The reaction $C H_{3} C O O H + C_{2} H_{5} O H ⇋ C H_{3} C O O C_{2} H_{5} + H_{2} O$ can be made to proceed in the forward direction by:

increasing the temperature

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sudden cooling of the reaction mixture

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conducting the reaction in the presence of a small quantity of

N a O H

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taking excess of

C_{2} H_{5} O H

and

C H_{3} C O O H

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Solution

The correct option is C taking excess of $C_{2} H_{5} O H$ and $C H_{3} C O O H$
The equilibrium reaction is $C H_{3} C O O H + C_{2} H_{5} O H ⇋ C H_{3} C O O C_{2} H_{5} + H_{2} O$ .
When an excess of either $C H_{3} C O O H$ or $C_{2} H_{5} O H$ is taken, its concentration increases. The equilibrium will shift in the forward direction so that the effect of the increased concentration of $C H_{3} C O O H$ or $C_{2} H_{5} O H$ is nullified as more and more $C H_{3} C O O H$ or $C_{2} H_{5} O H$ reacts to form products.

Suggest Corrections

Similar questions

C H_{3} C O O H + C_{2} H_{5} O H \Leftrightarrow C H_{3} C O O C_{2} H_{5} + H_{2} O

C H_{3} C O O C_{2} H_{5} + H_{2} O^{+} C H_{3} C O O H + C_{2} H_{5} O H

1.1 \times 10^{- 2}

1.5 \times 10^{- 3}

C H_{3} C O O C_{2} H_{5} + H_{2} O ⇌ C H_{3} C O O H + C_{2} H_{5} O H

1.1 \times 10^{- 2}

1.5 \times 10^{- 3}

C H_{3} C O O C_{2} H_{5} + H_{2} O ⇌ C H_{3} C O O H + C_{2} H_{5} O H

1.1 \times 10^{- 2}

1.5 \times 10^{- 3}

C H_{3} C O O C_{2} H_{5} + H^{+} ⇌ C H_{3} C O O H + C_{2} H_{5} O H

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