# How do I determine the equilibrium concentration given Kc and the concentrations of component gases?

I feel my method for solving it is correct and I've checked my calculations yet I can't get the right answer.

Nitrogen and oxygen gases may react to form nitrogen monoxide. At 1500 °C, $K_c$ equals 1.0E-5.

$\ce{N2_{(g)} + O2_{(g)} <=> 2 NO_{(g)}}$

If $2.75 \times 10^{-2}$ mol $N_2$ and $2.75 \times 10^{-2}$ mol $O_2$ are sealed in a $0.886 L$ flask at $1500 °C$, what is the concentration of $\ce{NO_{(g)}}$ when equilibrium is established?

Notes: $K_c$ = equilibrium constant

a) 1.38E-7 M
b) 4.35E-5 M
c) 3.85E-5 M
d) 8.70E-5 M
e) 2.30E+4 M


I set up the problem with $(2x)^2$ as a numerator and $(.03104 - x)^2$ as the denominator

I took square root of both sides and did algebra and got $4.9*10^{-5}$

For the record the answer is $8.7 \times 10^{-5}$

• Welcome to Chemistry.SE! We have the MathJax plugin installed here so you can format your chemistry and maths better. Could you be a little more specific as to which formulas you used - it would be best if you wrote down the whole calculation method you used. – Philipp Feb 16 '14 at 1:11
• I used k = [Product]/[Reactant] I'll write out all my calculations in a few minutes. – Jaco Feb 16 '14 at 1:55

For a reaction in the form $\ce{ \mathit{a}\,A + \mathit{b}\,B <=> \mathit{c}\,C}$, the quilibrium constant $K_c$ is given by
$$K_c = \frac{[C]^c}{[A]^a[B]^b}$$