A 23.0 g sample of I2(g) is sealed in a gas bottle having a volume of 500 mL. Some of the molecular I2(g) dissociates into iodine atoms and after a short time the following equilibrium is established. I2(g-->2I(g) For this system, Kc= 3.80 x 10^-5. What mass of I2(g) will be in the bottle when equilibrium is established?

O.K., this is actually a pretty common equilibrium calculation. What we are going to do is first write down the equilibrium expression, then rearrange it to solve for `I_2(g)`.

Then it gets interesting because we also have a condition that that total amount of iodine must equal 23.0 grams. So we are going to have to substitute a variable for the amount of `I_(g)` and then solve for that variable. Once we have that, we can go back and determine how much `I_2(g)` remains.

By the way, 23.0 grams of `I_2(g)_0` in 500 ml is a concentration of 0.18123869080419310683067349794692 moles per liter. (We always retain all the digits in a calculation until we get to the answer, then we round it off to, in this case, 3 significant figures.)

So: `K_c` = `[I_(g)]^2`/`[I_2(g)]` (1)

Great. Now we have to figure out how much `I_(g)` we have.

The balanced chemical equation is:

`I_2(g)` <==> `2I_(g)`

Let's say that "X" `I_2` molecules dissociate into "2X" iodine atoms:

Initial amounts: 0.18123869 M of `I_2(g)` and 0 M of `I_(g)`
Final Amounts: 0.18123869-X M of `I_2(g)` and 2X M of `I_(g)`
(that is, each iodine molecule will produce two iodine atoms)

So, substituting the final amounts into equation (1):

`(2X)^2/(0.1823869-X) = 3.80 × 10^-5`

Since `K_"c"` is such a small number, very little of the `"I"_2` will dissociate. We can assume that `X` is negligible in comparison with 0.1823869.

The equation then becomes

`(4X^2)/0.1823869 = 3.80 × 10^-5`

or

`4X^2 = 0.1823869 × 3.80 × 10^-5 = 6.930702 × 10^-6`

`X^2 = 1.732676 × 10^-6`

`X = 1.31631 × 10^-3`

So the final concentration of `I_2(g)` is going to be (0.18123869 - 1.31631 × 10⁻³ ) M = 0.1799224 M.

The mass of `"I"_2` at equilibrium is

0.500 L × `(0.1799224"mol I"_2)/(1"L") × (253.81"g I"_2)/(1"mol I"_2) = 22.8"g I"_2`

So your final answer is, 22.8 grams of `I_2(g)`

Cheers!