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13 Dec 2019

The value of Kc for the reaction of acetic acid with ethanol is 3.4 at 25∘C:

CH3CO2H(soln)+ C2H5OH(soln)⇌ CH3CO2C2H5(soln)+H2O(soln)Kc=3.4
Acetic acid Ethanol Ethyl acetate

Part A

How many moles of ethyl acetate are present in an equilibrium mixture that contains 3.7 mol of acetic acid, 6.4 mol of ethanol, and 12.5 mol of water at 25 ∘C?

Part B

Calculate the number of moles of all reactants and products in an equilibrium mixture prepared by mixing 1.00 mol of acetic acid and 10.00 mol of ethanol.

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Patrina Schowalter
Patrina SchowalterLv2
17 Dec 2019

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Related questions

Ethyl acetate, a solvent used in many fingernail-polish removers, is made by the reaction of acetic acid with ethanol: CH3CO2H(soln)Aceticacid +C2H5OH(soln)Ethanol⇌CH3CO2H(soln)Ethylacetate+H2O(soln)ΔH∘=−2.9kJ

Part A

Does the amount of ethyl acetate in an equilibrium mixture increase or decrease when the temperature is increased? Does the amount of ethyl acetate in an equilibrium mixture increase or decrease when the temperature is increased?

increase

decrease

Part B

How does Kc change when the temperature is decreased? Justify your answers using Le Chatelier's principle.

1- As the temperature is decreased, the reaction shifts from left to right. The product concentrations increase, and the reactant concentrations decrease. This corresponds to an increase in Kc.

2- As the temperature is decreased, the reaction shifts from right to left. The product concentrations decrease, and the reactant concentrations increase. This corresponds to an decrease in Kc.

3-As there is neither products no reactants in gas state the temperature does not shift the reaction. So decrease in temperature does not change Kc. SubmitMy AnswersGive Up

Ok this is a three part questions but I think i correctly answered#1 but i get lost after that..

1. An aqueous solution of ethyl acetate at room temperature has aninitial concentration of 0.15 M. Calculate the equilibriumconcentrations of ethyl acetate (CH3COOC2H5), ethanol (C2H5OH) andacetic acid if the equilibrium constant at this temperature is Kc =3.00.
CH3COOC2H5(aq) + H2O ? CH3COOH(aq) + C2H5OH(aq)

Here I set up an ice table an ice table and got a quadratic ofx^2+3.00x-.45=0
and i got x=.143.

Now here is what I get confused..

2) 2. A solution contains 0.15 M acetic acid and 0.15 ethanol. Ifthe equilibrium constant defined by the reaction in Problem 1 is Kc= 3.00, calculate the equilibrium concentrations of ethyl acetate,acetic acid and ethanol.

Im guessing you just do products/reactants?
so
[0.15 M] x [0.15 M] / [CH3COOC2H5] = 3.00
and CH3COO2H5 = .0075 M.

I think im right up to this point then im not sure what to donext?

then there's

3. A solution contains 0.01 M ethyl acetate, 0.10 M acetic acid and0.10 M ethanol. For the same equilibrium constant of 3.00,calculate the equilibrium concentrations of ethyl acetate, ethanol,and acetic acid in the solution.

The equilibrium constant for the esterification of ethanol and aceticacid in an aqueous solution CH3COOH(aq) + C2H5OH(aq) -----> CH3COOC2H5(aq) + H2O(l) amounts to 0.566 at 298 K. The reaction mixture initially contains 0.040 mol dm^−3 of acetic acid and 0.025 mol dm^−3 of ethanol. Find themolar concentration of ethyl acetate at the equilibrium. Assume that the pH is low enough that CH3COOH is not dissociated at all and thatthe activity of water equals 1 just like for pure liquid water (after all,the solution is quite dilute). Hint:This problem requires you to keep an extra large number ofsignificant digits in all intermediate calculations: otherwise, your finalanswer will be very inaccurate. Do not forget to check your answer in the end.

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