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Esterification and Saponification Reaction

Two moles of ...

Question

Two moles of an equimolar mixture of two alcohols $R_{1} - O H$ and $R_{2} - O H$ are esterified with one mole of acetic acid, If only 80% of the acid is consumed till equilibrium and the quantities of ester formed under equilibrium are in the ratio 3:2 what is the value of the equilibrium constant for the esterification of $R_{1} - O H ?$

A

0.48

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B

2.2

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C

0.32

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D

0.75

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Two moles of an equimolar mixture of two alcohols R1-OH and R2-OH are esterified with 1 mole of acetic acid .If 80% of the acid is comsumed and the quantities of ester formed under eqilibrium are in the ratio of3:2,what is the value of the equlibrium constant for the esterification of R1-OH ?

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:

Q. In the esterification

C_{2} H_{5} O H (l) + {C H}_{3 k e} C O O H (l) ⇌ {C H}_{3} C O O C_{2} H_{t a 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 an experiment starting with

1

mole of ethyl of alcohol,

1

mole of acetic acid and

1

mole of water at

100^{o} C

, the equilibrium mixture on analysis shows that

54.3

% of the acid is esterified. Calculate the equilibrium constant of this reaction.

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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