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Topic: Finding Equilibrium Concentrations--What did I do wrong?  (Read 33538 times)

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Offline o1ocups

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Finding Equilibrium Concentrations--What did I do wrong?
« on: February 15, 2009, 12:34:25 AM »
Problem
Reaction: aA(g) + bB(g) <-> cC(g)
Equilibrium Constant (Kc) = 4.0
Find the equilibrium concentrations of A, B, and C for a=1, b=1, and c=2. Assume that the initial concentrations of A and B are each 1.0 M and that no product is present at the beginning of the reaction.

What I did
ABC
Initial Concentration1.0M1.0M0.0M
Change in Concentration-x-x+x
Equilibrium Concentration1-x1-xx

4.0=x^2/[(1-x)^2]
4(1-x)^2=x^2
4-4x^2=x^2
5x^2=4
x^2=4/5
x=0.894

Equilibrium Concentrations
A: 1-x=1-0.894=0.106 M
B: 1-x=0.106 M
C: x=0.894 M

But it's wrong...  :(  Please tell me where I did wrong. Thanks!!

Offline plat_num

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #1 on: February 15, 2009, 12:41:45 AM »
I think it has to do with the fact that 2 moles of C are produced in the balanced equation.  However you have only single "x" values for all of A, B and C.  Either change the C one to 2x, or change the A and B ones to .5

caution: I'm also learning this stuff so I might not be correct.

Offline o1ocups

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #2 on: February 15, 2009, 12:46:50 AM »
That's it! I can't believe that I didn't even think about it. Let me try and see if I can get the correct answer. Thank you for the hint!!

MODIFY: I still can't get it  :-\

Offline Borek

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #3 on: February 15, 2009, 06:10:44 AM »
Hard to tell without looking at equations and numbrs. Please show your work now.
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Offline o1ocups

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #4 on: February 15, 2009, 04:06:34 PM »
Hi Borek!

Hi Borek!

OK. Here is my new work:

ABC
Initial Concentration1.0M1.0M0.0M
Change in Concentration-x-x+2x
Equilibrium Concentration1-x1-x2x

4.0 = C^2/(A)(B)
C=2x, A=B=1-x
4.0=[(2x)^2]/[(1-x)^2]
4(1-x)^2=4x^2
4-4x^2=4x^2
8x^2=4
x^2=4/8=1/2
x=0.707

Equilibrium Concentrations
A: 1-x=1-0.707=0.293 M
B: 1-x=0.293 M
C: x=1.414 M

Thanks!
« Last Edit: February 15, 2009, 04:30:18 PM by o1ocups »

Offline o1ocups

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #5 on: February 15, 2009, 04:39:14 PM »
Would my original answers be correct if a, b, and c all equal 1?

Offline Borek

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #6 on: February 15, 2009, 05:08:24 PM »
4(1-x)2 is not 4-4x2
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Offline o1ocups

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #7 on: February 15, 2009, 05:10:09 PM »
4(1-x)2 is not 4-4x2

You are right  :-[

4.0 = C^2/(A)(B)
C=2x, A=B=1-x
4.0=[(2x)^2]/[(1-x)^2]
4=[(2x)^2]/[x^2-2x+1]
4(x^2-2x+1)=4x^2
4x^2-8x+4=4x^2
4=8x
x=2??

Am I missing something again? Sorry, my math is not very good.
« Last Edit: February 15, 2009, 05:30:08 PM by o1ocups »

Offline Borek

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #8 on: February 15, 2009, 05:37:18 PM »
4=8x
x=2??

No, but you are verrrry close.

If x is 2, 8x is 16, not 4.

How do you get rid of the 8 on the right?
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Offline o1ocups

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Re: Finding Equilibrium Concentrations--What did I do wrong?
« Reply #9 on: February 15, 2009, 05:40:30 PM »
4=8x
x=2??

No, but you are verrrry close.

If x is 2, 8x is 16, not 4.

How do you get rid of the 8 on the right?

OMG! x=1/2
THANKS.

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