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Topic: Finding Equilibrium Constant (Kc) from Kp  (Read 8993 times)

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Offline kookaburra1701

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Finding Equilibrium Constant (Kc) from Kp
« on: June 02, 2014, 11:27:12 PM »
Here's the question from my homework:
Quote
At 850K, the value of the equilibrium constant Kp for the ammonia synthesis reaction:

N2(g) + H2(g)  ::equil:: N2H2(g)

is 0.2250. If a vessel contains an initial reaction mixture in which [N2]=0.0200M, [H2]=0.0200M, and [N2H2]=.000200M, what will the [N2H2] be when equilibrium is reached?

I know that to find Kc from Kp, we use this equation:

Kp=Kc(RT)Δn, which becomes .2250=Kc(.08206×850)-1
Solving for Kc, I get Kc=15.694.

Here is the ICE table:
N2H2N2H2
I.02M.02M.0002M
C-X-X+X
E.02-X.02-X.0002+X

I know that Kc=[N2H2]/[N2][H2], so plugging in the numbers from row E, I get...

15.694=(.0002+X)/(.02-x)2

This ends up being a quadratic equation, -15.694X2-1.62776X+.000428=0

When solved, X=2.6227×10-4 and -0.10398

The root that makes sense is the first one, which would make the ending concentration of N2H2 4.6227×10-4 M.

But...I can't the the computer to accept my answer. I've been over my calculations with a fine toothed comb, and I can't find any errors. (Well, ok, at first I found a bunch of errors, but after fixing them it still won't take it!) If anyone can tell me where I'm going wrong, I would really appreciate it!
« Last Edit: June 02, 2014, 11:59:30 PM by kookaburra1701 »

Offline curiouscat

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Re: Finding Equilibrium Constant (Kc) from Kp
« Reply #1 on: June 03, 2014, 01:49:06 AM »
I get roots as x=0.0039 & 0.099

Accepting first root, ending concentration of N2H2 becomes 4.1E-3 M

Does your computer overlord like this any better? :)

http://www.wolframalpha.com/input/?i=solve+15.694%3D%280.0002%2Bx%29%2F%28.02-x%29%5E2

Offline kookaburra1701

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Re: Finding Equilibrium Constant (Kc) from Kp
« Reply #2 on: June 03, 2014, 02:23:02 AM »
Yes! The Overlord has accepted my offering. :D I must have made a mistake when FOIL-ing out the equation. Thank you!

Offline mjc123

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Re: Finding Equilibrium Constant (Kc) from Kp
« Reply #3 on: June 03, 2014, 04:31:07 AM »
Just to be pedantic, N2H2 is diazine, not ammonia! (And can you really make it like that? C+W say it's unstable above -180°C.)

Offline kookaburra1701

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Re: Finding Equilibrium Constant (Kc) from Kp
« Reply #4 on: June 03, 2014, 05:13:18 PM »
Just to be pedantic, N2H2 is diazine, not ammonia! (And can you really make it like that? C+W say it's unstable above -180°C.)

Ha ha, good to know. I just copied down what was on my homework from the book. :)

Offline mjc123

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Re: Finding Equilibrium Constant (Kc) from Kp
« Reply #5 on: June 04, 2014, 08:51:06 AM »
A textbook gave you that equation? I wouldn't trust that textbook! (Which one, by the way?) The equation for ammonia is
N2 + 3H2  ::equil:: 2NH3
The expression for Kp will be quite different, and the calculation would give a very different answer. Does your answer agree with the book's?

Offline kookaburra1701

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Re: Finding Equilibrium Constant (Kc) from Kp
« Reply #6 on: June 04, 2014, 11:07:50 AM »
A textbook gave you that equation? I wouldn't trust that textbook! (Which one, by the way?) The equation for ammonia is
N2 + 3H2  ::equil:: 2NH3
The expression for Kp will be quite different, and the calculation would give a very different answer. Does your answer agree with the book's?

The book is published by W. W. Norton, "Chemisty, the Science in Context 3e." The questions are on the book's website and we do them as homework. The computer accepted my answer and said I got it right. Oh well -at least I know I have the concept down.

Offline kriggy

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Re: Finding Equilibrium Constant (Kc) from Kp
« Reply #7 on: June 07, 2014, 08:23:56 AM »
A textbook gave you that equation? I wouldn't trust that textbook! (Which one, by the way?) The equation for ammonia is
N2 + 3H2  ::equil:: 2NH3
The expression for Kp will be quite different, and the calculation would give a very different answer. Does your answer agree with the book's?

You are right, however N2H2 is to my knowledge one of the intermediates in amonia production so it might be meant this way. Its bit confusing to be honest.

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