CH_3COOH_{(l)} + C_2H_5OH_{(l)} rightleftharp | vedclass

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(D) The reaction is an esterification reaction: $CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O$. For this reaction,the equilibrium const

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$2/3$ mole of ethyl acetate is formed

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(D) The reaction is an esterification reaction: $CH_3COOH + C_2H_5OH ightleftharpoons CH_3COOC_2H_5 + H_2O$. For this reaction,the equilibrium constant $K_c$ is $4$. Let $x$ be the number of moles of ethyl acetate formed at equilibrium. Initial moles: $1$ mole of acid,$1$ mole of alcohol,$0$ mole of ester,$0$ mole of water. At equilibrium: $(1-x)$ moles of acid,$(1-x)$ moles of alcohol,$x$ moles of ester,$x$ moles of water. $K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]} = \frac{x 	imes x}{(1-x)(1-x)} = 4$. Taking the square root on both sides: $\frac{x}{1-x} = 2$. $x = 2 - 2x \implies 3x = 2 \implies x = 2/3$. Thus,$2/3$ mole of ethyl acetate is formed.

$CH_3COOH_{(l)} + C_2H_5OH_{(l)} \rightleftharpoons CH_3COOC_2H_{5_{(l)}} + H_2O_{(l)}$. In the above reaction,one mole of each of acetic acid and alcohol are heated in the presence of a little conc. $H_2SO_4$. On equilibrium being attained:

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