Problem 79 - Chemistry and Chemical Reactivity

Question

Calculate the enthalpies of solution for \(\mathrm{Li}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) Are the solution processes exothermic or endothermic? Compare them with LiCl and KCI. What similarities or differences do you find? $$\begin{array}{lll} \hline & \Delta H_{f}^{\circ}(\mathrm{s}) & \Delta H_{f}^{\circ}(\mathrm{aq}, 1 \mathrm{m}) \\ \text { Compound } & (\mathrm{kJ} / \mathrm{mol}) & (\mathrm{kJ} / \mathrm{mol}) \\ \hline \mathrm{Li}_{2} \mathrm{SO}_{4} & -1436.4 & -1464.4 \\\ \mathrm{K}_{2} \mathrm{SO}_{4} & -1437.7 & -1414.0 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

\( \text{Li}_2 \text{SO}_4 \) is exothermic, \( \text{K}_2 \text{SO}_4 \) is endothermic, mirroring LiCl and KCl trends.

Step 1: Understand the Concept

The enthalpy of solution is the heat change when a solute dissolves in a solvent. It can be determined using the formula: \( \Delta H_{solution} = \Delta H_{f}^{\circ}(\text{aq}) - \Delta H_{f}^{\circ}(\text{s}) \), where \( \Delta H_{f}^{\circ}(\text{aq}) \) is the enthalpy of formation of the compound in aqueous form and \( \Delta H_{f}^{\circ}(\text{s}) \) is the enthalpy of formation of the compound in solid form.

Step 2: Calculate for \( \text{Li}_2 \text{SO}_4 \)

Substitute the values given for \( \text{Li}_2 \text{SO}_4 \) into the formula: \( \Delta H_{solution} = -1464.4 \, \text{kJ/mol} - (-1436.4 \, \text{kJ/mol}) = -28.0 \, \text{kJ/mol} \). This shows the dissolution of \( \text{Li}_2 \text{SO}_4 \) is an exothermic process.

Step 3: Calculate for \( \text{K}_2 \text{SO}_4 \)

Similarly, substitute the values given for \( \text{K}_2 \text{SO}_4 \) into the formula: \( \Delta H_{solution} = -1414.0 \, \text{kJ/mol} - (-1437.7 \, \text{kJ/mol}) = 23.7 \, \text{kJ/mol} \). This indicates the dissolution of \( \text{K}_2 \text{SO}_4 \) is an endothermic process.

Step 4: Compare with LiCl and KCl

Typically, LiCl releases heat in water (exothermic) while KCl absorbs heat (endothermic). Similar patterns are observed in \( \text{Li}_2 \text{SO}_4 \) and \( \text{K}_2 \text{SO}_4 \) respectively. Both lithium compounds are exothermic and both potassium compounds are endothermic, showing a group trend across both chloride and sulfate compounds.

Key Concepts

Exothermic Process

An exothermic process is a chemical reaction or physical change that releases heat energy to its surroundings. In simpler terms, during an exothermic process, the temperature of the surrounding environment increases. This is because the energy required to break the bonds in the reactants is less than the energy released when new bonds are formed in the products.
This surplus energy is released as heat. When you dissolve substances like \( \text{Li}_2 \text{SO}_4 \) in water and observe heat being emitted, it means the process is exothermic, and the enthalpy of solution is negative.

• Examples include burning a candle or mixing acids and bases.
• In the context of enthalpy of solution, \( \text{Li}_2 \text{SO}_4 \) dissolving in water is exothermic with an enthalpy of solution of \(-28.0 \, \text{kJ/mol}\).

Endothermic Process

An endothermic process absorbs heat energy from its surroundings, causing the temperature of the surrounding environment to decrease. This occurs when the energy required to break the bonds in the reactants is greater than the energy released from forming new products.
If adding a substance to water makes the solution colder, it's undergoing an endothermic process. The enthalpy change is positive, reflecting the absorption of heat.

• Examples include melting ice or photosynthesis.
• In the case of \( \text{K}_2 \text{SO}_4 \) dissolving in water, the process is endothermic with an enthalpy of solution of \(23.7 \, \text{kJ/mol}\).

Enthalpy of Formation

The enthalpy of formation \( \Delta H_f^{\circ} \) is the heat change associated with forming a compound from its elements in their standard state. It's a fundamental concept in thermochemistry, helping predict the energy changes in chemical reactions and processes.
The enthalpy of formation can be presented in two forms: for a compound in solid form, \( \Delta H_f^{\circ} (\text{s}) \), and in aqueous form, \( \Delta H_f^{\circ} (\text{aq}) \).
To calculate the enthalpy of solution, you take the difference between the aqueous and solid enthalpies of formation:

• For \( \text{Li}_2 \text{SO}_4 \), \( \Delta H_{solution} = -1464.4 - (-1436.4) = -28.0 \, \text{kJ/mol} \).
• For \( \text{K}_2 \text{SO}_4 \), \( \Delta H_{solution} = -1414.0 - (-1437.7) = 23.7 \, \text{kJ/mol} \).

This comparison helps determine if the dissolution process is either exothermic or endothermic for those specific compounds.