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Find the equilibrium concentration

  • Thread starter jumbogala
  • Start date
  • #1
423
2

Homework Statement


At 300 K, kp = 1.5 x 1018 for the reaction 3NO --> N2O + NO2

If 0.030 mol of NO were placed in a 1.00 L vessel and and equilibrium was established, find the equilibrium concentrations of NO, N2O and NO2.


Homework Equations





The Attempt at a Solution


First, I converted kp to kc using the formula kc = kp (RT)(change in mols of gas)

I get kc = (1.5 x 10^18)((0.0821)(300K))^(-1) = 6.08 x 10^16

Then, using an ICE chart, I found the mass action expression is equal to:

x2 / (0.030 - 3x)3

kc should be equal to the mass action expression.

But I can't figure out how to solve for x. There don't seem to be any simplifications I can make.
 

Answers and Replies

  • #2
Borek
Mentor
28,327
2,714
There don't seem to be any simplifications I can make.
Looks to me like equilibrium is shifted faaaar to the right. What does it tell you about amount of products?
 
  • #3
423
2
That the amount of products is very big compared to the amount of reactants. The reaction goes pretty much to completion.

So, I shouldn't even do an ICE chart, I should just use stoich?

At what point is kc big enough that I can assume the reaction goes to completion?
 

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