The compound NOCl decomposes to nitric oxide and chlorine according to the following equation:
2 NOCl (g) → 2 NO (g) + Cl2 (g)
Suppose that 0.550 mol NOCl is placed in a 2.00-L flask at a given temperature. When equilibrium has been
established, it is found that the concentration of NO is 0.0778 M. Calculate the equilibrium constant for this
reaction.
So I got 2.39x10^-3, anyone else?
2 NOCl (g) → 2 NO (g) + Cl2 (g
John, Michael, you should make up your mind what screen name you will use and stick with it. You will get better help that way.
If you will post your work I'll check it for you.
So I used an ice table
and got the following equation; keq= (2x)^2(x)/(0.275-2x)^2. I plugged in x as 0.0778, and solved.
Isn't 2x = 0.0778?
..................2 NOCl (g) → 2 NO (g) + Cl2 (g
I.................0.275...............0...............0
C.............-2x.....................2x..............x
E...........0275-2x..............2x...............x
If (NO )at equilibrium is 0.0778 then that is 2x.
did you get 6.0315x10^-3?
Close. I got 0.00607.
wait but isnt NO 0.0778?Because it says NO is 0.0778. So 2NO would then be 2 times the molarity no?
No.
NO in the problem is 0.0778. In the ICE chart it is 2x. So x is 0.0389 which makes 2x 0.0778 and that's the number given in the problem.