Calculating equilibrium constant?

Question

A solution contains 0.01 M ethyl acetate, 0.10 M acetic acid and 0.10 M ethanol. For the same equilibrium constant of 2.00, calculate the equilibrium concentrations of ethyl acetate, ethanol, and acetic acid in the solution.

Answer 1

It depends on how the reaction is written. If it's for the reaction of water with ethyl acetate to produce acetic acid and ethanol, then:

K = [EtOH][CH3CO2H]/[EtOAc]

Given the numbers, you would actually see ethyl acetate react to produce acetic acid and ethanol, for the Q value will be 1.

2 = (0.10 + x)^2 / 0.01 - x
0.02 - 2x = 0.01 + 2x + x2
0 = x2 + 4x -0.01
x = 0.0025
[EtOH] = [CH3CO2H] = 0.10025, [EtOAc] = 0.0075

If it's the other way around, then ethyl acetate will be produced, as Q would be 1 as well.

K = [EtOAc]/[EtOH][CH3CO2H]
2 = 0.01 + x / (0.1 - x)^2
0.2 - 4x + 2x2 = 0.01 + x
0 = 2x2 - 5x + 0.19
x = 0.039
[EtOH] = [CH3CO2H] = 0.061, [EtOAc] = 0.049

So you need to know if the equilibrium constant is written for the formation of ethyl acetate or its reaction with water to produce acetic acid and ethanol.