[T]

Hello, this is a question I encountered that I don't know if my answer was right. (There was no answer)

Question: A + 3B  2C is a reaction. At the start the concentration of A was 2 M, B was 6 M and C was 0 M. At equilibrium the concentration of A is 1 M, B is 3 M and C is 2 M. If the concentration of A was then increased by 1 M then what is the concentration of A, B and C when the reaction has reached the new equilibrium.

My answer was A = 1.88 M, B = 2.64 M and C = 2.24

Please show working out thanks

[Borek]

Please show working out thanks

Its the other way around - we expect you to show your workings as a starting point.

But the numbers look OK.

[T]

Ok, I thought that people would not care about my working out, this is my working out:

Before I go on, all my equations are not written in fractions but in divisions. I tried using the LaTeX code \frac{numerator}{denominator} and others but it would not work.

The Kc is 2²/(1 × 3³) = 4/27

The molarity of each molecule immediately after concentration of A is increased is A = 2 M, B = 3 M and C = 2 M.  If x molarity of A is converted into reactants then 3x molarity of B would be converted and C would have an increase of 2x molarity.

So then if we put this into the Kc then:

4/27 = (2+2x)²/((2-x) × (3-3x)³)

x = 0.12 or 3.4

A cannot be decreased by 3.4 so x = 0.12

A = 2 - 0.12   B = 3 - 0.12 × 3   C = 2 + 0.12 × 2
   = 1.88           = 2.64                 = 2.24

Borek, in your reply, to my equilibrium constant post on equilibrium concentrations from initial concentrations you used a mixture of stoichiometry and relations between the molecules. I tried to solve the problem that way and got a very different result please check what I did wrong when I attempted this way.

This is my working

The M of A at the new equilibrium is 2 - (2 - [C])/2
The M of B at the new equilibrium is 3 - 3 × (2 - [C])/2

So 4/27 = [C]²/((2 - (2 - [C])/2)×(3 - 3 × (2 - [C])/2)³

From this [C] = 1.24

I don't know why the answer is different, It would be great if you can tell me.

I apologise again for the lack of fractions.

Thanks

[Borek]

The M of A at the new equilibrium is 2 - (2 - [C])/2

And not 2-([C]-2)/2?

[T]

Aha! Thanks Borek, now it works fine. So you would say [C] = 2.24 M is the correct answer? Also how do you do fractions on this?

[Borek]

Yes, 2.24 looks OK.

Not seeing what you exactly did it is hard to say why you have problem formatting equations, feel free to start another thread in feedback (http://www.chemicalforums.com/index.php?board=16.0). Have you read http://www.chemicalforums.com/index.php?topic=59314.0?

[T]

Ok thanks Borek