Posted by **Im so sick of this problem** on Monday, September 12, 2011 at 9:57pm.

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 -
**Don**, Monday, September 12, 2011 at 10:19pm
K = (Product/Reactant)

1.6*10^-5 = (x)^2 * 0 )/(.480)^2

x = 0.00191833261

Someone check this

- chemistry -
**it's not right**, Monday, September 12, 2011 at 10:25pm
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 -
**Dr Russ**, Tuesday, September 13, 2011 at 6:19am
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.

## Answer this Question

## Related Questions

- Ap Chemistry - Hi, I was wondering if you could possibly help me with this chem ...
- Chem II - 2 NOCl(g) --> 2 NO(g) + Cl2(g) Let’s assume that at a given ...
- Chemistry - 0.500 moles of NOCl is placed in a 1.00L vessel at 700K,and after ...
- Chemistry - NOCL(g) decomposes to from nitrogen monoxide gas and chlorine gas. ...
- chemistry - Hey Guys so I have this problem with ICE tables that is extremely ...
- CHEMISTRY - When NO(g) (0.06623 mol/L) and 422.6 grams of Cl2(g) in a 180.0 L ...
- Chemistry - At 35 degrees Celcius, K = 1.6x10^-5 for the reaction: 2NOCl<--&...
- Chemistry - Consider the following reaction: 2NOCl(g) 2NO(g) + Cl2(g) Initially...
- Chem II - 2.5 mol NOCl(g) was placed in a 2.50 L reaction vessel at 400 degrees ...
- Chemistry - The reaction 2NO + Cl2 <==> 2NOCl has Kc= 210 what is the ...

More Related Questions