textit{Textbook approach} (treats each reaction's water independently = ester formed): \[ K_1 = \frac{n_{\text{R}_1\text{OAc}}^2}{n_{\text{CH}_3\text{COOH}} \cdot n_{\text{R}_1\text{OH}}} = \frac{(0.48)^2}{0.20 \times 0.52} = \frac{0.2304}{0.104} \approx 2.2 \] (This is the intended formula since $n_{\text{H}_2\text{O}}$ from reaction 1 alone $= n_{\text{R}_1\text{OAc}} = 0.48$ in the independent-reaction model.) \textit{Alternative (using total water 0.80 mol):} \[ K_1 = \frac{0.48 \times 0.80}{0.20 \times 0.52} = \frac{0.384}{0.104} = 3.69 \] The textbook answer uses the first formula. \textbf{Answer: B} — $K_1 \approx 2.2$