[The Phil]

Question: Keq=0.914 for the reaction NO2(g) + NO(g) <---> N20(g) +O2(g)
A mixture was prepared containing 0.200 mol N02, 0.300 mol NO, 0.150 mol N2O and 0.250 mol O2 in a 4.00 L container. What will be the equilibrium concentration of each gas?

I've worked it down with a quadratic and the numbers I get are too large.

Here is my work so far:

0.914=(0.0375+x)(0.0625+x)/(0.05-x)(0.075-x)

0.914=x^2 + 0.1x + 0.00234/x^2 - 0.125x + 0.00375

0.914x^2 - 0.114x + 0.00343 = x^2 + 0.1x + 0.00234

0 = 0.086x^2 + 0.214x - 0.00109

-0.214 +/- root((0.214)^2 - 4(0.086)(-0.00109))/2(0.086)

My answer for that comes out to be -0.214 +/- 1.25 which appears to be too large to plug in for the x (change in the ICE table) value.

Any help will be greatly appreciated!