A quantity of 60 g CH3COOH and 46 g CH3CH2OH reacts in 5 L flask to form ethyl acetate: CH3COOH(l) + — Chemical Equilibrium Chemistry Question
Question
A quantity of 60 g CH3COOH and 46 g CH3CH2OH reacts in 5 L flask to form ethyl acetate: CH3COOH(l) + $C_2H_5OH$(l) ⇌ CH3COOC2H5(l) + $H_2O$(l). On taking 120 g CH3COOH and 46 g CH3CH2OH, CH3COOC2H5 formed at equilibrium is:
💡 Solution & Explanation
Reaction: $\text{CH}_3\text{COOH}(l) + \text{C}_2\text{H}_5\text{OH}(l) \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5(l) + \text{H}_2\text{O}(l)$ \textbf{Step 1 — Find $K_c$ from Case 1:} (60 g CH$_3$COOH = 1 mol, 46 g C$_2$H$_5$OH = 1 mol) For a liquid-phase esterification, $K_c$ is expressed in terms of moles (volume cancels): \[ K_c = \frac{n_{\text{ester}} \cdot n_{\text{H}_2\text{O}}}{n_{\text{acid}} \cdot n_{\text{alcohol}}} \] Standard result (or from equilibrium data): at equimolar start with $K_c = 1$: $n_{\text{ester}} = n_{\text{H}_2\text{O}} = x$, $n_{\text{acid}} = n_{\text{alcohol}} = 1-x$. $K_c = x^2/(1-x)^2 = 1 \Rightarrow x = 0.5\ \text{mol}$ (ester formed in Case 1). \textbf{Step 2 — Case 2:} (120 g CH$_3$COOH = 2 mol, 46 g C$_2$H$_5$OH = 1 mol) Let $x$ mol ester (and $x$ mol H$_2$O) form at equilibrium: \[ K_c = \frac{x \cdot x}{(2-x)(1-x)} = 1 \] \[ x^2 = (2-x)(1-x) = 2 - 3x + x^2 \implies 3x = 2 \implies x = \frac{2}{3}\ \text{mol} \] \textbf{Step 3 — Mass of ester:} $M_{\text{CH}_3\text{COOC}_2\text{H}_5} = 88$ g/mol \[ m = \frac{2}{3} \times 88 = 58.67\ \text{g} \] \textbf{Answer: D} — 58.67 g