I2(g) + Br2(g) ↔ 2 IBr(g) What is the equilibrium concentration of IBr (in M to 4 decimal places) if 0.200 mol of I2 and 0.100 mol Br2 in a 5.00 L vessel if Kc = 29 at 150°C?

The equilibrium concentration of IBr is 0.0360 M or 0.1032 M

The equilibrium concentration of IBr is 0.752 mol

Explanation:

Mathematically, concentration = number of moles/volume

For Iodine molecule, concentration = 0.2/5 = 0.04 M

For the bromine molecule, concentration = 0.1/5 = 0.02 M

We need to set up an ICE table ( Initial, change, equilibrium)  to get the concentrations of the molecules at equilibrium.

Please check this table in the attachment.

From the table, we can see that the equilibrium concentration of Iodine is (0.04-x), that of Bromine is (0.02-x) while that of IBr is 2x

Now, we need to get the equilibrium constant  using the equilibrium concentration of the molecules

With the given chemical reaction, we can represent the equilibrium constant in terms of pressure as follows;

Kc = [IBr]^2/[Br2][I2]

Thus;

Kc = (2x)^2/(0.04-x)(0.02-x)

Kc = 4x^2/(0.04-x)(0.02-x)

From the question, Kc = 29 , so substitute this;

29 = 4x^2/(x^2-0.06x+0.008)

Cross multiply

29((x^2-0.06x+0.008) = 4x^2

29x^2 - 1.74x +0.0232 = 4x^2

Thus;

25x^2 - 1.74x + 0.0232 = 0

Solving this quadratic equation, we get two values for x which are x = 0.0516 and 0.0180

The equilibrium concentration is either (2 * 0.018) or (2 * 0.0516) which is 0.0360 or 0.1032