1.00 mol of ethanoic acid and 2.00 mol of ethanol react in a rigid vessel at constant temperature to equilibrium: CH3COOH + C2H5OH CH3COOC2H5 + H2O, with K_c = 4.00 (the total volume cancels). The number of moles of ethyl ethanoate at equilibrium is closest to

JEE Advanced levelSingle correct optionDifficulty 3 of 3co213

1.00 mol of ethanoic acid and 2.00 mol of ethanol react in a rigid vessel at constant temperature to equilibrium: $CH X_{3} COOH + C X_{2} H X_{5} OH CH X_{3} COOC X_{2} H X_{5} + H X_{2} O$ , with $K_{c} = 4.00$ (the total volume cancels). The number of moles of ethyl ethanoate at equilibrium is closest to

Let

x

mol of ester form.

K_{c} = x^{2} / [(1 - x) (2 - x)] = 4

gives

3 x^{2} - 12 x + 8 = 0

, so

x = (12 - 48) /6 = 0.845

(the other root exceeds 1). This is closest to 0.85 mol.