For the esterification C2H5OH (l) + CH3COOH (l) CH3COOC2H5 (l) + H2O (l), an equimolar mixture of alcohol and acid is taken initially. At equilibrium the mole fraction of water is 0.333. The equilibrium constant Kc is approximately:

Question

For the esterification C2H5OH (l) + CH3COOH (l) <=> CH3COOC2H5 (l) + H2O (l), an equimolar mixture of alcohol and acid is taken initially. At equilibrium the mole fraction of water is 0.333. The equilibrium constant Kc is approximately:

Accepted Answer

4.0. With initial 1 mol each, x = 0.667 mol reacts (since x/2 = 0.333), leaving 0.333 mol of each reactant and 0.667 mol of each product, so Kc = (0.667/0.333)² ~= 4.

For the esterification C2H5OH (l) + CH3COOH (l) <=> CH3COOC2H5 (l) + H2O (l), an equimolar mixture of alcohol and acid is taken initially. At equilibrium the mole fraction of water is 0.333. The equilibrium constant Kc is approximately:

Solution

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