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Free Radical Halogenation

When ethyl al...

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When ethyl alcohol and acetic acid mixed together in equimolecular proportions, equilibrium is attained when two-third of the acid and alcohol are consumed. The equilibrium constant of the reaction will be

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Q. When ethyl alcohol and acetic acid mixed together in equimolecular proportions, equilibrium is attained when two - thirds of the acid and alcohol are consumed. The equilibrium constant of the reaction will be:

1.0 m o l e

of ethyl alcohol and

1.0 m o l e

of acetic acid are mixed. At equilibrium,

0.666 m o l e

of ester is formed. The value of equilibrium constant 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:

3 m o l e s

of ethyl alcohol are mixed with

3 m o l e s

of acetic acid,

2 m o l e s

of ester are formed at equilibrium according to the equation:

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)

The value of the equilibrium constant for the reaction is:

Q. When alcohol

(C_{2} H_{5} O H)

and acetic acid are mixed together in equimolar ratio at

27^{\circ} C

, 33% is converted into ester. Then the

K_{C}

for the equilibrium :

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

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