1
$\begingroup$

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 $8.7 \times 10^{-5}$

$\endgroup$
2
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.