Finding the composition of a mixture at equlibrium

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Homework Statement



Initially a mixture contains 0.723 mol each of N2 and O2 in an 7.20 L vessel. Find the composition of the mixture when equilibrium is reached at 3900°C. For the reaction below, the Kc = 0.0123 at 3900°C.
N2(g) + O2(g) 2 NO(g)



Homework Equations


I.C.E Tables (not an equation but relevant) and the quadtratic formula

x=-b +or- sq root(b^2 -4ac) / 2a


The Attempt at a Solution



I did my I.C.E Table
Conc.(M) N2 + O2 --> 2 NO
Start 0.6025 0.6025 0
Change -x -x 2x
Eq. 0.6025-x 0.6025-x 2x

Then I plugged that into my Kc formula
Kc=[products]/[reactants]
0.0123=[2x]^2/[0.6025-x][0.6025-x]

I then rearranged that to get a quadratic equation to solve for x (this is where I think I made a mistake)
-324x^2 -1.205x +0.3630=0
I plugged it into the quadratic formula and
got 0.03166 for x, which i plugged into the equations from the EQ. line in my I.C.E table, however when i plugged those values into the Kc formula, i didn't come out with the Kc value I was given (0.0123)

Can anyone help?
 

Answers and Replies

  • #2
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nevermind, I was (stupidly) using a value from another question, but my method was right.
 

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