Consider the following reaction for which K = 1.60 x 10 -7 at some temperature:
ID: 925733 • Letter: C
Question
Consider the following reaction for which K = 1.60 x 10-7 at some temperature:
Since K is very small, K = 1.60 x 10-7, very little product will be present at equilibrium, which tells us that very little NOCl will react to reach equilibrium. Because of this, the equilibrium concentration of NOCl will remain, essentially, at its initial concentration. Making this assumption, calculate the equilibrium concentration of NO. Note: use the ICE table from the previous problem to help you solve this problem. (Scientific notation can be entered using the following convention: 1.93 x 10-3 = 1.93E-3.)
[NO]eq =
Explanation / Answer
2 NOCl (g) <-> 2 NO (g) + Cl2(g)
I [NOCl]o 0 0
C - + [x] +[x/2]
E [NOCl]o x x/2
1.6 x 10-7 = [NO]2 [Cl2] / [NOCl]2
1.6 x 10-7 = [x]2 [x/2] / [NOClo]2
From there you'll need more data to obtain initial concentration of NOClo, so that you can isolate for x.
In case the concentration changes, the problem will be stated as follows:
2 NOCl (g) <-> 2 NO (g) + Cl2(g)
I [NOCl]o 0 0
C [NOCl]o - x + [x] +[x/2]
E [NOCl]o - x x x/2
1.6 x 10-7 = [NO]2 [Cl2] / [NOCl]2
1.6 x 10-7 = [x]2 [x/2] / [NOClo - x]2