Equilibrium constant problem. Did most of the work but cant find my error.?

Question

Iodine and Bromine react to give Ibr

I2 + Br2 <---------> 2IBr

What is the equilibrium composition of a mixture that initially contained .0015 mol each of iodine and bromine in a 5.0 L vessel. K=1.2 x 10*2

I and Br=.0003 mol

[2x]*2
[.0003-x]*2 = 120

Square root that….
2x/.0003-x = 10.95

Solve for x= 3.67 x 10*4

The correct answer is

I and Br at equil = 4.7 x 10*-5

What did I Miss the boat on? I know its hard to try to understand my work on this thing but please help.

Answer 1

Your first problem is that you need to be better with units. When you wrote .0003 mol, you should have wrote .0003mol/L, but that is just a clerical problem. Your work up to solve for x is correct, I just suspect that you made a small error while trying to solve for x. Lets go from the previous step together.

(2x)/(.0003-x) = 10.95 Good.
2x = (.0003-x) * 10.95
2x = .00329 - 10.95x
2x + 10.95x = .00329
12.95x = .00329
x = .000254 mol/L
I and Br = .0003 - x = .0003 - .000254 = 4.6*10^-5.

Your process and thinking were good, just be careful of the algebra, it can cause a lot of problems. :)

Answer 2

I2 + Br2 ↔ 2 IBr
Initial concentration 0.0003 M 0.0003 M 0

Equilibrium conc. ( 0.0003 – x )M ( 0.0003 – x ) 2x M

This was where you went wrong:

Solving [ (2x)*2]/(0.0003 - x)*2= 120 will on expansion and simplification result in quadratic function in which there are two possible values of x. One of the values would be too high so the second is ideal. Let us expand, simplify and solve for x:

120(0.0003*2 - 2(0.0003)x + x*2) = 4x*2
0.0000108 - 0.072x + 116x*2

x = 3.67 x 10*-4 or x = 2.54 x 10*-4

3.67 x 10*-4 will cause the concentration or I2 or Br2 to be negative since it is too high.
[I2] = [Br2] = (0.0003 - 3.67 x 10*-4) = - 6.7 x 10*-5

x = 2.54 x 10*-4 is the better choice

so [I2] = [Br2] = (0.0003 - 2.54 x10*-4) = 4.6 x 10*-5

This I am absolutely sure is the correct answer.