Question

$1.0$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:

• A. $1/4$
• B. $1/2$
• C. $4$
• D. $3$

Answer 1

The correct option is C $4$

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

$1mole$ $1mole$ $\to$ At $t=0$

$1-\alpha$ $1-\alpha$ $\alpha$ $\alpha$ $\to$ At equilibrium

At equilibrium, moles of ester=$0.666$

$\therefore \alpha =0.666$

$1-0.666$ $1-0.666$ $0.666$ $0.666$

$0.334$ $0.334$ $0.666$ $0.666$

If $V$ is volume of vessel, concentrations are,

$\frac{0.334}{V}$ $\frac{0.334}{V}$ $\frac{0.666}{V}$ $\frac{0.666}{V}$

Equilibrium constant= $\frac{\left[H_{2}O\right]\left[C{H}_{3}COO{C}_2{H}_5\right]}{\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}=\frac{\left(\frac{0.666}{V}\right)\left(\frac{0.666}{V}\right)}{\left(\frac{0.334}{V}\right)\left(\frac{0.334}{V}\right)}$

$=\frac{(0.666{)}^2}{(0.334{)}^2}=3.97$

$\therefore$ Equilibrium constant= $4$ (approx)