chemistry
posted by Im so sick of this problem on .
I have been at this question for at least three hours now.
Consider the chemical system below.
2 NOCl(g) 2 NO(g) + Cl2(g) K = 1.6 105
What is the equilibrium concentration of nitrogen monoxide, [NO], given each of these initial conditions?
(a) [NOCl]0 = 0.480 M; [Cl2]0 = 0.0 M
0.00975949307 M <this is the answers I keep getting, but it's wrong I guess
(b) [NOCl]0 = 1.490 M; [Cl2]0 = 1.99 M
M
(c) [NOCl]0 = 2.13 M; [Cl2]0 = 1.06 M
M
The formula uses ICEA and for the first one at least, I have to use 1.6x10^5= x(2x)^2/0.480^2
I plug in the numbers, WRONG. Please help!

K = (Product/Reactant)
1.6*10^5 = (x)^2 * 0 )/(.480)^2
x = 0.00191833261
Someone check this 
If this is ICEA, you forgot the 2x^2.....should be getting something like 4x^3 on the top w/ the products.
I'm actually getting the same number as the person asking this question 
2 NOCl(g)> 2 NO(g) + Cl2(g) K = 1.6 105
note that units are missing for K, which I assume as mol dm^3
The equilibrium constant is given by
K=[NO]^2[Cl2]/[NOCl]^2
start with [NOCl]= 0.480 mol dm^3
at equilibrium
[NOCl]=0.480x
[NO] = x
[Cl2] =x/2 (because half number of moles wrt NOCl.)
hence
(x)^2(x/2)/(0.480x)^2= 1.6 x 10^5
assume x is small
x^3/0.960 = 1.6 x 10^5
x^3=0.00001536
x=0.0249
part b)
start with [NOCl]= 0.480 mol dm^3
and [Cl2]=1.99 mole dm^3
at equilibrium
[NOCl]=0.480x
[NO] = x
[Cl2] =1.99+x/2 (because half number of moles wrt NOCl.)
hence
(x)^2(1.99+x/2)/(0.480x)^2= 1.6 x 10^5
assume x is small
x^2(1.99)/0.960 = 1.6 x 10^5
x^2=7.72E6
x=0.00278
x=0.0249
(which is the result expected as higher [Cl2] will drive the equilibrium to the left and hence lower [NO] then in a).
c) repeat the process of b)
I have left out units which will need to be inserted. Also please check my calculator maths.