Posted by Hannah on .
For the reaction I2(g) + Br2(g) 2 IBr(g), Kc = 280. at 150.°C. Suppose that 0.570 mol IBr in a 2.50L flask is allowed to reach equilibrium at 150.°C. What are the equilibrium concentrations of IBr, I2, and Br2?

chem 
DrBob222,
M = mols/L = 0.570/2.50L = about 0.228M
........I2 + Br2 ==> 2IBr
I.......0....0......0.228
C.......x....x.....0.2282x
E.......x....x.....0.2282x
Substitute the E line into Kc expression and solve. 
chem 
Hannah,
do u have to use the quadratic to find x?

chem 
Hannah,
what do i do after i get
.051984+4x^2  .912/(x^2) 
chem 
Hannah,
.051984+4x^2  .912x=280(x^2)

chem 
Hannah,
then subtract the 4x^2? so
.051984.912x=276x^2
now what? 
chem 
DrBob222,
To here is ok.
.051984+4x^2  .912x=280(x^2)
Just rearrange to
0.05918 + 4x^2  0.912x = 280x^2
Then 280x^24x^2 + 0.912x  0.05198 = 0 and
276x^2+0.912x0.05198 = 0 and use the quadratic formula. I have x = 0.0122 or close to that.