Final answer:
To find the equilibrium concentration of H2O in the reaction, the equilibrium constant formula Kc = [CH3CO2C2H5][H2O] / [CH3CO2H][C2H5OH] is used. Since the concentration of ethanol (C2H5OH) is not given, it is assumed constant, and the calculation shows that the equilibrium concentration of H2O is a function of the ethanol concentration.
Step-by-step explanation:
The student's question is about calculating the equilibrium concentration of H2O for an esterification reaction. We have the equilibrium constant (Kc=4.0) and the concentrations of acetic acid (CH3CO2H) and ethyl acetate (CH3CO2C2H5) at equilibrium. To find the concentration of H2O at equilibrium, we can use the expression for the equilibrium constant of the reaction:
Kc = [CH3CO2C2H5][H2O] / [CH3CO2H][C2H5OH]
Since the concentration of ethanol (C2H5OH) was not provided, it is assumed that it remains relatively constant because it is the solvent and its concentration is much larger than that of the other components in the reaction mixture.
Let's solve for [H2O] at equilibrium:
4.0 = (2.2)[H2O] / (0.75)[C2H5OH]
We can rearrange this to solve for [H2O]:
[H2O] = 4.0(0.75)[C2H5OH] / 2.2
Without the concentration of C2H5OH, the solution is expressed in terms of this unknown concentration. Therefore, the equilibrium concentration of H2O is dependent on the concentration of ethanol in the solvent mixture.