Question

At \(800 \mathrm{~K},\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5}\). If an equilibrium mixture in a 5.00-L vessel contains \(30.5 \mathrm{mg}\) of \(\mathrm{I}(g)\), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

Short answer

In the equilibrium mixture, there are 0.01624 grams of \(I_2\).

Step-by-step solution

Write the balanced equation

The balanced equation for the dissociation of I2 into I atoms is given as: \(I_2 (g) \rightleftharpoons 2\; I(g)\)

Convert the mass of I to moles

We are given that there are 30.5 mg of I(g) in the equilibrium mixture. To convert this into moles, use the molecular weight of I (which is 126.90 g/mol): \(n_{I} = \frac{30.5\; \text{mg}}{126.90\; \frac{\text{g}}{\text{mol}}} \times \frac{1\; \text{g}}{1000\; \text{mg}} = 2.403 \times 10^{-4}\; \text{mol}\)

Determine the initial moles of each species

Since the reaction has not yet occurred, we know that there are initially no I atoms. Let's assume there are y moles of I2: Initial moles of I2: \(y\; \text{mol}\)

Set up the ICE table

An ICE (Initial, Change, Equilibrium) table can be used to track the changes in the moles of each species throughout the reaction: \[ \begin{array}{c|c|c|c} & \text{[I}_2\text{]} & \text{[I]} \\ \hline \text{Initial} & y & 0 \\ \text{Change} & -x & +2x \\ \text{Equilibrium} & y-x & 2x \\ \end{array} \] Since we know the moles of I at equilibrium (2.403 x 10^{-4} mol), we can express the moles of I2 at equilibrium in terms of x: \(y - x = 2x\) We are given the equilibrium constant Kc: \(K_c = 3.1 \times 10^{-5}\)

Use the equilibrium constant and the ICE table

From the balanced equation and the ICE table, we can write the expression for the equilibrium constant Kc: \(K_c = \frac{[\text{I}]^2}{[\text{I}_2]} = \frac{(2x)^2}{y-x}\) Now substitute the given value of Kc and the moles of I at equilibrium: \(3.1 \times 10^{-5} = \frac{(2x)^2}{y-x}\) We can use the expression we derived for the moles of I2 at equilibrium in terms of x: \(3.1 \times 10^{-5} = \frac{(2x)^2}{3x}\) Solve this equation for x: \(x = 6.397 \times 10^{-5}\; \text{mol}\)

Convert the moles of I2 to mass

We found that at equilibrium, there are 6.397 x 10^{-5} mol of I2. To convert this to grams, use the molecular weight of I2 (which is 253.8 g/mol): Mass of I2 = \(6.397 \times 10^{-5}\; \text{mol} \times 253.8\; \frac{\text{g}}{\text{mol}} = 0.01624\; \text{g}\) Therefore, there are 0.01624 grams of I2 in the equilibrium mixture.

Key concepts

Equilibrium Constant

In the study of chemical equilibrium, the equilibrium constant (K_c) plays a vital role. It provides us with a quantitative measure of the concentrations of reactants and products at equilibrium. For gaseous reactions, we often represent the equilibrium constant using concentrations in mol/L. The formula to calculate K_c is derived from the balanced chemical equation. Here, for the dissociation of I_2 into 2 Ig, the expression for K_c is given by:

• \(\ K_c = \frac{[\text{I}]^2}{[\text{I}_2]} \)

If the equilibrium constant is small, like in our case, it suggests the reaction exothermic and favors the reactants.
We use this constant to predict the concentration of a substance at equilibrium once we know the other concentrations in the mixture.

ICE Table

An ICE Table is a useful tool in chemistry to track the Initial concentrations, Changes, and Equilibrium concentrations of reactants and products. When using an ICE table:

• Initial: Start by writing the initial amounts of reactants and products. If a product is not initially present, its initial value will be zero.
• Change: Determine the change needed to reach equilibrium. It is expressed as a variable (often 'x') that facilitates the calculation of concentrations throughout the reaction.
• Equilibrium: Add or subtract the change from the initial concentration to find the equilibrium state.

For example, in the reaction I_2(g) \(\rightleftharpoons\) 2 I(g), starting with 'y' moles of I_2 and none of I, the changes can be determined based on stoichiometry.
The ICE Table helps us to visualize these changes clearly, making the determination of unknown concentrations easier.

Mole Calculations

Calculating moles is a fundamental concept in chemistry, necessary for converting between mass and amount of substance. Using the molar mass of an element or compound allows us to perform this conversion with ease:

• Convert the mass of a substance to moles by dividing by its molar mass.
• Convert moles back to mass by multiplying by molar mass.

In our given problem, we first converted 30.5 mg of I(g) to moles, using the atomic weight of iodine (126.90 g/mol). Similarly, to find the mass of I_2 in the equilibrium mixture, we calculated the moles from the stoichiometry of the reaction and then converted it back to grams using the molar mass of I_2 (253.8 g/mol).
This process ensures accurate measurement of the substances involved in the reaction.