Question

Acetic acid, CH3CO2H, reacts with ethanol, C2H5OH, to form water and ethyl acetate, CH3CO2C2H5. The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations for a mixture that is initially 0.15 M in CH3CO2H, 0.15 M in C2H5OH, 0.40 M in CH3CO2C2H5, and 0.40 M in H2O?

Answer 1

Explanation:

The initial concentrations for a mixture :

Acetic acid at equilibrium = 0.15 M

Ethanol at equilibrium = 0.15 M

Ethyl acetate at equilibrium = 0.40 M

Water at equilibrium = 0.40 M

$CH3COOH + C_2H_5OH\rightleftharpoons CH_3CO_2C_2H_5+H_2O$

Initially:

0.15 M            0.15 M            0.40 M   0.40 M

At equilibrium

(0.15-x)M       (0.15-x) M     (0.40+x) M   (0.40+x) M

The equilibrium constant is given by expression

$K_c=\frac{[CH_3CO_2C_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$

$4.0=\frac{(0.40-x)\times (0.40-x)}{(0.15+x)\times (0.15+x)}$

Solving for x:

x = 0.0333

The equilibrium concentrations for a mixture :

Acetic acid at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M

Ethanol at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M

Ethyl acetate at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M

Water at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M