In the esterification C 2 H 5 O H l + C H 3 C O O H l ⇌ C H 3 C O O C 2 H 5 l + H 2 O l an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction

Question

In the esterification C 2 H 5 O H l + C H 3 C O O H l ⇌ C H 3 C O O C 2 H 5 l + H 2 O l an equimolar mixture of alcohol and acid taken initially yields under equilibrium, the water with mole fraction = 0.333. The equilibrium constant is.

TestHub · Accepted Answer

As alcohol and acid are equimolar so the initial concentration would also be same. Suppose the initial concentration is 'C'.

$\frac{x}{(C - x) + (C - x) + x + x} = 0.33 = \frac{1}{3}$
$3 x = 2 C$
$x = \frac{2 C}{3}$
$K_{c} = \frac{\frac{2 C}{3} \times \frac{2 C}{3}}{(C - \frac{2 C}{3}) \times (C - \frac{2 C}{3})} = 4$

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