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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 constant-pressure process, \(\Delta H=0 .\) What can you conclude about \(\Delta E, q,\) and \(w ?\)

Short Answer

Expert verified
(a) The enthalpy change, \(\Delta H\), equals the amount of heat transferred, \(q\), at constant pressure (\(\Delta H = q_P\)). (b) In a constant-pressure process with heat release, the enthalpy decreases, as both \(q\) and \(\Delta H\) are negative. (c) When \(\Delta H = 0\) in a constant-pressure process, the heat exchanged (\(q\)) is equal to the difference between the work done by the gas on its surroundings (\(w\)) and the pressure-volume work done (\(P\Delta V\)).

Step by step solution

01

(a) Condition when enthalpy change equals heat transferred

The enthalpy change of a process, \(\Delta H\), is equal to the amount of heat transferred into or out of the system (\(q\)), under the condition of constant pressure. Mathematically, this can be written as: \[ \Delta H = q_P, \] where \(q_P\) is the heat transferred at constant pressure.
02

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

During a constant-pressure process, the system releases heat to the surroundings. This means that the heat transfer is negative (\(q < 0\)). According to the equation stated earlier, \(\Delta H = q_P\), the enthalpy change will also be negative (\((螖H < 0)\)) during the process. Therefore, the enthalpy of the system decreases during the process.
03

(c) Analysing the relationship between \(\Delta H\), \(\Delta E\), \(q\), and \(w\) when \(\Delta H=0\)

In a constant-pressure process, the enthalpy change is related to the internal energy change, heat, and work as follows: \[ \Delta H = \Delta E + P\Delta V, \] where \(\Delta E\) is the change in internal energy, \(P\) is the constant pressure, and \(\Delta V\) is the change in volume. When \(\Delta H = 0\), the equation becomes: \[ 0 = \Delta E + P\Delta V. \] Considering the first law of thermodynamics, which relates internal energy change, heat, and work, we have: \[ \Delta E = q - w, \] where \(w\) is the work done by the gas on its surroundings. By substituting this expression for \(\Delta E\) into the equation for constant pressure, we obtain: \[ 0 = (q - w) + P\Delta V. \] Solving for \(q\), we find: \[ q = w - P\Delta V. \] This indicates that when the enthalpy change is 0 during a constant-pressure process, the heat exchanged between the system and its surroundings (\(q\)) is equal to the difference between the work done by the gas on its surroundings (\(w\)) and the pressure-volume work done (\(P\Delta V\)).

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Key Concepts

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

constant pressure
Constant pressure is a condition in thermodynamics where the pressure of a system does not change during a process. This situation is particularly useful when analyzing chemical reactions and physical changes because it simplifies the calculations involved. At constant pressure, the relationship between enthalpy change (\(\Delta H\)) and heat transfer is straightforward:
  • Enthalpy change equals the heat transferred to or from the system.
  • This relationship is expressed mathematically by the equation \(\Delta H = q_P\), where \(q_P\) is the heat transferred at constant pressure.

During a constant-pressure process, if the system releases heat to its surroundings, the heat transfer is negative (\(q < 0\)). This means the system's enthalpy decreases, as indicated by a negative enthalpy change (\(\Delta H < 0\)). It's a simple yet powerful concept often encountered in thermodynamic studies.
heat transfer
Heat transfer involves the movement of thermal energy from one object or system to another. This can occur in several ways, such as conduction, convection, and radiation. In the context of thermodynamics and constant pressure, understanding heat transfer is crucial.
  • At constant pressure, the entire heat transfer during a process is equated to the change in enthalpy (\(\Delta H\)).
  • When heat is released by the system, the value of \(q\) is negative, indicating energy flow out of the system.
  • Conversely, when heat is absorbed, \(q\) is positive, and the system gains energy.

Heat transfer is a core concept in understanding how energy transactions affect the state and phase of a system. It is essential for analyzing processes in chemistry and engineering.
first law of thermodynamics
The first law of thermodynamics is a fundamental principle stating that energy cannot be created or destroyed, only transferred or transformed. Mathematically, it is expressed as:\[\Delta E = q - w,\]where \(\Delta E\) is the change in internal energy, \(q\) is the heat transferred to the system, and \(w\) is the work done by the system.
In constant-pressure processes, this law interconnects with enthalpy changes:
  • When \(\Delta H = 0\), energy balance becomes apparent: the heat added equals the work done minus any pressure-volume work.
  • Substituting \(\Delta E = q - w\) into the enthalpy equation \(\Delta H = \Delta E + P\Delta V\), we find \(q = w - P\Delta V\).
  • This showcases the intricate balance between heat, work, and internal energy in constant-pressure scenarios.

Understanding this law is crucial for grasping how energy conservation applies in real-world thermodynamic systems, ensuring energy changes are accurately accounted for across processes.

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Most popular questions from this chapter

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s) \quad \Delta H=-65.5 \mathrm{kJ}$$ (a) Calculate \(\Delta H\) for the production of 0.450 mol of AgCl by this reaction. (b) Calculate \(\Delta H\) for the production of 9.00 \(\mathrm{g}\) of AgCl. (c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{mol}\) of AgCl dissolves in water.

(a) Which of the following cannot leave or enter a closed system: heat, work, or matter? (b) Which cannot leave or enter an isolated system? (c) What do we call the part of the universe that is not part of the system?

During a normal breath, our lungs expand about 0.50 L against an external pressure of 1.0 atm. How much work is involved in this process (in J)?

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g)$$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$\begin{array}{ll}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-890.3 \mathrm{kJ}} \\ {\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)} & {\Delta H^{\circ}=-136.3 \mathrm{kJ}} \\ {2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-571.6 \mathrm{kJ}} \\ {2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-3120.8 \mathrm{kJ}}\end{array}$$

Ozone, \(\mathrm{O}_{3}(g),\) is a form of elemental oxygen that plays an important role 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 \(\mathrm{O}_{3}(g) ?\) (b) Which has the higher enthalpy under these conditions, 2 \(\mathrm{O}_{3}(g)\) or 3 \(\mathrm{O}_{2}(g) ?\)
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