Consider the following reaction in the solvent CCl4 in a 1-L flask:

2IBr(g)<->Br2(g)+I2(g)
What is the equilibrium moles of IBr, Br2 and I2 are 0.2, 0.1 and 0.1, calculate the equilibrium constant Kc.

I assume you mean the equilibrium mols are as given and you want to know Kc.

(IBr) = 0.2/1 L - 0.2M
(Br2) = 0.1/1) = 0.1M
(I2) = 0.1/1) = 0.1M
Then Kc = (Br2)*(I2)/(IBr)^2
Substitute and solve for Kc.

To calculate the equilibrium constant, Kc, for the given reaction, you need to use the balanced equation and the initial and equilibrium concentrations of the species involved.

Given:
Initial moles of IBr = 0.2
Initial moles of Br2 = 0.1
Initial moles of I2 = 0.1

Let's assume that at equilibrium, the moles of IBr, Br2, and I2 are x, y, and z, respectively.

The balanced equation for the reaction is:
2IBr(g) <-> Br2(g) + I2(g)

According to the balanced equation, for every 2 moles of IBr, 1 mole of Br2 and 1 mole of I2 are produced.
Therefore, at equilibrium, the moles of Br2 and I2 will be equal.

The expression for the equilibrium constant, Kc, is given by:
Kc = ([Br2] * [I2]) / ([IBr]²)

Now, let's plug in the values:

Initial moles of IBr = 0.2
Moles of IBr at equilibrium = x

Initial moles of Br2 = 0.1
Moles of Br2 at equilibrium = y

Initial moles of I2 = 0.1
Moles of I2 at equilibrium = z

Since the moles of Br2 and I2 are equal at equilibrium, we can substitute y for z.

Kc = ([Br2] * [I2]) / ([IBr]²)
= (y * y) / (x * x)

Given that x = 0.2, y = 0.1, and z = 0.1, we can calculate Kc:

Kc = (0.1 * 0.1) / (0.2 * 0.2)
= 0.01 / 0.04
= 0.25

Therefore, the equilibrium constant Kc for the given reaction is 0.25.