What fraction of the alcohol would be esterified if the initial mole fraction of the alcohol is 0.33 — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Let initial total moles = 1. Initial alcohol = 1/3, acid = 2/3. Let y moles react. Equilibrium: alcohol = 1/3 - y, acid = 2/3 - y, ester = y, water = y. K_c = 4 = y² / [(1/3 - y)(2/3 - y)]. 4(2/9 - y + y²) = y² => 8/9 - 4y + 4y² = y² => 3y² - 4y + 8/9 = 0 => 27y² - 36y + 8 = 0. Solving for y using the quadratic formula: y = [36 - √(1296 - 864)] / 54 = [36 - √432] / 54 = [36 - 12√3] / 54 = [3 - √3] / 4.5. Using √3 = 1.732, y = (3 - 1.732) / 4.5 = 1.268 / 4.5 ≈ 0.2817. Fraction of alcohol esterified = y / (1/3) = 3y = 3 × 0.2817 = 0.8451 ≈ 0.85. Therefore, correct answer is B.