The esterification of acetic acid and ethanol is given by the reaction below: C₂H₅OH(aq) + CH₃COOH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l)
When 1.00 mol of ethanol was mixed with 2.00 mol of acid in a 1.00 L flask,0.86 mol of ester was formed at room temperature.What is the value of the equilibrium constant,K_c?

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction is:
CH3CH2OH + CH3COOH ⇌ CH3COOCH2CH3 + H2O

The equilibrium constant expression is given by:
Kc = ([CH3COOCH2CH3] / [CH3CH2OH] [CH3COOH])

At equilibrium, the concentration of ethanol is (1 - 0.86) = 0.14 mol/L, the concentration of acetic acid is (2 - 0.86) = 1.14 mol/L, and the concentration of ester is 0.86 mol/L.

Substituting the values into the equilibrium constant expression, we get:
Kc = (0.86 / (0.14 x 1.14))
Kc = 5.351

Therefore, the value of the equilibrium constant is 5.4 (rounded to one significant figure).

The correct answer is D.

The Esterification of Acetic Acid and Ethanol Is Given by the Reaction