For the reaction I2(g) + Br2(g) 2 IBr(g), Kc = 280. at 150.°C. Suppose that 0.590 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

I would hope that Bob Pursley's answer to your previous post cleared up some of your trouble you were having. What is your trouble with this problem?

Answer 2

i don't understand how to do it
I thought i'd use I.C.E. but somehow it didn't work
Please help?

Answer 3

It works.By the way, look around your keyboard to find the arrows. You can't do these problems if you can't tell the difference between products and reactants and you can't do that without arrows.
0.590moles/2.50L =0.236 (I'm willing to bet that this is the step you didn't do but maybe not.)
...........I2 + Br2 ==> 2IBr
initial.....0....0.......0.236M
change......+x....x..... -2x
equil........x....x......0.236-2x

Substitute the ICE values into the Kc expression and solve for x.
Post your work if you get stuck.

Answer 4

Sorry I forgot the arrow
but I got it thnx a lot

Answer 5

and yeah i forgot about dividing 2.5L

Answer 6

I've got another question. it's similar to this but this time with Kp. Here's the question:
For the equilibrium
2 IBr(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