Question

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Be sure to answer all parts. The dissociation of molecular iodine into iodine atoms is represented as I2(g) - 21g) At 1000 K, the equilibrium constant Kc for the reaction is 3.80 × 10-5. Suppose you start with 0.0458 mol of I2 in a 2.33-L flask at 1000 K. What are the concentrations of the gases at equilibrium? What is the equilibrium concentration of I2? What is the equilibrium concentration of I?

Answer 1

The given chemical reaction is

$I_2 _{(g)} \rightleftharpoons 2I_{(g)}$

Equilibrium constant $K_c$ of a reaction is given by

$K_c = \frac{\prod [Product]^a}{\prod [Reactant]^b}$

Where a and b are the stoichiometric coefficient in the balanced reaction.

Hence, $K_c$ for our reaction is written as

$K_c = \frac{[I]^2}{I_2} = 3.80 \times 10^{-5}$

where the concentrations are their equilibrium values.

The initial amount of I2 is 0.0458 mol.

Volume of the flask = 2.33 L

Hence, initial concentraion of the I2 gas is

$\frac{0.0458 \ mol}{2.33 \ L} \approx 0.020 \ M$

Each mole of $I_2$ dissociates to give 2 moles of .

Hence, we can write the ICE(Initial, Change, Equilibrium) chart as follows:

	$[I_2]$
Initial	0.020	0
Change	-x	+2x
Equilibrium	0.020-x	2x

Hence, using the dissociation constant equation

$K_c = \frac{[I]^2}{I_2} = 3.80 \times 10^{-5} \\ \Rightarrow 3.80 \times 10^{-5} = \frac{(2x)^2}{0.020-x} \\ \Rightarrow x = - 4.4 \times 10^{-4} \ or \ x = 4.31 \times 10^{-4}$

Since, x is a concentration and cannot be negative, $x = 4.31 \times 10^{-4} \ M$ .

Equilibrium concentration of $I_2$ = 0.020 M -x.

But since , we can say that the concentration of $I_2$ does not change.

Hence,

$[I_2]_{(eq)} = 2.0 \times 10^{-2} \ M - 4.31 \times 10^{-4} \ M \approx 2.0 \times 10^{-2} \ M$

Now,

$[I]_{eq} = 2x = 2 \times 4.31 \times 10^{-4} \ M = 8.62 \times 10^{-4} \ M$

Hence, equilibrium concentration of $I_2$ is 0.020 M.

Equilibrium concentraion of I is 0.000862 M.