Question

In the esterification $C_2{H}_{5}OH\left(l\right)+C{H}_{3}COOH\left(l\right)\rightleftharpoons C{H}_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$ an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction $=\mathrm{0.333.}$ Calculate the equilibrium constant :

• A. $K=2$
• B. $K=4$
• C. $K=8$
• D. None of these

Answer 1

The correct option is B $K=4$

Let initially, $1$ mole of ethanol and $1$ mole of acetic acid is present.
Let x moles of ethanol reacts with $x$ moles of acetic acid to reach equilibrium.

	Ethanol	Acetic acid	Ethyl acetate	Water
Equilibrium number of moles	$1-x$	$1-x$	$x$	$x$

The total number of moles $=1-x+1-x+x+x=2$
The mole fraction of water at equilibrium is $\frac{x}{2}=0.333$
Hence, $x=0.666$ and $1-x=0.334$
The equilibrium constant expression is
$K=\frac{\left[\text{Ethyl acetate}\right[\text{Water}\left]\right]}{\left[\text{ethanol}\right]\left[\text{acetic acid}\right]}$
$K=\frac{0.666\times 0.666}{0.334\times 0.334}$
$K=4$