Kc and equillibrium amount?

[ChemSiddiqui]

Hi all!

 

I was doing some past papers and I have some problem. The question was:

 

Q. Alcohols and esters are important organic compounds which are widely used as solvents.

 

Esters such as ethyl ethanoate can be formed by reacting carboxylic acids with alcohols.

 

CH3CO2H + C2H5OH <------> CH3CO2C2H5 + H20

 

For this equilibrium, the value of Kc is 4.0 at 298K.

 

A mixture containing 0.5 mol of ethanoic acid , 0.5 mol ethanol, 0.1 mol ethyl ethanoate and 0.1 mol water was set up and allowed to come to equilibrium at 298K. The final volume of the solution was V dm3.

 

Calculate the amount, in moles, of each substance present at equilibrium.

 

My try at the solution:

CH3CO2H + C2H5OH <------> CH3CO2C2H5 + H20

Initial amounts:

0.5 0.5 0.1 0.1

 

Equilibrium amount:

 

0.5 – X 0.5 – X 0.1 – X 0.1 – X

 

Concentrations at equilibrium:

 

0.5 – X 0.5 –X 0.1 –X 0.1 -X

V V V V

 

Kc = [CH3CO2C2H5] [H20]

[CH3CO2H] [C2H5OH]

 

 

Putting values in the formula makes:

 

4 = (0.1 –X)^2

(0.5 – X) ^2

 

4(0.5-X) = 0.1 –X

X = 0.63Then evaluating each amount in equilibrium using x.

 

But the mark scheme says that the value of x is 0.3. Also mark scheme says that amount in moles in equilibrium of water and ethyl ethanoate will be (1 + X) than (1-x)? That’s my question why is it?

 

Any help will be appreciated!

[zule]

Firstly, you have to think that if the reactives decrease, the products of the reaction have to increase.

 

On the other hand, if you make square roots in one equation, you have to be sure that you do it to both sides of the equal, so the equation remains the same.

[ChemSiddiqui]

Firstly, you have to think that if the reactives decrease, the products of the reaction have to increase.

 

 

Thank you for the reply. But could you elaborate this point.

[zule]

Thank you for the reply. But could you elaborate this point.

 

You have ten raw apples and two roast apples. You roast three of the raw apples. How many raw and roast apples do you have?

 

Raw apples 10-3=7

Roast apples 2+3=5

 

The Kc is calculated multiplying the concentration in the equilibrium of the products and dividing this result by the multiplication of the reactive concentrations in the equilibrium.

[ChemSiddiqui]

Ok. But what I wanted to ask was why it is 1 +X than 1-X amount after the equillibrium? although I understood what you had to say in your post but I am actually confused about the equillirium amounts? Sorry if it is annoying to you!

[hermanntrude]

your first step should have been to find the reaction quotient, Q. that would tell you which direction to expect a shift, which would in turn tell you if it was 1+x or 1-x

[zule]

Ok. But what I wanted to ask was why it is 1 +X than 1-X amount after the equillibrium? although I understood what you had to say in your post but I am actually confused about the equillirium amounts? Sorry if it is annoying to you!

 

Don’t worry; you are not annoying at all.

 

x is the amount of raw apples you roast. The raw apples are the reactives. The roast apples are the products. After “roasting” x reactives, you will have the initial amount of reactives minus x. Those x reactive amount have been transformed in products, so you will have the initial amount of products plus x.

 

Kc is bigger than one and the number of moles than react are the same than the number of moles than are produced. As we start with more moles in the left part of the reation than in the right one, the reaction will have to go towards the right part: the amount of ethanoic acid an ethanol are going to decrease because they are going to be converted on ethyl ethanoate and water. So both the ethyl ethanoate and the water will increase its number of moles in the same amount than they decrease in ethanoic acid an ethanol.

[ChemSiddiqui]

O... I see you have to apply Le Chateleir's principal here then to see which direction the equillibrium shift and then determine the concentration. I understand completely now thank you very much zule. Good chatting with yeh.

 

One more thing to hermanntrude. when you find the reaction quotient to determine the direction of shift of equillibrium, does the equillibrium amount of reactant remain the same all the time regardless of the direction of the shift?

[hermanntrude]

I'm not really sure i understand your question

 

For a given value of K, the amounts of products and reactants at equilibrium will always be the same, yes. If the equilibrium is changed, however, by addition of a product or a reactant, or by a change in temperature or pressure or volume, then the quantity of the reactants and products will change to partially offset the change, according to le chatelier's principle, and the direction indicated by a calculation of the reaction quotient.

[ChemSiddiqui]

Using the same reaction in equillibrium:

 

CH3CO2H + C2H5OH <-------> CH3CO2C2H5 + H20

 

The reaction quotient will be 2/2 = 1 right (correct me if I am wrong). This is less than the kc of the reaction so the reaction should yeild less products and more reactants. But the mark scheme says the reaction sould yeild more products because the equillibrium amount is 0.1 + X and not 0.1 –X . So firtly clarify this point to me.

 

Also say the RQ is less than kc i.e. the reaction will have the left shift, then sould the equillibrium amount be 0.5 + x in this case?

[hermanntrude]

what are the phase symbols in your equation?

 

pure liquids should be left out of the equilibrium expression. that should help you get x

[ChemSiddiqui]

O.alright Thank you very much hermantrude. I very much appreciate your help!

[hermanntrude]

did you get the right value of x?

[ChemSiddiqui]

Yes I did. Thank you again ! good day!