# chemistry

posted by 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 10-5
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!

• chemistry - ,

K = (Product/Reactant)
1.6*10^-5 = (x)^2 * 0 )/(.480)^2
x = 0.00191833261

Someone check this

• chemistry - ,

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

• chemistry - ,

2 NOCl(g)-> 2 NO(g) + Cl2(g) K = 1.6 10-5

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]=0.480-x
[NO] = x
[Cl2] =x/2 (because half number of moles wrt NOCl.)

hence

(x)^2(x/2)/(0.480-x)^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]=0.480-x
[NO] = x
[Cl2] =1.99+x/2 (because half number of moles wrt NOCl.)

hence

(x)^2(1.99+x/2)/(0.480-x)^2= 1.6 x 10^-5

assume x is small

x^2(1.99)/0.960 = 1.6 x 10^-5

x^2=7.72E-6

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.

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