Problem 105 - General Chemistry

Question

In a calorimetric experiment, \(6.48 \mathrm{~g}\) of lithium hydroxide, LiOH, was dissolved in water. The temperature of the calorimeter rose from \(25.00^{\circ} \mathrm{C}\) to \(36.66^{\circ} \mathrm{C}\). What is \(\Delta H\) for the solution process? $$ \mathrm{LiOH}(s) \longrightarrow \mathrm{Li}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ The heat capacity of the calorimeter and its contents is \(547 \mathrm{~J} /{ }^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

\(\Delta H = 23.57 \text{ kJ/mol}\) for the LiOH solution process.

Step 1: Calculate the Temperature Change

Determine the change in temperature of the calorimeter during the experiment. To find this change, subtract the initial temperature from the final temperature: \[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 36.66^{\circ}\text{C} - 25.00^{\circ}\text{C} = 11.66^{\circ}\text{C} \]

Step 2: Calculate the Heat Absorbed by the Calorimeter

Use the formula \( q = C \cdot \Delta T \) to calculate the heat absorbed by the calorimeter, where \( C \) is the heat capacity. Substituting the known values gives: \[ q = 547 \frac{\text{J}}{^{\circ}\text{C}} \times 11.66^{\circ}\text{C} = 6379.82 \text{ J} \]

Step 3: Calculate Moles of LiOH Dissolved

Calculate the number of moles of \(\text{LiOH}\) using its mass and molar mass. The molar mass of \(\text{LiOH}\) is approximately 23.95 \text{g/mol} (6.94 \text{g/mol for Li, 16.00 \text{g/mol for O, and 1.01 \text{g/mol for H}). \[ \text{Moles of LiOH} = \frac{6.48 \text{ g}}{23.95 \text{ g/mol}} = 0.2707 \text{ moles} \]

Step 4: Calculate ΔH for the Solution Process

Calculate \( \Delta H \) for the dissolution process. Since \( q = \Delta H_{\text{solution}} \) and \( q \) was found to be 6379.82 J,\[ \Delta H = \frac{q}{\text{moles of LiOH}} = \frac{6379.82 \text{ J}}{0.2707 \text{ moles}} = 23571.55 \frac{\text{J}}{\text{mol}} \]Convert to kJ/mol by dividing by 1000:\[ \Delta H = 23.57 \text{ kJ/mol}\]

Step 5: Conclusion

The enthalpy change (\(\Delta H\)) for the dissolution of lithium hydroxide in water is \(23.57\text{ kJ/mol}\), indicating the process is endothermic as the temperature of the surroundings increased.

Key Concepts

Enthalpy Change

Enthalpy change (\( \Delta H \)) is a key concept in chemistry that measures the heat exchange at constant pressure during a chemical process. When lithium hydroxide (LiOH) dissolves in water, it either absorbs or releases heat. \( \Delta H \) tells us whether the process is endothermic (absorbing heat) or exothermic (releasing heat).In calorimetry, the enthalpy change is calculated by measuring the heat absorbed or released by the solution. We express this measurement in kilojoules per mole (kJ/mol). For instance, in the cited experiment, LiOH dissolution in water showed a\( \Delta H \) of 23.57 kJ/mol, indicating an endothermic process.Understanding\( \Delta H \) helps predict how substances react, being crucial for thermodynamics and many practical chemical applications.

Lithium Hydroxide

Lithium hydroxide, or LiOH, is a chemical compound used extensively for its reactivity and absorption properties. It appears as a white crystalline solid and is known for influencing the temperature of a solution when dissolved in water. In the discussed calorimetry experiment, LiOH was used to study its enthalpy change during dissolution.LiOH reacts with water and breaks down into lithium ions (Li\(^+\)) and hydroxide ions (OH\(^-\)), releasing or absorbing heat in the process.Important characteristics of Lithium Hydroxide:

• Formula: LiOH
• Molar Mass: 23.95 g/mol
• Highly soluble in water
• Used in air purification and chemical manufacturing

Recognizing the properties of LiOH is essential for understanding its role in calorimetric experiments and other chemical processes.

Heat Capacity

Heat capacity is a fundamental property that describes how much heat energy a substance can absorb before its temperature changes. In a calorimetry experiment, the heat capacity allows us to calculate the heat exchanged in the system.For our calorimeter, the heat capacity was given as 547 J/°C. This means that for each degree Celsius of temperature change, the calorimeter absorbs or loses 547 joules of energy.Using this heat capacity, we find the heat change:- Formula: \( q = C \times \Delta T \)
- Calculated heat change: 6379.82 J from a temperature change of 11.66°CHeat capacity enables precise calculations by accounting for the energy needed to alter the temperature of the calorimeter and its contents.

Temperature Change

Temperature change (\( \Delta T \)) is the difference between the final temperature and the initial temperature of a substance after a heat exchange has occurred. It's a crucial measurement in calorimetry that affects the calculation of the heat absorbed or released.In the experiment, the initial temperature was 25.00°C, and the final temperature was 36.66°C. Thus, the calculated temperature change was:\[ \Delta T = 36.66°C - 25.00°C = 11.66°C \]This temperature change is directly used alongside the system's heat capacity to determine the heat absorbed or released (\( q \)). The magnitude of \( \Delta T \) reveals the degree of thermal energy involved, providing insight into the process's energetics.