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## Homework Statement

At 300 K, kp = 1.5 x 10

^{18}for the reaction 3NO --> N

_{2}O + NO

_{2}

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, N

_{2}O and NO

_{2}.

## 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:

x

^{2}/ (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.