The esterification of acetic acid and ethanol is given by the reaction below:

C2H5OH(aq) + CH3COOH(aq)===>CH3COOC2H5(aq) + H2O(l)

When 1.00 mol of ethanol was mixed with 2.00 mol of acid in a 1.00 L flask, 0.86 mol of ester was formed at room temperature. What is the value of the equilibrium constant, Kc?

Write the Kc expression.

Set up an ICE chart.
Plug into Kc expression and solve for Kc.

I 1 2 0

C -x -x +x
E 1-x 2-x 0.86

(.86)/[(1-.86)(2-.86)]

^^^ that is wrong

To determine the value of the equilibrium constant, Kc, we need to use the stoichiometry of the reaction and the given amount of reactants and products at equilibrium.

The balanced equation for the reaction is:

C2H5OH(aq) + CH3COOH(aq) <=> CH3COOC2H5(aq) + H2O(l)

From the stoichiometry of the balanced equation, we can see that 1 mol of ethanol reacts with 1 mol of acetic acid to produce 1 mol of the ester. Therefore, the initial moles of ethanol and acetic acid are both 1.00 mol.

At equilibrium, we are given that 0.86 mol of ester is formed. This implies that 0.86 mol of ethanol and acetic acid have reacted.

To find the concentrations of ethanol, acetic acid, and ester at equilibrium, we need to divide the number of moles by the volume of the flask, which is 1.00 L:

[ethanol] = [acetic acid] = (0.86 mol) / (1.00 L) = 0.86 M
[ester] = (0.86 mol) / (1.00 L) = 0.86 M

Now, we can substitute these concentrations into the equilibrium expression for Kc:

Kc = [ester] / ([ethanol] * [acetic acid])
= (0.86 M) / ((0.86 M) * (0.86 M))
= 1.00

Therefore, the value of the equilibrium constant, Kc, is 1.00.

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