Equilibrium problem, need help please?

Question

Consider the reaction N2(g) + O2(g) 2NO(g), for which Kc = 0.10 at 2,000ºC. Starting with initial concentrations of 0.040 M N2 and 0.040 M O2, determine the equilibrium concentration of NO.

a) 5.4 × 10–3 M
b) 0.0096 M
c) 0.011 M
d) 0.080 M
e) 0.10 M

okay I keep getting 0.013 which isnt even on here, maybe im doing a significant digit error, and its C?

I set the problem up like this [NO]^2/[N2][O2]=.10

I know N2 and O2 are .040 so multiply .040 with .040 = .0016, multiple .0016 with .10 to get rid fo the denominator so the problem is now. [NO]^2 = .00016 take the square of both sides to get the concentration of NO = 0.0126 = 0.013 M??

hey kb, Im not sure if you will get this, but I figured it out with your help. I was using the wrong formula to solve it. But im not sure if your numerator should be 2x^2 when you move the 2 to the exponet shouldnt it get rid of the 2 in the front? since the problem says 2NO it should be [NO]^2 with the 2 in the front gone? so the top # should just be x^2 and not 2x^2? I did it it like that and it came out as answer B, but maybe you leave the 2 in there for some reason? just making sure

nvm, I see that it is A, thanks!

Answer 1

N2 /O2 /NO
Begin 0.04 0.04 0
Change -x -x +2x (maybe you forgot the 2 here?)

K = 0.1 ={NO}^2/ {N2}{O2} (or maybe you forgot to square it?)

edit: WTH let me just do it for you (and get a tutor later or see your prof, k?)

0.1 = (2x)^2/(0.04-x)(0.04-x)
thus take the sq rt both sides
0.316227766 = 2x/0.04-x
swing 0.04 - x to the left and the constant to the right....
0.04-x = 2x/0.316227766
0.04-x = 6.32455532 x
0.04 = 7.32455532 x
x = 0.005461

the answer is a) :)

Answer 2

You need to work on your balance... Get an exercise ball and kneel on it for a while, this will strengthen your core, improve balance, improve neural links to your stabilising muscles, and therefore help retain equilibrium while sitting, walking, playing sport... whatever.

Oh, it was chemistry? Chem sux. Physics roolz