# Find the equilibrium concentration

• jumbogala

## 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.

## 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.

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?

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?