Question

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

Answer 7

[AI-generated answer — Explain Bot]

To find the equilibrium concentrations of IBr, I2, and Br2, you can use the given information about the reaction and the initial conditions in the 2.50-L flask.

We are given that the reaction is: I2(g) + Br2(g) ↔ 2 IBr(g), and the equilibrium constant (Kc) at 150.°C is 280.

Let's assume that x mol of I2 and x mol of Br2 react to form 2x mol of IBr at equilibrium.

Based on the stoichiometry of the reaction, the change in the concentrations of I2, Br2, and IBr are as follows:

[I2] changes by -x (concentration decreases by x).
[Br2] changes by -x (concentration decreases by x).
[IBr] changes by +2x (concentration increases by 2x).

At equilibrium, the concentrations of I2, Br2, and IBr will be the sum of the initial concentration and the change in concentration.

Given that the initial concentration of IBr is 0.590 mol and the total volume is 2.50 L, we can determine the initial concentration as follows:

[IBr]initial = (0.590 mol)/(2.50 L) = 0.236 M

Since the change in concentration of I2, Br2, and IBr are -x, -x, and +2x respectively, we can express their equilibrium concentrations as:

[I2]equilibrium = [I2]initial - x
[Br2]equilibrium = [Br2]initial - x
[IBr]equilibrium = [IBr]initial + 2x

At equilibrium, the equilibrium constant expression (Kc) is given by:

Kc = ([IBr]equilibrium)^2 / ([I2]equilibrium * [Br2]equilibrium)

Substituting the expressions for the equilibrium concentrations, we have:

Kc = ([IBr]initial + 2x)^2 / ([I2]initial - x) * ([Br2]initial - x)

Plugging in the given Kc value of 280, we can solve for x. However, this requires solving a quadratic equation.

By substituting the initial concentrations, values for x can be found and then plugged back into the equilibrium expressions to find the equilibrium concentrations [I2], [Br2], and [IBr].