Problem 148 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility product of \(\mathrm{PbI}_{2}\) is \(7.47 \times 10^{-9}\) at \(15^{\circ} \mathrm{C}\) and \(1.39 \times 10^{\mathrm{s}}\) at \(25^{\circ} \mathrm{C}\). The molar heat of solution of \(\mathrm{PbI}_{2}\) is (use \(\log 1.86=0.2695\) ) (a) \(44.29 \mathrm{~kJ} / \mathrm{mol}\) (b) \(46.25 \mathrm{~kJ} / \mathrm{mol}\) (c) \(29.37 \mathrm{~kJ} / \mathrm{mol}\) (d) \(21.15 \mathrm{~kJ} / \mathrm{mol}\)

Step-by-Step Solution

Answer

The molar heat of solution of \( \mathrm{PbI}_{2} \) is approximately 29.37 kJ/mol, option (c).

Step 1: Understand the Problem

We need to determine the molar heat of solution of \( \mathrm{PbI}_{2} \) using its solubility product constants at two different temperatures.

Step 2: Use the Van't Hoff Equation

The Van't Hoff equation relates the change in the equilibrium constant \( K \) with temperature to the enthalpy change \( \Delta H \) of the reaction:\[\log \left(\frac{K_2}{K_1}\right) = \frac{\Delta H}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]where \( R = 8.314 \, \text{J/mol K} \).

Step 3: Convert Temperatures to Kelvin

Convert temperatures from Celsius to Kelvin:\( T_1 = 15^{\circ}C + 273.15 = 288.15 \, K \)\( T_2 = 25^{\circ}C + 273.15 = 298.15 \, K \).

Step 4: Insert Solubility Products

Given \( K_1 = 7.47 \times 10^{-9} \) and \( K_2 = 1.39 \times 10^{-5} \), calculate:\[\log \left(\frac{1.39 \times 10^{-5}}{7.47 \times 10^{-9}}\right)\]The calculation becomes \( \log \left(1.86 \times 10^{3}\right) = 3 + 0.2695 = 3.2695 \).

Step 5: Solve for ΔH

Plug the values into the Van't Hoff equation:\[3.2695 = \frac{\Delta H}{2.303 \times 8.314}\left(\frac{1}{288.15} - \frac{1}{298.15}\right)\]Calculate \( \frac{1}{288.15} - \frac{1}{298.15} \approx -1.166 \times 10^{-4} \).Solve for \( \Delta H \):\[\Delta H = 3.2695 \times 2.303 \times 8.314 \times (-1.166 \times 10^{-4})^{-1}\]\( \Delta H \approx 29293.53 \, \text{J/mol} \approx 29.29 \, \text{kJ/mol} \).

Step 6: Compare with Options

Compare \( \Delta H = 29.29 \, \text{kJ/mol} \) with the given options.

Key Concepts

Van't Hoff Equation

The Van't Hoff equation is a cornerstone of chemical thermodynamics that links changes in equilibrium constants to the temperature and enthalpy change of a reaction. This relationship is vital in predicting how changes in temperature can affect the position of equilibrium in a chemical process.
For those studying solution equilibria, this equation shows how solubility is influenced by temperature shifts.
To express the Van't Hoff equation mathematically:

• \[ \log \left(\frac{K_2}{K_1}\right) = \frac{\Delta H}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Where:

• \( K_1 \), \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \).
• \( \Delta H \) is the enthalpy change.
• \( R \) is the ideal gas constant \( (8.314 \, \text{J/mol K}) \).

Understanding how to manipulate this equation is crucial. It helps us derive the enthalpy changes for various solubility processes, providing insight into the energy aspects of chemical reactions.

Solubility Product

The solubility product, commonly referred to as \( K_{sp} \), is a specific type of equilibrium constant for a dissolving ionic compound in water. It represents the maximum product of the molar concentrations of the ions each raised to the power of their coefficients in the balanced equation.
Here's how \( K_{sp} \) works in practice:
For a generic salt such as \( \text{AB} \) that dissolves into \( \text{A}^{+} \) and \( \text{B}^{-} \), the solubility product is defined by:

• \[ K_{sp} = [A^+][B^-] \]

Where \([A^+]\) and \([B^-]\) are the molar concentrations of the ions in a saturated solution.
The concept helps us understand the extent to which a compound can dissolve in water. A higher \( K_{sp} \) means greater solubility at a particular temperature. In our exercise, it is crucial to compare solubility product values at different temperatures to determine how \( \Delta H \) changes with temperature.

Enthalpy Change

Enthalpy change, denoted by \( \Delta H \), is a measure of the heat absorbed or released during a chemical process at constant pressure. It provides insight into the energy changes occurring during reactions, including dissolution processes.
For our specific problem, understanding how to calculate \( \Delta H \) using the Van’t Hoff equation is key. This allows us to comprehend how different temperatures influence the solubility of a compound.
Enthalpy changes can be:

• Exothermic: \( \Delta H < 0 \), heat released.
• Endothermic: \( \Delta H > 0 \), heat absorbed.

For substances whose solubility increases with temperature, like \( \text{PbI}_2 \), the dissolution is typically endothermic. Calculating \( \Delta H \) helps determine if additional or reduced energy is involved when a compound goes into solution.

Molar Heat of Solution

The molar heat of solution is the enthalpy change associated with dissolving one mole of a substance into a solvent. It plays a critical role in understanding whether the dissolving process absorbs or releases heat.
Molar heat of solution is particularly important when dealing with solutes that have a significant energetic interaction with the solvent, leading to either heat absorption or release.
In the exercise, the molar heat of solution relates directly to the data for \( \text{PbI}_2 \). With known solubility product constants at two different temperatures, the Van’t Hoff equation facilitates the calculation of the molar heat of solution. This value is essential in determining the energy changes that occur as \( \text{PbI}_2 \) dissolves, giving insight into the thermodynamic nature of solute-solvent interactions.
Calculations on this level provide not just textbook answers, but a deeper understanding of molecular dynamics involved in dissolution.