Question

One mole of ethanol is treated with one mole of ethanoic acid at ${25}^{o}C$. Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be :

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

One mole of ethanol is treated with one mole of ethanoic acid at 25oC. Half of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be 1.
$C_2{H}_{5}OH+C{H}_{3}COOH\to C{H}_{3}COOH+H_{2}O$

	Ethanol	Ethanoic acid	Ethyl acetate	Water
Intial moles	1	1	0	0
Change in moles	-0.5	-0.5	0.5	0.5
Equilibrium moles	1-0.5=0.5	1-0.5=0.5	0.5	0.5

The equilibrium constant expression is
$K=\frac{\text{[ethyl acetate]}\text{[water]}}{\text{[ethanol]}\text{[ethanoic acid]}}$
$K=\frac{\frac{0.5}{\text{Volume}}\times \frac{0.5}{\text{Volume}}}{\frac{0.5}{\text{Volume}}\times \frac{0.5}{\text{Volume}}}=1$