Question

One mole of ethyl alcohol was treated with one mole of acetic acid at ${25}^o$C. $2/3$ 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 D $4$

Let us assume that the one mole of $C{H}_{3}COOH$ and $C_2{H}_5{O}_5$ are present and $\frac{2}{3}rd$ of $C{H}_{3}COOH$ and $C_2{H}_{5}OH$ is consumed . It gives $C{H}_{3}COOH$

$left=1-\frac{2}{3}$

$C_2{H}_{5}OH=1-\frac{2}{3}$

Hence the number of moles of water and $C{H}_{3}COO{C}_2{H}_5$ are $\frac{2}{3}$ each

$K_a=\frac{(\frac{2}{3}\times \frac{2}{3})}{(\frac{1}{3}\times \frac{1}{3})}=4$