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Chapter 5: Problem 34

(a) Under what condition will the enthalpy change of a process equal the amount of heat transferred into or out of the system? (b) During a constant- pressure process, the system releases heat to the surroundings. Does the enthalpy of the system increase or decrease during the process? (c) In a constantpressure process, \(\Delta H=0 .\) What can you conclude about \(\Delta E, q\) and \(w ?\)

Short Answer

Expert verified
(a) The enthalpy change (\(\Delta H\)) is equal to the heat transfer (\(q\)) under constant pressure conditions, such as \(\Delta H = q_P\). (b) When the system releases heat to the surroundings during a constant-pressure process, the enthalpy of the system decreases (\(\Delta H < 0\)). (c) For a constant-pressure process with \(\Delta H = 0\), we can conclude that \(\Delta E = 0\), \(q = 0\), and \(w = 0\).

Step by step solution

01

a) Condition for enthalpy change equal to heat transfer

The enthalpy change (\(\Delta H\)) is equal to the heat transfer (\(q\)) under constant pressure conditions. Mathematically, this can be expressed as follows: \[\Delta H = q_P\] where \(P\) denotes constant pressure.
02

b) Enthalpy change during heat release in a constant-pressure process

When the system releases heat to the surroundings during a constant-pressure process, the heat transfer (\(q\)) is negative (i.e., \(q < 0\)), as energy is flowing out of the system. According to the relationship in part (a), we have \(\Delta H = q_P\), so if \(q_P < 0\), then \(\Delta H < 0\). This means that the enthalpy of the system decreases during the process.
03

c) Conclusion for \(\Delta H = 0\) in a constant-pressure process

When \(\Delta H = 0\) in a constant-pressure process, the heat transfer (\(q_P\)) equals zero as well, since \(\Delta H = q_P\). We can use this information and the first law of thermodynamics to determine the values or relationships of \(\Delta E\) and \(w\). The first law of thermodynamics can be expressed as: \[\Delta E = q + w\] Since \(\Delta H = 0\) and we know that \(\Delta H = q_P\), we have: \[q + w = 0\] Since \(q = 0\) (no heat transfer), the only way for the equation to be satisfied is when \(w = 0\). Therefore, for a constant-pressure process with \(\Delta H = 0\), we can conclude that \(\Delta E = 0\), \(q = 0\), and \(w = 0\).

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Key Concepts

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

Enthalpy
Enthalpy is a foundational concept in thermodynamics related to the heat of a system. It is a property that combines the system's internal energy with the work done by the system due to pressure and volume changes. Enthalpy, represented as \(H\), is defined as:\[ H = E + PV \]where \(E\) is the internal energy, \(P\) is the pressure, and \(V\) is the volume of the system. This means enthalpy is a measure of energy in the form of heat within a system under constant pressure conditions. A change in enthalpy, \(\Delta H\), is crucial during chemical reactions where heat exchange with the surroundings occurs. For instance, when a reaction takes place at constant pressure, the enthalpy change equals the heat transferred, denoted by:\[ \Delta H = q_P \]"\(q_P\)" represents the heat transferred at constant pressure. Thus, enthalpy change indicates whether a process absorbs or releases heat.
Constant-pressure process
In thermodynamics, a constant-pressure process is a process where the pressure within the system remains unchanged even as other parameters, such as temperature or volume, might vary. This is particularly relevant in many natural and industrial processes, where pressure equilibrium is maintained with the surroundings. During a constant-pressure process, the enthalpy change \((\Delta H)\) directly corresponds to the heat exchanged \((q_P)\). If heat is released to the surroundings, as with exothermic reactions, \(q_P\) is negative, indicating a decrease in enthalpy:
  • \(\Delta H < 0\) 鈥� The system's enthalpy decreases when it releases heat.
  • \(\Delta H > 0\) 鈥� The system's enthalpy increases if it absorbs heat.
Understanding constant-pressure processes is essential for predicting system behavior during heat exchange, especially in chemical reactions and phase changes.
First law of thermodynamics
The first law of thermodynamics is a principle of conservation of energy, stating that energy cannot be created or destroyed, only transformed or transferred. In formula terms, this is expressed as:\[ \Delta E = q + w \]where \(\Delta E\) represents a change in internal energy, \(q\) is the heat added to the system, and \(w\) is the work done by the system. In constant-pressure processes where \(\Delta H = 0\), the heat change \(q_P\) is zero, leading to:\[ q + w = 0 \]This demonstrates that, in such cases, any work done by or on the system is internally compensated by changes, keeping the internal energy \(\Delta E\) unchanged. Consequently, for a process with no net heat exchange and no work done \((w = 0)\), both \(\Delta E\) and \(q\) are zero, illustrating energy balance in thermodynamic processes.

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Most popular questions from this chapter

Using values from Appendix \(\mathrm{C},\) calculate the standard enthalpy change for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)+4 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{SiCl}_{4}(l)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{SiO}_{2}(s)+4 \mathrm{HCl}(g)\)

(a) When a \(0.235-\mathrm{g}\) sample of benzoic acid is combusted in a bomb calorimeter (Figure 5.19 ), the temperature rises \(1.642{ }^{\circ} \mathrm{C}\). When a 0.265 -g sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~N}_{4}\), is burned, the temperature rises \(1.525^{\circ} \mathrm{C}\). Using the value \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to 0.001 \(\mathrm{g}\), what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{~g} \mathrm{CO}_{2}(g), 4.47 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}(g),\) and \(311 \mathrm{~kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empirical-formula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

Ozone, \(\mathrm{O}_{3}(g)\), is a form of elemental oxygen that is important in the absorption of ultraviolet radiation in the stratosphere. It decomposes to \(\mathrm{O}_{2}(g)\) at room temperature and pressure according to the following reaction: $$ 2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) \quad \Delta H=-284.6 \mathrm{~kJ} $$ (a) What is the enthalpy change for this reaction per mole of \(\mathrm{O}_{3}(g) ?(\mathbf{b})\) Which has the higher enthalpy under these condi- $$ \text { tions, } 2 \mathrm{O}_{3}(g) \text { or } 3 \mathrm{O}_{2}(g) ? $$

Using values from Appendix \(\mathrm{C},\) calculate the value of \(\Delta H^{\circ}\) for each of the following reactions: (a) \(\mathrm{CaO}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{CaCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(4 \mathrm{FeO}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) (c) \(2 \mathrm{CuO}(s)+\mathrm{NO}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{NO}_{2}(g)\) (d) \(4 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)
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