Q. In the esterification $C_{2}H_{5}OH\left(l\right)+CH_{3}COOH \, \left(l\right)\rightleftharpoons CH_{3}COOC_{2}H_{5}\left(l\right)+H_{2}O\left(l\right)$ an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction $0.333$ . The equilibrium constant is.
NTA AbhyasNTA Abhyas 2022
Solution:
As alcohol and acid are equimolar, so the initial concentration would also be the same. Suppose the initial concentration is 'C'.
$\frac{x}{\left(C - x\right) + \left(C - x\right) + x + x}=0.33=\frac{1}{3}$
$3x=2C$
$x=\frac{2 C}{3}$
$K_{c}=\frac{\frac{2 C}{3} \times \frac{2 C}{3}}{\left(C - \frac{2 C}{3}\right) \times \left(C - \frac{2 C}{3}\right)}=4$
$\frac{x}{\left(C - x\right) + \left(C - x\right) + x + x}=0.33=\frac{1}{3}$
$3x=2C$
$x=\frac{2 C}{3}$
$K_{c}=\frac{\frac{2 C}{3} \times \frac{2 C}{3}}{\left(C - \frac{2 C}{3}\right) \times \left(C - \frac{2 C}{3}\right)}=4$