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?

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

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

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

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explanation:

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the answer is symmetrical : )