1 mole of ethyl alcohol and 1 mole of acetic are mixed at equilibrium 0.666 mole of the ester is present then kc is
Answers
The way this question is phrased makes me very uneasy. The given information only refers to the stoichiometry of the reaction
[math]HOAc + EtOH → EtOAc + H_2O[/math]
But it says nothing about the concentrations or quantities of the reactants prior to the reaction. (I’m assuming the concentrations of the products were both zero to start with.)
If we make the unwarranted inference that we the reaction mixture consists entirely of 1 mol of HOAc and 1 mol of EtOH and no inert solvent, then it is possible to calculate a Kc. But I want to be clear that the inference is in no way justified by the problem statement.
At the end of the reaction, we will have 2/3 mol of EtOAc, 2/3 mol of H2O, 1/3 mol of HOAc, and 1/3 mol of EtOH. They are all contained within the same volume, and since there are equal numbers of constituents on reactant and product sides, we don’t have to bother converting quantity to concentration.
[math]K_c = \frac{[EtOAc][H_2O]}{[HOAc][EtOH]} = \frac{\frac{2}{3}^2}{\frac{1}{3}^2} = 4[/math]