Question

When alcohol $\left(C_2{H}_{5}OH\right(l\left)\right)$ and acetic acid $\left(C{H}_{3}COOH\right(l\left)\right)$ are mixed together in equimolar ratio at ${27}^o$C. 33% of each is converted into ester. Then the $K_c$ for the equilibrium
$C_2{H}_{5}OH\left(l\right)+C{H}_{3}COOH\left(l\right)\rightleftharpoons C{H}_{3}COOH{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$ is:

• A. 4
• B. $\frac{1}{4}$
• C. 9
• D. $\frac{1}{9}$

Answer 1

The correct option is B $\frac{1}{4}$

The equivalent constant $K_c$ is the ratio of the concentration of products to that of reactants.

$C_2{H}_{5}OH+C{H}_{3}COOH\leftrightharpoons C{H}_{3}COO{C}_2{H}_5+H_{2}O$

$t=0$ $1$ $1$ $0$ $0$

$t=equilibrium$ $\frac{2}{3}$ $\frac{2}{3}$ $\frac{1}{3}$ $\frac{1}{3}$

$K_c=\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H_{2}O\right]}{\left[C_2{H}_{5}OH\right]\left[C{H}_{3}COOH\right]}$

$=\frac{\frac{1}{3}\times \frac{1}{3}}{\frac{2}{3}\times \frac{2}{3}}$

$K_c=\frac{1}{4}$

Hence, option (B) is correct.