Posted by **Hannah** on Wednesday, February 20, 2013 at 9:03pm.

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

- chem -
**DrBob222**, Wednesday, February 20, 2013 at 9:48pm
M = mols/L = 0.570/2.50L = about 0.228M

........I2 + Br2 ==> 2IBr

I.......0....0......0.228

C.......x....x.....0.228-2x

E.......x....x.....0.228-2x

Substitute the E line into Kc expression and solve.

- chem -
**Hannah**, Wednesday, February 20, 2013 at 9:59pm
do u have to use teh quadratic to find x?

- chem -
**Hannah**, Wednesday, February 20, 2013 at 10:01pm
what do i do after i get

.051984+4x^2 - .912/(x^2)

- chem -
**Hannah**, Wednesday, February 20, 2013 at 10:02pm
.051984+4x^2 - .912x=280(x^2)

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**Hannah**, Wednesday, February 20, 2013 at 10:03pm
then subtract the 4x^2? so

.051984-.912x=276x^2

now what?

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**DrBob222**, Wednesday, February 20, 2013 at 10:22pm
To here is ok.

.051984+4x^2 - .912x=280(x^2)

**Just rearrange to
**

0.05918 + 4x^2 - 0.912x = 280x^2

Then 280x^2-4x^2 + 0.912x - 0.05198 = 0 and

276x^2+0.912x-0.05198 = 0 and use the quadratic formula. I have x = 0.0122 or close to that.

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