1.0 mole of ethyl alcohol and 1.0 mole of acetic acid are mixed. At equilibrium, 0.666 mole of ester is formed. The value of equilibrium constant is

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\frac{1}{4}

\frac{1}{2}

answer is C.

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Detailed Solution

$large {C_2}{H_5}OH + C{H_3}COOH rightleftharpoons C{H_3}COO{C_2}{H_5} + {H_2}O$

$large {left( {{n_{{C_2}{H_5}OH}}} right)_i} = 1$

$large {left( {{n_{{CH_3}COOH}}} right)_i} = 1$

1 mole of C₂H₅OH react with 1 mole of CH₃COOH gives 1 mole of CH₃COOC₂H₅ and 1 mole of H₂O

'X₁' mole of C₂H₅OH react with 'X₂' mole of CH₃COOH gives 0.666 mole of COOC₂H₅ and X₃moles of H₂O

X₁ = X₂ = X₃ = 0.666

Stoichiometry	$large mathop {{C_2}{H_5}OH}limits^{1,mole} +$	$large mathop {C{H_3}COOH}limits^{1,mole}$	$large rightleftharpoons$	$large mathop {C{H_3}COO{C_2}{H_5}}limits^{1,mole} +$	$large mathop {{H_2}O}limits^{1,mole}$
Initial moles	1	1		0	0
Moles at equilibrium	(1 - 0.666)	(1 - 0.666)		0.666	0.666