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Suppose that the gas-phase reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow\) 2 \(\mathrm{NO}_{2}(g)\) were carried out in a constant-volume container at constant temperature. (a) Would the measured heat change represent \(\Delta H\) or \(\Delta E ?\) (b) If there is a difference, which quantity is larger for this reaction? (c) Explain your answer to part (b).

Short Answer

Expert verified
(a) The measured heat change in a constant-volume container at constant temperature represents the internal energy change, ΔE. (b) In this specific reaction, ΔH and ΔE are equal as there is no difference between them. (c) Since the reaction is carried out in a constant-volume container at constant temperature, and the change in volume (ΔV) is zero, it results in enthalpy change (ΔH) being equal to internal energy change (ΔE). Therefore, neither ΔH nor ΔE is larger, as they are equal in this situation.

Step by step solution

01

(a) Identifying the Measured Heat Change

: The heat change measured in a constant-volume container is equivalent to the change in internal energy (ΔE) of the reaction. Therefore, in this case, the measured heat change would represent ΔE.
02

(b) Determining the Larger Quantity Between ΔH and ΔE

: The relationship between the enthalpy change (ΔH) and internal energy change (ΔE), at constant temperature, is given by: \[\Delta H = \Delta E + P\Delta V\] Where P is the pressure and ΔV is the change in volume. Since the volume is constant in this case, ΔV = 0, which means that ΔH = ΔE. Therefore, there is no difference between ΔH and ΔE in this reaction, and they are equal.
03

(c) Explaining the Answer to Part (b)

: Since the reaction is carried out in a constant-volume container at constant temperature, the change in volume (ΔV) is zero. This means that the enthalpy change (ΔH) is equal to the internal energy change (ΔE) for this reaction. As a result, neither ΔH nor ΔE is larger, as they are equal in this specific situation.

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Key Concepts

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

Understanding Enthalpy Change
Enthalpy change, often denoted as \( \Delta H \), is a fundamental concept in thermodynamics. It represents the heat exchanged by a system under constant pressure conditions.Unlike internal energy, enthalpy takes into account the work done by the pressure-volume work associated with expansion or compression. This can be understood through the equation:\[ \Delta H = \Delta E + P\Delta V\]Where:
  • \(\Delta E\) is the change in internal energy.
  • \(P\) is the pressure of the system.
  • \(\Delta V\) is the change in volume.
In situations where the volume remains constant, \(\Delta V = 0\), making the enthalpy change equal to the internal energy change.This scenario is typical in reactions that occur in a constant-volume container. Here, no volume work is performed, thus simplifying the relationship between enthalpy and internal energy.
Explaining Internal Energy
Internal energy, symbolized by \(\Delta E\), is the total energy contained within a system. This includes both the kinetic and potential energy of all its particles.In thermodynamic processes, changes in internal energy are crucial as they indicate the system’s energy gain or loss due to heat transfer and work done.For a chemical reaction carried out in a constant-volume container, the only way the system can exchange energy with its surroundings is through heat. As such, the internal energy change \(\Delta E\), in this case, aligns with the measured heat.Internal energy changes reflect fundamental shifts:
  • Heat absorbed or released by the system.
  • Energy required or released during bond breakage and formation.
Understanding \(\Delta E\) helps scientists predict and measure the energy balance of reactions under various conditions.
The Concept of Constant-Volume
The constant-volume condition is vital in understanding energy changes in reactions.When a reaction takes place at constant volume, it means that the volume of the system does not change throughout the process. This condition significantly simplifies thermodynamic calculations.In this scenario:
  • No pressure-volume work is done.
  • The energy change of the system is directly related to heat transfer.
Since \(\Delta V = 0\), the relationship between enthalpy and internal energy simplifies to \(\Delta H = \Delta E\).Conducting reactions at constant volume is common in laboratory settings. This approach minimizes the need for corrections due to volume work, allowing scientists to accurately measure internal energy changes during chemical reactions.

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Most popular questions from this chapter

Meals-ready-to-eat (MREs) are military meals that can be heated on a flameless heater. The heat is produced by the following reaction: $$\mathrm{Mg}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{H}_{2}(g)$$ (a) Calculate the standard enthalpy change for this reaction. (b) Calculate the number of grams of Mg needed for this reaction to release enougy energy to increase the temperature of 75 mL of water from 21 to \(79^{\circ} \mathrm{C}\) .

In a thermodynamic study, a scientist focuses on the properties of a solution in an apparatus as illustrated. A solution is continuously flowing into the apparatus at the top and out at the bottom, such that the amount of solution in the apparatus is constant with time. (a) Is the solution in the apparatus a closed system, open system, or isolated system? (b) If the inlet and outlet were closed, what type of system would it be?

A magnesium ion, Mg \(^{2+},\) with a charge of \(3.2 \times 10^{-19} \mathrm{Cand}\) an oxide ion, \(\mathrm{O}^{2-},\) with a charge of \(-3.2 \times 10^{-19} \mathrm{C}\) , are separated by a distance of 0.35 \(\mathrm{nm}\) . How much work would be required to increase the separation of the two ions to an infinite distance?

The complete combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l),\) to form \(\mathrm{H}_{2} \mathrm{O}(g)\) and \(\mathrm{CO}_{2}(g)\) at constant pressure releases 1235 \(\mathrm{kJ}\) of heat per mole of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction.

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.

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