Question

When 1 mol each of C2H5OH and CH3CO2H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when 13 mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)

Answer 1

Answer:

• Keq = 4

Explanation:

The correct statement is:

When 1 mol each of C₂H₅OH and CH₃CO₂H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when 1/3 mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is a solute in this reaction.)

1. Equilibrium equation

• C₂H₅OH + CH₃CO₂H     ⇄    CH₃CO₂C₂H₅ +   H₂O

               ↑                  ↑                          ↑                    ↑

          ethanol     acetic acid        ethyl acetate       water

2. Equilibrium constant

• Keq = [Products] / [Reactants], each raised to tis stoichiometrical coefficient.

Since water is also a solute in this reaction (the solvent is dioxane) its concentration will appear in the equilibrium constant.

• $Keq=\frac{[CH_3CO_2C_2H_5]\cdot [H_2O]}{[C_2H_5OH]\cdot [CH_3CO_2H]}$

3. Equlibrium concentrations:

Moles

                  C₂H₅OH + CH₃CO₂H     ⇄    CH₃CO₂C₂H₅ +   H₂O

Initial                1                  1                                 0                  0

Change          -2/3            -2/3                            +2/3             +2/3

End                  1/3               1/3                              2/3              2/3

Since the volume is 1 liter, the concentration is equal to the number of moles

4. Calculations:

$Keq=\frac{[CH_3CO_2C_2H_5]\cdot [H_2O]}{[C_2H_5OH]\cdot [CH_3CO_2H]}=\frac{2/3\cdot 2/3}{1/3\cdot 1/3}=4$