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Does a measurement carried out in a bomb calorimeter give a value for \(\Delta H\) or \(\Delta E\) ? Explain.

Short Answer

Expert verified
Bomb calorimeter measures \( \Delta E \), the change in internal energy.

Step by step solution

01

Understanding the Problem

To determine whether a bomb calorimeter measures \( \Delta H \) or \( \Delta E \), we need to understand what each symbol represents. \( \Delta H \) is the enthalpy change, whereas \( \Delta E \) is the change in internal energy. These are thermodynamic quantities measured in different conditions.
02

Identifying Conditions in a Bomb Calorimeter

A bomb calorimeter measures the heat released or absorbed during a chemical reaction at constant volume. It is a closed system where no work is performed because the volume does not change.
03

Thermodynamic Relationship

According to thermodynamics, the change in internal energy \( \Delta E \) at constant volume can be measured as the heat exchanged, \( q_v \). This relationship is expressed as: \( \Delta E = q_v \). In contrast, \( \Delta H = q_p \), where \( q_p \) is heat exchanged at constant pressure.
04

Conclusion

Since a bomb calorimeter operates at constant volume, it measures the change in internal energy, \( \Delta E \), not \( \Delta H \).

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), is a fundamental concept in thermodynamics that describes the heat content change of a system when a reaction occurs at constant pressure. This is particularly significant in chemical reactions occurring in open systems, such as beakers or flasks, where atmospheric pressure remains constant.

In a practical sense, enthalpy is often easy to measure because many reactions occur at constant atmospheric pressure. The equation used to express this thermodynamic quantity is:\[\Delta H = \Delta E + P\Delta V\]where \( \Delta E \) is the change in internal energy, \( P \) is the pressure, and \( \Delta V \) is the change in volume.

If no work is done by expanding or compressing gas, the volume change \( \Delta V \) is zero, meaning that enthalpy change is equal to the heat exchanged, \( q_p \) at constant pressure. Therefore, measuring \( \Delta H \) gives insights not only into heat transfers but also into work done by or on the system. This is why understanding enthalpy is crucial for processes involving gases and heat exchanges commonly seen in laboratories.
Internal Energy
Internal energy, noted as \( \Delta E \), is a core thermodynamic quantity representing the total energy contained within a system. This encompasses both the kinetic energy of movement of molecules and the potential energy associated with the forces between them.

In a bomb calorimeter, the measurement focuses solely on internal energy changes because the system operates at constant volume. This ensures no work is done via volume change, as the equation:\[\Delta E = q_v\]shows the relationship at constant volume, where \( q_v \) is the heat exchanged.

Measuring internal energy is particularly important when the system is isolated and cannot exchange matter or do pressure-volume work with its surroundings. The ability of a bomb calorimeter to measure \( \Delta E \) precisely makes it valuable in the fields of chemistry and biochemistry where accurate energy changes in reactions are crucial.
Thermodynamic Quantities
Thermodynamic quantities help us understand the exchange and transformation of energy within systems. The main quantities include enthalpy \( \Delta H \), internal energy \( \Delta E \), entropy, and free energy, each offering detailed insight into different aspects of heat and work.

These quantities are interrelated and often depend on the conditions of the process, such as constant volume or constant pressure. For example:
  • Enthalpy \( \Delta H \): Measured at constant pressure; useful for understanding heat transfers when gases are evolved or absorbed.
  • Internal Energy \( \Delta E \): Measured at constant volume, offering insights into total energy changes within a system.

Understanding these quantities is key in fields like thermodynamics and chemistry, providing necessary guidelines for predicting reactions and energy flow. By knowing when and how to measure each thermodynamic quantity, scientists can design better experiments and develop new technologies effectively.

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Most popular questions from this chapter

We said in Section \(9.1\) that the potential energy of water at the top of a dam or waterfall is converted into heat when the water dashes against rocks at the bottom. The potential energy of the water at the top is equal to \(E_{p}=m g h\), where \(m\) is the mass of the water, \(g\) is the acceleration of the falling water due to gravity \(\left(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\), and \(h\) is the height of the water. Assuming that all the energy is converted to heat, calculate the temperature rise of the water in degrees Celsius after falling over California's Yosemite Falls, a distance of \(739 \mathrm{~m}\). The specific heat of water is \(4.18 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K})\)

Assume that the kinetic energy of a \(1400-\mathrm{kg}\) car moving at \(115 \mathrm{~km} / \mathrm{h}\) (Problem \(9.42\) ) could be converted entirely into heat. What amount of water could be heated from \(20^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) by the car's energy? \(4.184 \mathrm{~J}\) are required to heat \(1 \mathrm{~g}\) of water by \(1{ }^{\circ} \mathrm{C}\).

Instant hot packs contain a solid and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, increasing the temperature because of the exothermic reaction. The following reaction is used to make a hot pack: $$\mathrm{LiCl}(s) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{Li}^{+}(a q)+\mathrm{Cl}^{-}(a q) \quad \Delta H=-36.9 \mathrm{~kJ}$$ What is the final temperature in a squeezed hot pack that contains \(25.0 \mathrm{~g}\) of \(\mathrm{LiCl}\) dissolved in \(125 \mathrm{~mL}\) of water? Assume a specific heat of \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) for the solution, an initial temperature of \(25.0{ }^{\circ} \mathrm{C}\), and no heat transfer between the hot pack and the environment.

Acetic acid \(\left(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\), whose aqueous solutions are known as vinegar, is prepared by reaction of ethyl alcohol ( \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) ) with oxygen: $$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)+\mathrm{H}_{2} \mathrm{O}(l)$$ Use the following data to calculate \(\Delta H^{\circ}\) in kilojoules for the reaction: $$\begin{aligned}&\Delta H_{\mathrm{f}}\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)\right]=-277.7 \mathrm{~kJ} / \mathrm{mol} \\ &\Delta H^{\circ} \mathrm{f}\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)\right]=-484.5 \mathrm{~kJ} / \mathrm{mol} \\ &\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{O}(l)\right]=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$

When \(1.045 \mathrm{~g}\) of \(\mathrm{CaO}\) is added to \(50.0 \mathrm{~mL}\) of water at \(25.0^{\circ} \mathrm{C}\) in a calorimeter, the temperature of the water increases to \(32.3^{\circ} \mathrm{C}\). Assuming that the specific heat of the solution is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) and that the calorimeter itself absorbs a negligible amount of heat, calculate \(\Delta H\) in kilojoules \(/ \mathrm{mol} \mathrm{Ca}(\mathrm{OH})_{2}\) for the reaction $$\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)$$

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