Problem 24 - General Chemistry: Principles and Modern Applications

Question

For the dissociation of \(\mathrm{I}_{2}(\mathrm{g})\) at about \(1200^{\circ} \mathrm{C}\) \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}), K_{\mathrm{c}}=1.1 \times 10^{-2} .\) What volume flask should we use if we want 0.37 mol I to be present for every \(1.00 \mathrm{mol} \mathrm{I}_{2}\) at equilibrium?

Step-by-Step Solution

Answer

A volume of 80.4 Liters is needed to reach equilibrium under the given conditions.

Step 1: Understanding the Chemical Equilibrium

Write the balanced chemical equation. \[\mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g)\]This represents the equilibrium of the dissociation of Iodine gas.

Step2: Equilibrium Expression and Substituting given Values

For this reaction, \(Kc\) is given by \[Kc = {[I]^2 \over [I_2]} \] Substitute the given values into the equilibrium expression, we have: \[1.1 * 10^{-2} = {([0.37]^2) \over [I_2]} \] Solve for \([I_2]\), we find \([I_2] = 12.43 M \]

Step 3: Recipe Interpretation

The given recipe states that at equilibrium, for every 1.00 mol \(\mathrm{I}_2\), there are 0.37 mol I, therefore the concentration of \(\mathrm{I}_2\) is \([I_2] = \frac{1.00 mol}{Volume}\). Therefore we can set equal the two equations for \([I_2]\) and solve for the Volume : \[Volume = \frac{1.00 mol}{12.43 M} = 0.0804 m^3 = 80.4 L \]

Key Concepts

Dissociation Reactions

In chemistry, dissociation reactions refer to the process in which a compound breaks down into its individual components. These reactions are often reversible, meaning the products can recombine to form the original compound. The dissociation of iodine gas, as given by the equation \(\mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g)\), is a classic example of such a reaction. At high temperatures, iodine molecules split into iodine atoms. This dynamic nature of dissociation reactions is fundamental to understanding chemical equilibria, which is the point at which the forward and reverse reactions occur at the same rate.

• In the iodine dissociation reaction, \(\mathrm{I}_2\) is the "parent" molecule and \(2\mathrm{I}\) are the resulting products.
• The reactions are reversible, so at equilibrium, there is a constant maximum concentration of both molecules and atoms present.
• Studying these reactions helps in determining the proportions of reactants and products in a chemical equilibrium state.

Understanding these basics will help you grasp how such reactions reach equilibrium and thus influence the calculation of equilibrium constants.

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), plays a major role in describing the balance between products and reactants in a chemical reaction at equilibrium. For a given reaction, the expression for \(K_c\) is determined based on the balanced chemical equation. It relates the concentrations of the products to the reactants at equilibrium. In the iodine dissociation reaction \(\mathrm{I}_2(g) \rightleftharpoons 2\mathrm{I}(g)\), the expression is given as \(K_c = \frac{[I]^2}{[I_2]}\).

• \(K_c\) values indicate the extent of a reaction: a large \(K_c\) suggests that the products are favored, while a small \(K_c\) indicates the reactants are mostly present.
• For the reaction provided, \(K_c = 1.1 \times 10^{-2}\), which is relatively small, implying that the concentration of iodine molecules \(\mathrm{I}_2\) will be higher compared to iodine atoms \(2\mathrm{I}\) at equilibrium.
• The value of \(K_c\) is constant at a given temperature, so any change in temperature will alter the position of equilibrium, thereby affecting \(K_c\).

Whether dealing with forward or reverse reactions, \(K_c\) is crucial for understanding the concentrations of different species at equilibrium.

Molar Concentration Calculation

Calculating molar concentrations is key to predicting the state of equilibrium in chemical reactions. It involves determining the moles of solute per liter of solution. In the context of our iodine dissociation reaction, knowing the concentration of \(\mathrm{I}_2\) and \(\mathrm{I}\) is necessary to solve for the volume of the flask.

• The formula used for the molar concentration of a substance \([\text{Substance}] = \frac{\text{moles}}{\text{Volume}}\).
• Given in the exercise, there is 0.37 moles of \(\mathrm{I}\) for every 1.00 mole of \(\mathrm{I}_2\) at equilibrium, guiding us to plug these values into the equilibrium expression to find \([I_2]\).
• By substituting \(\frac{1.00 \text{mol}}{\text{Volume}}\) for \([I_2]\), and solving for Volume, the exercise leads to the calculation of 80.4 L for the flask needed to maintain the equilibrium concentrations.

Understanding how to perform these calculations is vital for predicting how much of each substance will be present under set conditions.