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 $8.7 \times 10^{-5}$

  • $\begingroup$ 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. $\endgroup$
    – Philipp
    Feb 16, 2014 at 1:11
  • $\begingroup$ I used k = [Product]/[Reactant] I'll write out all my calculations in a few minutes. $\endgroup$
    – Jaco
    Feb 16, 2014 at 1:55

1 Answer 1


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

$$K_c = \frac{[C]^c}{[A]^a[B]^b}$$

You might want to check yours.

  • $\begingroup$ I agree, that's how I attacked the problem I got (2x)^2 / (.03104 - x)^2 = K ;; 2x/(.03104 - x) = sqrt(k) ;;; 2x = sqrt(k)(.03104 - x) ;;;; 9.8157e-5 - .00316x = 2x ;;;;; 9.8157e-5 = 2.00316x ;;; x = 4.9e-5 $\endgroup$
    – Jaco
    Feb 16, 2014 at 16:11
  • $\begingroup$ It's .03104 and not .0275 because its .0275 moles in a .886 L container so .0275/.886 = .03104 M $\endgroup$
    – Jaco
    Feb 16, 2014 at 16:15

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