[nawinince]

1.0 mol of acetic acid and 1.0 mol of ethanol was mixed to react at a constant temperature to reach the equilibrium, and 0.75 mol of water was found in the product. Write the correct answer to two significant figures.

(1) Calculate the equilibrium constant of this reaction at the above temperature.
(2) At the same temperature, 1.0 mol of acetic acid, 1.0 mol of ethanol and 4.0 mol of water were initially mixed to react. Calculate the amount of ethylacetate produced after the mixture has reached the equilibrium.


my attempt :

(1)  CH3COOH + CH3CH2OH -----> CH3COOCH3CH2 + H2O

I :       1 mol              1 mol
C :      0.75 mol         0.75 mol              0.75 mol         0.75 mol
E :      0.25                0.25                    0.25                  0.25

I have found that if I calculate K without using H2O, then the answer will be wrong. But if I use H2O then it will be correct.
How come?   I thought H2O is pure no need to use it.

(2) For this question, I don't understand what question mean. Do I have to use 0.25 mol left from question1? if so ethylacetate will be reactant and it looks strange.

[Borek]

(1)  CH3COOH + CH3CH2OH -----> CH3COOCH3CH2 + H2O

I :       1 mol              1 mol
C :      0.75 mol         0.75 mol              0.75 mol         0.75 mol
E :      0.25                0.25                    0.25                  0.25

Perhaps that's just a typo, but you were specifically said there were 0.75 found in the final mixture, yet you wrote 0.25 moles of water at equilibrium.

How come?   I thought H2O is pure no need to use it.

You are confusing something. Perhaps you think about a situation when water is a solvent and its concentration is almost constant, that's not the situation here.

For this question, I don't understand what question mean. Do I have to use 0.25 mol left from question1? if so ethylacetate will be reactant and it looks strange.

No, if you just mixed ethanol, acetic acid and water, initial concentration of ethylacetate is zero.

[nawinince]

Sorry, 0.25 was a typo.

So here is my attempt to question 2 :

     CH3COOH + CH3CH2OH + H2O ---------> CH3COOCH3CH2 + 2 H2O
 
I        1 mol         1 mol           4 mol
C       1 mol          1mol            1mol                      1 mol              2 mol
E         0               0                 3                           1                    2

the correct answer was 0.5 mol, what did I miss again?

[Borek]

It is an equilibrium problem, in the first step you have calculated K, why do you assume now that the reaction went to completion?

[nawinince]

It is an equilibrium problem, in the first step you have calculated K, why do you assume now that the reaction went to completion?

I calculated K in the first question because it was asked in the question1.

I didn't know how to do question2, so I just assumed that it happend like what I wrote. what's the correct reaction?

[Borek]

Reaction equation is OK, but to calculate the outcome you have to calculate concentrations from the equilibrium (using K calculated in the first step).

That's actually what ICE table are most often used for. You just need to express one of the concentrations using an unknown (say x).

[nawinince]

Reaction equation is OK, but to calculate the outcome you have to calculate concentrations from the equilibrium (using K calculated in the first step).

That's actually what ICE table are most often used for. You just need to express one of the concentrations using an unknown (say x).

Did u mean this ok? because I haven't seen any reaction having H2O at both sides

CH3COOH + CH3CH2OH + H2O ---------> CH3COOCH3CH2 + 2 H2O