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.