$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$

In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of

little conc. $H_{2} S O_{4}$ . On equilibrium being attained

1 mole of ethyl acetate is formed

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2 mole of ethyl acetate are formed

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$\frac{1}{2}$ moles of ethyl acetate is formed

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$\frac{2}{3}$ moles of ethyl acetate is formed

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Solution

The correct option is D
$\frac{2}{3}$ moles of ethyl acetate is formed

In presence of little $H_{2} S O_{4}$ (as catalyst) about $\frac{2}{3} m o l e$ of each of

$C H_{3} C O O H$ and $C_{2} H_{5} O H$ react to form $\frac{2}{3} m o l e$ of the product at

equilibrium.

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Similar questions

$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l); K_{c} = 4$

In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little conc. $H_{2} S O_{4}$ . On equilibrium being attained

Q. In the esterfication

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

an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction = 0.333. Calculate the equilibrium constant.

Q. In the esterification

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

an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction

= 0.333.

Calculate the equilibrium constant :

Q. Acids of boron on heating with alcohol in presence of conc.

H_{2} S O_{4}

gives:

Q. Acetic acid on warming with hydrazoic acid in presence of conc.

H_{2} S O_{4}

gives:

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CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l) In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little conc. H2SO4. On equilibrium being attained

The correct option is D 23 moles of ethyl acetate is formed In presence of little H2SO4 (as catalyst) about 23 mole of each of CH3COOH and C2H5OH react to form 23 mole of the product at equilibrium.

$C H_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$

In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of

little conc. $H_{2} S O_{4}$ . On equilibrium being attained

The correct option is D
$\frac{2}{3}$ moles of ethyl acetate is formed

In presence of little $H_{2} S O_{4}$ (as catalyst) about $\frac{2}{3} m o l e$ of each of

$C H_{3} C O O H$ and $C_{2} H_{5} O H$ react to form $\frac{2}{3} m o l e$ of the product at

equilibrium.