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?

Answer 1

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.

Answer 2

do u have to use the quadratic to find x?

Answer 3

what do i do after i get

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

Answer 4

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

Answer 5

then subtract the 4x^2? so

.051984-.912x=276x^2

now what?

Answer 6

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.