$1.0 m o l e$ of ethyl alcohol and $1.0 m o l e$ of acetic acid are mixed. At equilibrium, $0.666 m o l e$ of ester is formed. The value of equilibrium constant is:

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Solution

The correct option is C $4$
${C H}_{3} C O O H + C_{2} H_{5} O H ⇌ {C H}_{3} {C O}_{2} C_{2} H_{5} + H_{2} O$
$1.0$ $1.0$ $0$ $0$
Eqn $(1 - 0.666)$ $(1 - 0.666)$ $0.666$ $0.666$
$∴$ $K_{C} = \frac{[{C H}_{3} {C O}_{2} C_{2} H_{5}] [H_{2} O]}{[{C H}_{3} {C O}_{2} H] [C_{2} H_{5} O H]}$
$= \frac{(0.666) (0.666)}{(0.333) (0.333)}$
$= 4$

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