At high temperature, 0.500 mol of HBR was placed in a 1.00L container where it decomposed to give the equilibrium:

Question

2HBr <==> H2 + Br2

At equilibrium, the [Br2] is 0.0855 mol/L. What is the value of the equilibrium constant?

My work:
..2HBr <==> H2 + Br2
I 0.500M...---..---
C -0.171..-0.0855..+0.0855
E 0.500-0.171..-0.855..0.0855

Keq = 0.0855/(0.500-0.171)^2(-0.0855)
= 0.0855/-0.009254605
= -9.24

I'm not even sure if I'm doing this right, please help!

Answer 1

You went wrong on the C step and E step.
(HBr) = mols/L = 0.5/1 L = 0.5M which you hvave correctly.

..............2HBr <==> H2 + Br2
I..............0.5.......0....0
C..............-2x.......x....x
E............0.5-2x......x....x

The problem states that (Br2) is 0.0855 (which is x); therefore, (H2) must be 0.0855 and (HBr) must be 0.5-(2*0.0855)
Plug those values into Keq expression and calculate Keq.
Note that mols don't add up; remember mols are not conserved in a reaction. Grams yes but not mols.

Answer 2

Keq = (0.0855)^2/0.5-0.171
= 0.00731025/0.329 = 0.0222
Keq = 0.0222

is this correct?

Answer 3

Close but no cigar.
You didn't square the (0.5-0.171) which makes the 0.329 squared also. I have 0.0675

Answer 4

ooh, thank you!