Problem 86 - 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. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaBr}(\mathrm{aq}) \rightarrow \mathrm{PbBr}_{2}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})\) $$ \Delta_{\mathrm{r}} H^{\circ}=? $$ 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 \(1.5 \mathrm{M} \mathrm{NaBr}(\mathrm{aq})\) 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 \(\mathrm{k} \mathrm{J} / \mathrm{mol}\). (Assume the density of the solution is \(1.0 \mathrm{g} / \mathrm{mL},\) and its specific heat capacity is \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\).

Step-by-Step Solution

Answer

The molar enthalpy change is \(-27.39\, \text{kJ/mol}\).

Step 1: Determine Total Heat Absorbed by Solution

First, we calculate the total heat, \(q\), absorbed by the solution. 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.- Mass \(m = 200\, \text{mL} + 200\, \text{mL} = 400\, \text{g}\) (since density is \(1.0 \text{g/mL}\)).- Specific heat capacity \(c = 4.2 \text{ J/g} \cdot \text{K}\).- Temperature change \(\Delta T = 2.44\, ^{\circ}\text{C}\).Thus, \(q = 400\, \text{g} \times 4.2\, \text{J/g} \cdot \text{K} \times 2.44\, \text{K} = 4108.8\, \text{J}\).

Step 2: Calculate Moles of PbBr2 Formed

The reaction is a 1:1 mole ratio for \(\text{Pb}^2+\) ions with \(\text{PbBr}_2\), based on the balanced chemical equation. Find the limiting reagent and calculate the moles of product. - Moles of \(\text{Pb(NO}_3\text{)}_2 = 0.75\, \text{mol/L} \times 0.200\, \text{L} = 0.150\, \text{mol}\).- Moles of \(\text{NaBr} = 1.5\, \text{mol/L} \times 0.200\, \text{L} = 0.300\, \text{mol}\).The limiting reagent is \(\text{Pb(NO}_3\text{)}_2\), so moles of \(\text{PbBr}_2\) formed = 0.150 mol.

Step 3: Calculate Molar Enthalpy Change

Convert the heat absorbed to the enthalpy change per mole of precipitate formed. - The heat absorbed \(q = 4108.8\, \text{J} = 4.1088\, \text{kJ}\).- Molar enthalpy change \( \Delta_rH^\circ = \frac{q}{\text{moles of } \text{PbBr}_2}=\frac{4.1088\, \text{kJ}}{0.150\, \text{mol}} = 27.39\, \text{kJ/mol}\).

Step 4: Report Result Appropriately

Based on the molar enthalpy change calculated, you state the reaction enthalpy as \( \Delta_rH^\circ = 27.39\, \text{kJ/mol}\) for the precipitation of \(\text{PbBr}_2\). Always consider the sign of \( \Delta_rH^\circ \). Since the temperature increased, the reaction is exothermic, so report \( \Delta_rH^\circ = -27.39\, \text{kJ/mol}\).

Key Concepts

Precipitation Reaction

Precipitation reactions occur when two aqueous solutions combine to form an insoluble solid, known as the precipitate. In the given example, when solutions of lead(II) nitrate, \( \text{Pb(NO}_3\text{)}_2 \), and sodium bromide, \( \text{NaBr} \), are mixed, the ions rearrange to produce lead(II) bromide, \( \text{PbBr}_2 \), as a solid precipitate. The equation for this reaction is: \[ \text{Pb(NO}_3\text{)}_2(\text{aq}) + 2\, \text{NaBr}(\text{aq}) \rightarrow \text{PbBr}_2(\text{s}) + 2\, \text{NaNO}_3(\text{aq}) \]. This reaction illustrates the simple exchange of ions between two compounds, forming new products. Lead(II) bromide is not soluble in water, which is why it precipitates out of the solution as a solid.

Coffee-Cup Calorimeter

A coffee-cup calorimeter is a simple device used to measure heat changes in chemical reactions occurring at constant pressure, such as the enthalpy changes. It is a practical setup where the reaction takes place in a cup that serves as the calorimeter. The calorimeter captures the heat exchanged in the reaction that causes temperature changes in the surrounding solution. It assumes all heat exchange happens within the solution and any heat loss to its surroundings is negligible. This type of calorimeter is suitable for reactions like precipitations, where the heat released or absorbed results in sensible temperature changes.

Specific Heat Capacity

Specific heat capacity is a property of a substance that indicates the amount of heat required to change the temperature of one gram of the substance by one degree Celsius. For water and dilute aqueous solutions, this value is generally considered to be approximately \(4.18\, \text{J/g}\, \cdot\, \text{K}\), but in this particular problem, it is given as \(4.2\, \text{J/g}\, \cdot\, \text{K}\). Using this value allows the calculation of the total heat absorbed by a solution during a reaction based on the formula: \[ q = m \cdot c \cdot \Delta T \] where \( q \) is the heat absorbed, \( m \) is the mass of the solution, \( c \) is its specific heat capacity, and \( \Delta T \) is the change in temperature. This relationship helps in understanding how energy transfers during chemical processes within the solution.

Limiting Reagent

A limiting reagent is a reactant that is completely consumed in a chemical reaction, limiting the amount of product that can be formed. In a given reaction, identifying the limiting reagent is crucial for determining the theoretical yield of the products. In the precipitation of \( \text{PbBr}_2 \), we calculate each reactant's moles:

• \( \text{Pb(NO}_3\text{)}_2: 0.75\, \text{mol/L} \times 0.200\, \text{L} = 0.150\, \text{mol} \)
• \( \text{NaBr}: 1.5\, \text{mol/L} \times 0.200\, \text{L} = 0.300\, \text{mol} \)

With these calculations, it is evident that \( \text{Pb(NO}_3\text{)}_2 \) is used up first, making it the limiting reagent. Since the product \( \text{PbBr}_2 \) forms in a 1:1 stoichiometry with \( \text{Pb(NO}_3\text{)}_2 \), the reaction constraints the amount of \( \text{PbBr}_2 \) formed to 0.150 moles.