CH 3 COOH l - C 2 H 5 OH l - CH 3 COOC 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 SO 4- On equilibrium being attainedA- 1 mole of ethyl acetate is formedB- 2 mole of ethyl acetate are formedC- 1 - 2 moles of ethyl acetate is formedD- 2 - 3 moles of ethyl acetate is formed

The correct option is D 2/3 moles of ethyl acetate is formed In presence of little H_2SO_4 (as catalyst) about 2/3 mole of each of CH_3COOH and C_2H_5OH react ...

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Byju's Answer

Standard XII

Chemistry

Introduction to Reversible and Irreversible Reactions

CH 3 COOH l +...

Question

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

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

1 mole of ethyl acetate is formed

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2 mole of ethyl acetate are formed

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1/2 moles of ethyl acetate is formed

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2/3 moles of ethyl acetate is formed

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Solution

The correct option is D i.e 2/3 moles of ethyl acetate is formed.
$C H_{3} C O O H (l) + C_{2} H_{5} O H (l)$ ⇌ $C H_{3} C O O C_{2} H_{5} (l) + H_{2} O (l)$
1 1 0 0
1-x 1-x x x
$K_{c} = \frac{[C H_{3} C O O C_{2} H_{5}] [H_{2} O]}{[C H_{3} C O O H] [C_{2} H_{5} O H]}$
$4 = \frac{x^{2}}{[1 - x]^{2}}$
$2 = \frac{x}{1 - x}$
x = 2-2x
x = $\frac{2}{3}$
On equilibrium being attained 2/3 moles of ethyl acetate is formed

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Similar questions

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

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

What happens when acetic acid reacts with ethanol in presence of conc. H₂SO₄?

Q. The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and their equilibrium is represented as:

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 (a q)} + H_{2} O_{(l)}

(a) Write the concentration ratio (reaction quotient),

Q_{e}

, for this reaction. Note that water is not in excess and is not a solvent in this reaction.

(b) At

293 K

, if one starts with

1.00

mole of acetic acid and

0.180

mole of ethanol, there is

0.171

mole of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

0.500

mole of ethanol and

1.000

mole of acetic acid and maintaining it at

293 K

0.214

mole of ethyl acetate is found after some time. Has equilibrium been reached?

2 moles of acetaldehyde are catalyzed by aluminium ethoxide to yield one mole of ethyl acetate. This is:

1.0

mole of ethyl alcohol and

1.0

mole of acetic acid are mixed. At equilibrium,

0.666

mole of ester is formed. The value of equilibrium constant is:

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CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l);Kc=4 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 i.e 2/3 moles of ethyl acetate is formed.CH3COOH(l)+C2H5OH(l) ⇌ CH3COOC2H5(l)+H2O(l) 1 1 0 01-x 1-x x xKc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]4=x2[1−x]2 2=x1−xx = 2-2xx = 23On equilibrium being attained 2/3 moles of ethyl acetate is formed

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

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