Question

I've got another question. it's similar to this but this time with Kp. Here's the question:

For the equilibrium
2IBr(g) <-> I2(g) + Br2(g)
Kp = 8.5 10-3 at 150.°C. If 0.016 atm of IBr is placed in a 2.5-L container, what is the partial pressure of all substances after equilibrium is reached?

I used I.C.E

2IBr <--> I2 + Br2
I .016 0 0

C x -x -x

E .016+x -x -x

and i got the quadratic equation as:
-0.0085x^2 + .032x + 2.56*10^-4 =0

Please Help

Answer 1

..........2IBr ==> I2 + Br2
initial....0.016.....0....0
change......-2x......x.....x
equil......0.016-2x...x.....x

Kp = 8.5E-3 = (x)(x)/(0.016-2x)^2
8.5E-3(0.016-2x)^2 = x^2
8.5E-3*(2.56E-4 - 0.064x + 4x^2) = x^2
etc.
I think your 8.5E-3 is in the wrong place.

Answer 2

thnx!!!