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!

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 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.

To solve this question, we can use the ICE table (initial, change, equilibrium) along with the expression for the equilibrium constant (K).

Let's break down the steps to solve this problem:

Step 1: Set up the ICE table.
The initial concentrations are given, so we can fill in the initial column:
```
NOCl NO Cl2
Initial: 0.480M 0M 0M
Change: -2x +2x +x
Equilibrium: 0.480M-2x 2x x
```
We use 'x' to represent the change in concentration for each species.

Step 2: Write the expression for the equilibrium constant.
The given equilibrium constant expression is: K = [NO]^2[Cl2] / [NOCl]^2

Step 3: Substitute the equilibrium concentrations in the expression.
For part (a), the equilibrium concentration of NOCl is 0.480M - 2x, and the equilibrium concentrations of NO and Cl2 are 2x and x, respectively. Substituting these values into the expression, we get:
K = (2x)^2 * x / (0.480M - 2x)^2

Step 4: Solve for x.
We solve the equation to find x by rearranging and substituting values as needed. However, this may lead to a high-degree polynomial equation, which can be challenging to solve.

It seems like you have already attempted step 4, but keep in mind that solving the equation with a quadratic term can be challenging. It's possible that a small mistake was made in the calculation.

To help verify your calculations, I recommend using a graphing calculator or software to solve the equation or approximate the value of x. This will ensure that you get accurate results for the equilibrium concentration of NO.

Once you find the value of x, substitute it back into the equilibrium concentrations to find the equilibrium concentration of NOCl, as stated in the ICE table.

Repeat the above steps (2-4) for parts (b) and (c) of the question to find the equilibrium concentrations of NO.

I hope this explanation helps you approach the problem and find the correct answers.