If 1 mole of acetic acid and 1 mole of ethyl alcohol are mixed and reaction proceeds to equilibrium, the concentration of acetic acid and water are found 1/3 and 2/3 mole respectively. If 1 mole of ethyl acetate and 3 mole of water are mixed,how much ester is present when equilibrium is reached?

Chemistry, rate of reactions - help needed

I looked at this earlier. My problem with this problem is that I don't believe the first equilibrium can be right; therefore, I just decided to leave it.

To determine the amount of ester present when equilibrium is reached, we need to use the principle of equilibrium constants and apply it to the given reaction.

First, let's write the balanced equation for the reaction between acetic acid (AA) and ethyl alcohol (EA) to form ethyl acetate (EA) and water (H2O):

AA + EA ⇌ EA + H2O

The given information tells us that at equilibrium, the concentrations of acetic acid and water are found to be 1/3 mole and 2/3 mole, respectively.

Now, we need to find the equilibrium constant (Kc) for this reaction using the concentrations of acetic acid and water:

Kc = [EA][H2O] / [AA][EA]

Given that the concentration of acetic acid ([AA]) is 1/3 mole and the concentration of water ([H2O]) is 2/3 mole, we can substitute these values into the equation:

Kc = (1/3)(2/3) / (1)(1)

Simplifying, we get:

Kc = 2/9

Now, let's consider the second scenario given where 1 mole of ethyl acetate and 3 moles of water are mixed. We need to determine the amount of ester present at equilibrium.

Similar to before, we write the balanced equation for the reaction:

EA + H2O ⇌ AA + EA

Using the equilibrium constant (Kc) that we calculated earlier (Kc = 2/9), we can determine the concentration of ester at equilibrium.

Let's assume the concentration of ester at equilibrium is x. Then, the concentration of water will be 3 moles, and the concentration of acetic acid will also be x (since it is formed in a 1:1 ratio with ester).

Applying the equilibrium constant expression:

Kc = [AA][EA] / [EA][H2O]

Substituting the known values:

2/9 = x * x / (1) * (3)

Simplifying:

2/9 = x^2 / 3

Cross-multiplying:

6 = 9x^2

Dividing both sides by 9:

2/3 = x^2

Taking the square root of both sides:

√(2/3) = x

Therefore, when equilibrium is reached, the concentration of ester (EA) will be √(2/3) moles.