When ethyl alcohol C2H5OH(l) and acetic acid CH3COOH (l)) are mixed together in equimolar ratio at 27°C, 33% of each is converted into ester. Then the KC for the equilibrium,C2H5OH(l)+CH3CHOOH(l) ⇌CH3COOC2H5(l)+H2O(l)

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When ethyl alcohol $(C_{2} H_{5} OH (l))$ and acetic acid $({CH}_{3} COOH (l))$ are mixed together in equimolar ratio at 27°C, 33% of each is converted into ester. Then the $K_{C}$ for the equilibrium,

$C_{2} H_{5} OH (l) + {CH}_{3} CHOOH (l)$ $⇌ {CH}_{3} {COOC}_{2} H_{5} (l) + H_{2} O (l)$

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$\frac{1}{9}$

$\frac{1}{4}$

answer is B.

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Detailed Solution

$C_{2} H_{5} OH (l) + {CH}_{3} COOH (l)$

$a \times \frac{2}{3} a \times \frac{2}{3}$

$⇌ {CH}_{3} {COOC}_{2} H_{5} (l) + H_{2} O (l)$

$\frac{a}{3} \frac{a}{3}$

$K = \frac{\frac{a}{3} \times \frac{a}{3}}{\frac{2 a}{3} \times \frac{2 a}{3}} = \frac{1}{4}$

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