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

• Keq = 4

Step-by-step 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=([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=([CH_3CO_2C_2H_5]\cdot [H_2O])/([C_2H_5OH]\cdot [CH_3CO_2H])=(2/3\cdot 2/3)/(1/3\cdot 1/3)=4`