Problem 96 - Chemistry and Chemical Reactivity

Question

Insoluble \(\mathrm{PbBr}_{2}(\mathrm{s})\) precipitates when solutions of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(\mathrm{NaBr}(\mathrm{aq})\) are mixed. $$\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaBr}(\mathrm{aq}) \rightarrow \mathrm{PbBr}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})$$ To measure the enthalpy change, \(200 .\) mL of \(0.75 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})\) and \(200 . \mathrm{mL}\) of \(1.5 \mathrm{M}\) NaBr(aq) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by \(2.44^{\circ} \mathrm{C}\) Calculate the enthalpy change for the precipitation of \(\mathrm{PbBr}_{2}(\mathrm{s}),\) in \(\mathrm{kJ} / \mathrm{mol}\). (Assume the density of the solution is \(1.0 \mathrm{g} / \mathrm{mI}_{7}\) and its specific heat capacity is \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .)\)

Step-by-Step Solution

Answer

The enthalpy change is 27.328 kJ/mol of PbBr2.

Step 1: Calculate the Mass of the Solution

The total volume of the mixed solutions is 200 mL + 200 mL = 400 mL. Assuming the density of the solution is 1.0 g/mL, the mass of the solution is 400 g. This is because mass = volume × density = 400 mL × 1.0 g/mL.

Step 2: Calculate the Heat Absorbed by the Solution

The heat absorbed by the solution can be calculated using the formula \( q = m \cdot c \cdot \Delta T \), where \( m \) is the mass of the solution, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Thus, \( q = 400 \, \text{g} \times 4.2 \, \text{J/g}^\circ\text{C} \times 2.44^\circ\text{C} = 4099.2 \, \text{J} \).

Step 3: Convert Heat to Kilojoules

Since 1 kJ = 1000 J, convert the heat absorbed into kilojoules: \( 4099.2 \, \text{J} = 4.0992 \, \text{kJ} \).

Step 4: Determine the Limiting Reactant

Calculate the moles of each reactant: For \( \text{Pb(NO}_3\text{)}_2 \), \( 0.200 \text{ L} \times 0.75 \text{ M} = 0.15 \text{ moles} \).For \( \text{NaBr} \), \( 0.200 \text{ L} \times 1.5 \text{ M} = 0.30 \text{ moles} \).The reaction requires 1 mole of \( \text{Pb(NO}_3\text{)}_2 \) for every 2 moles of \( \text{NaBr} \). The limiting reactant is \( \text{Pb(NO}_3\text{)}_2 \) because 0.30 moles of \( \text{NaBr} \) would require 0.15 moles of \( \text{Pb(NO}_3\text{)}_2 \), which are available.

Step 5: Calculate Enthalpy Change per Mole of PbBr2

The moles of \( \text{PbBr}_2 \) formed are equal to the moles of limiting reactant \( \text{Pb(NO}_3\text{)}_2 \), which is 0.15 moles. To find the enthalpy change per mole, use the formula: \( \Delta H = \frac{q}{\text{moles of } \text{PbBr}_2} = \frac{4.0992 \, \text{kJ}}{0.15 \, \text{moles}} = 27.328 \, \text{kJ/mol} \)Thus, the enthalpy change for the precipitation is 27.328 kJ/mol of \( \text{PbBr}_2 \).

Key Concepts

Precipitation Reaction

When two aqueous solutions are mixed and an insoluble solid—known as a precipitate—forms, this is called a precipitation reaction. In this scenario, the reaction between lead(II) nitrate \( \text{Pb(NO}_3\text{)}_2 \) and sodium bromide (NaBr) creates an insoluble compound, lead(II) bromide \( \text{PbBr}_2 \), which settles out of the solution. Such reactions are a subset of double displacement reactions, where the ions in the reactants switch partners to form new compounds.

Precipitation reactions are significant in various fields, including chemistry and environmental science, because they are often used to remove unwanted ions from a solution or to collect useful substances.

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first, limiting the amount of products formed. In our example, we mix solutions of \( \text{Pb(NO}_3\text{)}_2 \) and NaBr. To find the limiting reactant, we calculate the number of moles from each solution.

For \( \text{Pb(NO}_3\text{)}_2 \), the concentration is 0.75 M and the volume is 0.200 L, resulting in 0.15 moles. Similarly, NaBr has a concentration of 1.5 M and a volume of 0.200 L, giving 0.30 moles. According to the reaction equation, 1 mole of \( \text{Pb(NO}_3\text{)}_2 \) reacts with 2 moles of NaBr, meaning the sodium bromide is in excess and \( \text{Pb(NO}_3\text{)}_2 \) is the limiting reactant.

Identifying the limiting reactant is crucial because it dictates the maximum yield of the reaction.

Calorimetry

Calorimetry is a method used to measure the heat change in a chemical reaction. In this exercise, we use a coffee-cup calorimeter to determine the enthalpy change during the precipitation of \( \text{PbBr}_2 \). When the two solutions are mixed, energy is released in the form of heat, causing a measurable temperature increase.

The amount of heat absorbed or released by the solution (q) is calculated using the formula: \( q = m \cdot c \cdot \Delta T \), where \( m \) is the mass of the solution, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change. This experiment is a simple yet effective way to study energetics in chemical reactions.

Mole Calculation

Mole calculation serves as the backbone for stoichiometric assessments in chemistry, helping us quantify reactants and products. In the exercise, it is pivotal to find the moles of each reactant to determine which is the limiting reactant. The number of moles is derived using the formula: \( \text{moles} = \text{molarity} \times \text{volume} \).

This step ensures you know precisely how much of each substance is available, which is essential for calculating quantities under the reaction's stoichiometry. Already calculated, \( \text{Pb(NO}_3\text{)}_2 \) has 0.15 moles, and NaBr has 0.30 moles, clarifying that there is an excess of NaBr.

Accurately determining moles is essential not only in academic settings but also in real-world applications, such as calculating reagent needs for large-scale production in industries.