Question

One mole of ${CH}_{3}COOH$ and one mole of $C_2{H}_{5}OH$ reacts to produce $\frac{2}{3}$ mol of ${CH}_{3}COO{C}_2{H}_5$. The equilibrium constant is:

• A. $2$
• B. $+2$
• C. $-4$
• D. $4$

Answer 1

The correct option is D $4$

${CH}_{3}COOH+C_2{H}_{5}OH\to {CH}_{3}COO{C}_2{H}_5+H_{2}O$
At start $1$ $1$ $0$ $0$

At eqbm. $1-\frac{2}{3}$ $1-\frac{2}{3}$ $\frac{2}{3}$ $\frac{2}{3}$

Let the total volume $=VL$

$\therefore$ $\left[{CH}_{3}OOH\right]=\frac{1}{3}Vmol/L$

$\left[C_2{H}_{5}OH\right]=\frac{1}{3}Vmol/L$

$\left[{CH}_{3}COO{C}_2{H}_5\right]=\frac{2}{3}Vmol/L$

$\left[H_{2}O\right]=\frac{2}{3}Vmol/L$

$\therefore$ $K=\frac{\frac{2}{3}V\times \frac{2}{3}V}{\frac{1}{3}V\times \frac{1}{3}V}=4$

Hence, option D is correct