Question

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, Kc?

Answer 1

Write the Kc expression.
Set up an ICE chart.
Plug into Kc expression and solve for Kc.

Answer 2

I 1 2 0
C -x -x +x
E 1-x 2-x 0.86

(.86)/[(1-.86)(2-.86)]

Answer 3

5.39

Answer 4

^^^ that is wrong