AP CHEMISTRY
posted by Stan3000 .
For the reaction I2(g) + Br2(g) 2 IBr(g), Kc = 280. at 150.°C. Suppose that 0.590 mol IBr in a 2.50L flask is allowed to reach equilibrium at 150.°C. What are the equilibrium concentrations of IBr, I2, and Br2?

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?

i don't understand how to do it
I thought i'd use I.C.E. but somehow it didn't work
Please help? 
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......0.2362x
Substitute the ICE values into the Kc expression and solve for x.
Post your work if you get stuck. 
Sorry I forgot the arrow
but I got it thnx a lot 
and yeah i forgot about dividing 2.5L

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 = 8.5 103 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
Respond to this Question
Similar Questions

Chemistry
The decomposition of IBr(g)into I2 (g) and Br2 (g) is first order in IBr with k = 0.00255/sec. (a)Starting with [IBr] = 1.50M, what will [IBr] become after 2.50 minutes? 
Chemistry
The decomposition of IBr(g)into I2 (g) and Br2 (g) is first order in IBr with k = 0.00255/sec. (a)Starting with [IBr] = 1.50M, what will [IBr] become after 2.50 minutes? 
Chemistry
The decomposition of IBr(g)into I2 (g) and Br2 (g) is first order in IBr with k = 0.00255/sec. (a)Starting with [IBr] = 1.50M, what will [IBr] become after 2.50 minutes? 
ChemistryDrBob222
Arrows disappeared from below added them back to the equations. The decomposition of IBr(g)into I2 (g) and Br2 (g) is first order in IBr with k = 0.00255/sec. (a)Starting with [IBr] = 1.50M, what will [IBr] become after 2.50 minutes? 
ChemistryDrBob222
Arrows disappeared from below added them back to the equations. ? 
ChemistryDrBob222
Arrows disappeared from below added them back to the equations. ? 
AP CHEMISTRY
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 … 
chem
For the reaction I2(g) + Br2(g) 2 IBr(g), Kc = 280. at 150.°C. Suppose that 0.570 mol IBr in a 2.50L flask is allowed to reach equilibrium at 150.°C. What are the equilibrium concentrations of IBr, I2, and Br2? 
Chemistry
A sealed 1.0L flask is charged with .500mol I2 and .500mol Br2 and an equilibrium reaction esues: I2(g)+Br2(g)=2IBr(g) When the container contents achieve equilibrium, the flask contains .84 mol of IBr. What is the value of the eqilibrium … 
chemistry
At a certain temperature, Keq is 4.13 x 10^5 for the equilibrium: 2IBr(g) ↔ I2(g) + Br2(g) Assume that the equilibrium is established at the above temperature by adding only the reactant to the reaction flask. What are the concentrations …