In the esterification C _ 2 H _ 5 OH (l) + C H _ 3 COOH (l) ⇌ C H _ 3 COO 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 =

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Correct answer: 4

As alcohol and acid are equimolar so the initial concentration would also be same. Suppose the initial concentration is 'C'.
In the esterification C _ 2 H _ 5 OH (l) + C H _ 3 COOH (l) ⇌ C H _ 3 COO C _ 2 H _ 5 (l) + H _ 2 O(l) an equimolar

In the esterification C _ 2 H _ 5 OH (l) + C H _ 3 COOH (l) ⇌ C H _ 3 COO C _ 2 H _ 5 (l) + H _ 2 O(l) an equimolar

\dfrac{x}{\left(C - x\right) + \left(C - x\right) + x + x} = 0.33 = \dfrac{1}{3}

3x = 2C

x = \dfrac{2C}{3}

{K}_{c} = \dfrac{\dfrac{2C}{3} \times \dfrac{2C}{3}}{\left(C - \dfrac{2C}{3}\right) \times \left(C - \dfrac{2C}{3}\right)} = 4

In the esterification C _ 2 H _ 5 OH (l) + C H _ 3 COOH (l) ⇌ C H _...

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