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Chapter 6: Problem 105

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 \(547 \mathrm{~J} /{ }^{\circ} \mathrm{C}\).

Short Answer

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

Step by step solution

01

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: \[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 36.66^{\circ}\text{C} - 25.00^{\circ}\text{C} = 11.66^{\circ}\text{C} \]
02

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: \[ q = 547 \frac{\text{J}}{^{\circ}\text{C}} \times 11.66^{\circ}\text{C} = 6379.82 \text{ J} \]
03

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}). \[ \text{Moles of LiOH} = \frac{6.48 \text{ g}}{23.95 \text{ g/mol}} = 0.2707 \text{ moles} \]
04

Calculate 螖H 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:\[ \Delta H = 23.57 \text{ kJ/mol}\]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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/掳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掳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掳C, and the final temperature was 36.66掳C. Thus, the calculated temperature change was:\[ \Delta T = 36.66掳C - 25.00掳C = 11.66掳C \]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.

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Most popular questions from this chapter

Hydrogen, \(\mathrm{H}_{2}\), is used as a rocket fuel. The hydrogen is burned in oxygen to produce water vapor. $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-484 \mathrm{~kJ} $$ What is the enthalpy change per gram of hydrogen?

You have two samples of different metals, metal \(\mathrm{A}\) and metal \(\mathrm{B}\), each having the same mass. You heat both metals to \(95^{\circ} \mathrm{C}\) and then place each one into separate beakers containing the same quantity of water at \(25^{\circ} \mathrm{C}\). a. You measure the temperatures of the water in the two beakers when each metal has cooled by \(10^{\circ} \mathrm{C}\) and find that the temperature of the water with metal \(\mathrm{A}\) is higher than the temperature of the water with metal \(\mathrm{B}\). Which metal has the greater specific heat? Explain. b. After waiting a period of time, the temperature of the water in each beaker rises to a maximum value. In which beaker does the water rise to the higher value, the one with metal \(\mathrm{A}\) or the one with metal \(\mathrm{B} ?\) Explain.

The specific heat of copper metal was determined by putting a piece of the metal weighing \(35.4 \mathrm{~g}\) in hot water. The quantity of heat absorbed by the metal was calculated to be \(47.0 \mathrm{~J}\) from the temperature drop of the water. What was the specific heat of the metal if the temperature of the metal rose \(3.45^{\circ} \mathrm{C}\) ?

A 29.1-mL sample of \(1.05 \mathrm{M}\) KOH is mixed with \(20.9 \mathrm{~mL}\) of \(1.07 M \mathrm{HBr}\) in a coffee-cup calorimeter (see Section \(6.6\) of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest wholenumber coefficients, is \(-55.8 \mathrm{~kJ} .\) Both solutions are at \(21.8^{\circ} \mathrm{C}\) prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and volumes are additive.

Ammonia will burn in the presence of a platinum catalyst to produce nitric oxide, \(\mathrm{NO}\). $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ What is the heat of reaction at constant pressure? Use the following thermochemical equations: $$ \begin{gathered} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) ; \Delta H=180.6 \mathrm{~kJ} \\ \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) ; \Delta H=-91.8 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-483.7 \mathrm{~kJ} \end{gathered} $$
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