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?

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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?

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Answer 1 · 2014-07-22T19:30:46

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 $I_{2} (g)$ $I_2(g)$ .

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 $I_{g}$ $I_(g)$ and then solve for that variable. Once we have that, we can go back and determine how much $I_{2} (g)$ $I_2(g)$ remains.

By the way, 23.0 grams of $I_{2} {(g)}_{0}$ $I_2(g)_0$ 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: $K_{c}$ $K_c$ = ${[I_{g}]}_{g}^{2}$ $[I_(g)]^2$ / $[I_{2} (g)]$ $[I_2(g)]$ (1)

Great. Now we have to figure out how much $I_{g}$ $I_(g)$ we have.

The balanced chemical equation is:

$I_{2} (g)$ $I_2(g)$ <==> $2 I_{g}$ $2I_(g)$

Let's say that "X" $I_{2}$ $I_2$ molecules dissociate into "2X" iodine atoms:

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

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

$\frac{{(2 X)}^{2}}{0.1823869 - X} = 3.80 \times 10^{- 5}$ $(2X)^2/(0.1823869-X) = 3.80 × 10^-5$

Since $K_{c}$ $K_"c"$ is such a small number, very little of the $I_{2}$ $"I"_2$ will dissociate. We can assume that $X$ $X$ is negligible in comparison with 0.1823869.

The equation then becomes

$\frac{4 X^{2}}{0.1823869} = 3.80 \times 10^{- 5}$ $(4X^2)/0.1823869 = 3.80 × 10^-5$

or

$4 X^{2} = 0.1823869 \times 3.80 \times 10^{- 5} = 6.930702 \times 10^{- 6}$ $4X^2 = 0.1823869 × 3.80 × 10^-5 = 6.930702 × 10^-6$

$X^{2} = 1.732676 \times 10^{- 6}$ $X^2 = 1.732676 × 10^-6$

$X = 1.31631 \times 10^{- 3}$ $X = 1.31631 × 10^-3$

So the final concentration of $I_{2} (g)$ $I_2(g)$ is going to be (0.18123869 - 1.31631 × 10⁻³ ) M = 0.1799224 M.

The mass of $I_{2}$ $"I"_2$ at equilibrium is

0.500 L × $\frac{0.1799224 {mol I}_{2}}{1 L} \times \frac{253.81 {g I}_{2}}{1 {mol I}_{2}} = 22.8 {g I}_{2}$ $(0.1799224"mol I"_2)/(1"L") × (253.81"g I"_2)/(1"mol I"_2) = 22.8"g I"_2$

So your final answer is, 22.8 grams of $I_{2} (g)$ $I_2(g)$

Cheers!

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