When \(1.00 \mathrm{~g}\) of gaseous \(\mathrm{I}_{2}\) is heated to \(1000 . \mathrm{K}\) in a \(1.00-\mathrm{L}\) sealed container, the resulting equilibrium mixture contains \(0.830 \mathrm{~g}\) of \(\mathrm{I}_{2}\). Calculate \(K_{c}\) for the dissociation equilibrium \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\).

The equilibrium constant expression is a vital concept in chemical equilibrium, reflecting the ratio of product and reactant concentrations at equilibrium for a reversible reaction. For the dissociation of iodine, \text{I}_2(g) \rightleftharpoons 2 \text{I}(g)\, the equilibrium constant (\(K_c\)) is calculated using the formula:

\[K_c = \frac{[\text{I}]^2}{[\text{I}_2]}\]
This expression denotes that the equilibrium constant is the square of the molar concentration of iodine atoms divided by the molar concentration of iodine molecules at equilibrium. The square arises due to the stoichiometry of the dissociation, as one molecule of \(\text{I}_2\) yields two atoms of I. It's crucial for students to grasp that only substances in the gaseous or aqueous phases are included in the expression and that pure solids or liquids are omitted because their concentrations do not change.

Molar concentration, also known as molarity, measures the number of moles of a solute present in one liter of solution. The steps to determine the molarity of iodine atoms and molecules at equilibrium involve a straightforward calculation:

\[\text{Molar concentration} = \frac{\text{number of moles}}{\text{Volume of solution in liters}}\]
In the context of the given exercise, molar concentrations of both \(\text{I}_2(g)\) and \(\text{I}(g)\) in a 1.00 L container are crucial to calculate the equilibrium constant. It's essential to remind students to ensure that the volume is in liters for the molarity calculation to be correct.

Dissociation equilibrium pertains to a specific type of reversible reaction where a compound breaks apart into two or more components. In our example, solid iodine (\text{I}_2) dissociates to form iodine atoms (\text{I}). Equilibrium is achieved when the rate of the forward dissociation reaction equals the rate of the reverse association reaction, leading to a constant ratio of products to reactants, signified by the equilibrium constant, \(K_c\). Reflecting on our dissocation equilibrium:

\text{I}_2(\text{g}) \rightleftharpoons 2 \text{I}(\text{g})
Students should note that only changes in temperature can affect the value of \(K_c\), whereas changes in concentration, pressure, or volume will only shift the position of equilibrium without altering \(K_c\)'s value.

This concept is crucial in the preparation to calculate equilibrium constants, as chemical reactions are often described in terms of mass. The conversion between moles and mass involves the molar mass of the substance, which corresponds to the mass per mole of particles. The formula used is:

\[\text{Mass} = \text{Moles} \times \text{Molar Mass}\]
In the given exercise, converting the mass of iodine (\(\text{I}_2\)) into moles is crucial for finding the molar concentration at equilibrium. It's important to accentuate to the students the necessity of using the correct molar mass and ensure it's in grams per mole (\text{g/mol}). Furthermore, it's beneficial to verify that they are using the mass in grams to avoid any errors in calculations.