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Topic: Find the equilibrium concentrations of N2 , O2, and NO at equilibrium. :/  (Read 6047 times)

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Offline awooooooooga

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The reaction N2(g)+O2(g)⇌2NO(g) is carried out at a temperature at which Kc = 0.040. The reaction mixture starts with only the product, [NO] = 0.0500M , and no reactants.
I did the thing where its
((Ci)(SqrtKc) = (1+2(sqrtKc))x
X=.00714
So I tried to find the concentrations, and I get the concentrations of NO .036M and N2 and O2 .0071M, but they aren't right??
This is due tommorrow morning and I don't know what to do!

Offline awooooooooga

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Re: Find the equilibrium concentrations of N2 , O2, and NO at equilibrium. :/
« Reply #1 on: February 24, 2014, 10:25:44 AM »
I figured it out.
I had the products and reactants backwards

Offline Corribus

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Re: Find the equilibrium concentrations of N2 , O2, and NO at equilibrium. :/
« Reply #2 on: February 24, 2014, 10:28:42 AM »
So I tried to find the concentrations, and I get the concentrations of NO .036M and N2 and O2 .0071M, but they aren't right??
I see you've figured it out, but just as a general rule there are easy ways to check whether your answers are likely to be right. Case in point: If these were your concentrations at equilibrium, your equilibrium constant would be ~25.  Your equilibrium constant given is 0.04. This should tell you that you will have considerably more reactants than products at equilibrium, generally speaking. You should be able to recognize immediately that your answer isn't correct just by this fact alone.

(Also the fact that the 25 is exactly the reciprocal of 0.04 should have also told you something.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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