Posted by Stan3000 on Monday, February 6, 2012 at 9:28pm.
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 = 8.5 103 at 150.°C. If 0.016 atm of IBr is placed in a 2.5L 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

AP CHEMISTRY  DrBob222, Monday, February 6, 2012 at 10:27pm
..........2IBr ==> I2 + Br2
initial....0.016.....0....0
change......2x......x.....x
equil......0.0162x...x.....x
Kp = 8.5E3 = (x)(x)/(0.0162x)^2
8.5E3(0.0162x)^2 = x^2
8.5E3*(2.56E4  0.064x + 4x^2) = x^2
etc.
I think your 8.5E3 is in the wrong place. 
AP CHEMISTRY  Stan3000, Tuesday, February 7, 2012 at 9:06pm
thnx!!!