Ethyl acetate is synthesized from ethanol and acetic acid in a nonreacting solvent by the following reaction:
$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l); K = 2.2$
If the reaction is set up taking $[C H_{3} C O_{2} C_{2} H_{5}]$ = 0.22 M, $[H_{2} O]$ = 0.10 M,
$[C H_{3} C O_{2} H]$ = 0.010 M, $[C_{2} H_{5} O H]$ = 0.010 M, the changes which take place as the system attains equilibrium is/are

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Concentrations of both ethyl acetate and water decrease.

Concentrations of both ethanol and acetic acid increase.

Concentration of ethyl acetate decreases while that of water remains constant.

No change in concentrations of reactants as well as product as the given concentrations correspond to their equilibrium values.

answer is A, C.

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Detailed Solution

$Q = \frac{0.22 \times 0.1}{0.01 \times 0.01} = 220$

As Q>K, the reaction moves in the backward direction to attain equilibrium.

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