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
3 answers
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
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
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
1 answer