Sorry for postiong twice. But I really need to know if the answers I am getting are correct; Nitrogen and oxygen react to produce nitric oxide according to the following equation:

N2 (g) + O2 (g) → 2 NO (g)
The equilibrium constant for this reaction is 1.70 x 10-3. Suppose that 0.110 mol N2 and 0.330 mol O2 are
mixed in a 2.00-L reaction vessel. What will be the concentrations of N2, O2, and NO when equilibrium is
established? (Hint: assume that the amounts of N2 and O2 that react are small
So I got N2=0.052266,O2=0.162266,NO=0.005468

I ran through it very quickly but didn't get that number. Post your work and let me check it. Same for the previous post

So first I found the Conc. of N2 and O2, and got 0.0550 and 0.165 respectively. I then made an ice table and got the following equation; 1.7x10^-3=(2x)^2/(0.0550-x)(0.165-x). Once I got X i plugged it into the equilibrium expressions derived from the ice table to get my answer.

My eq expressions were; N2=0.0550-x,O2=0.165-x, NO=2x

So far you're ok but you didn't say what x was.

Your expression for N2 = 0.055-x is OK and I neglected x when solving.
Same for O2 = 0.165 -x is OK and I neglected x when solving.
Same for NO = 2x is OK.
We may differ on x. I found x to be 1.96E-3 .
Then N2 = 0.0530
O2 = 0.163
NO = 3.92E-3
Your N2 and O2 are close which makes me think x is close to mine but I can't explin how you obtained the 0.005468 for NO. If my x = 1.96E-3 then 2x = 3.92E-3

To determine the concentrations of N2, O2, and NO at equilibrium, we'll need to set up an ICE (Initial, Change, Equilibrium) table and use the given information.

1. Set up the ICE table:
N2 (g) + O2 (g) → 2 NO (g)

| | N2 (g) | O2 (g) | NO (g) |
| ------- | -------- | -------- | -------- |
| Initial | 0.110 M | 0.330 M | 0 |
| Change | -x | -x | +2x |
| Equilib.| 0.110-x | 0.330-x | 2x |

2. Use the equilibrium constant expression to relate the equilibrium concentrations:
Kc = [(NO)^2] / [(N2)(O2)]
1.70 x 10^(-3) = [(2x)^2] / [(0.110-x)(0.330-x)]

3. Solve the equation for x:
(2x)^2 = 1.70 x 10^(-3) * (0.110-x)(0.330-x)
4x^2 = 1.70 x 10^(-3) * (0.0363 - 0.44x + x^2)
4x^2 = 6.171 x 10^(-5) - 7.38 x 10^(-4) x + 1.70 x 10^(-3) x^2
1.70 x 10^(-3) x^2 - 7.38 x 10^(-4) x + 6.171 x 10^(-5) - 4x^2 = 0
-2.30 x 10^(-3) x^2 + 7.38 x 10^(-4) x + 6.171 x 10^(-5) = 0

Solving this quadratic equation will yield two possible values for x. We'll consider the positive value since the concentrations cannot be negative.

4. Calculate the concentrations at equilibrium:
Substitute the positive value of x back into the equilibrium expressions from the ICE table to find the concentrations.

[N2] = 0.110 - x
[O2] = 0.330 - x
[NO] = 2x

Plug in the positive value of x obtained from the quadratic equation and calculate the concentrations.

Without solving the quadratic equation, it is not possible to confirm if your calculated values for N2, O2, and NO are correct.