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What is the difference between \(\Delta H\) and \(\Delta E\) ?

Short Answer

Expert verified
The difference between ΔH (enthalpy change) and ΔE (internal energy change) lies in the way energy is exchanged with the surroundings during a reaction or process. ΔH accounts for energy exchange in the form of both heat and pressure-volume work (ΔH = ΔE + PΔV), while ΔE focuses on the internal energy of the system, which includes kinetic and potential energies of particles but excludes energy transfer in the form of work done by or on the system. Both quantities are essential in understanding the energetics of chemical reactions and thermodynamic processes.

Step by step solution

01

Define ΔH and ΔE

ΔH, or enthalpy change, is the difference in enthalpy (total energy in a system) between the final state and initial state of a process. It is an important parameter in thermodynamics, and it accounts for energy exchanged by the system in the form of heat and work. ΔE, or internal energy change, is the difference in internal energy between the final state and initial state of a process. It represents the energy within the system, excluding any energy exchange with the surroundings.
02

Identify the difference in energy exchange

The main difference between ΔH and ΔE is the form of energy exchange with the surroundings. In ΔH, energy is exchanged in the form of both heat (q) and pressure-volume work (pV work, w = -PΔV), where P is pressure and ΔV is the change in volume of the system. It is applicable to both constant pressure and constant volume processes. In ΔE, energy is exchanged in the form of internal energy, which is the sum of kinetic and potential energies of the particles in the system. It does not include any energy transfer in the form of work done by or on the system.
03

Relating ΔH and ΔE

For constant pressure processes, we can relate ΔH and ΔE using the following equation: ΔH = ΔE + PΔV This equation shows that the enthalpy change is equal to the internal energy change plus the pressure-volume work done on or by the system.
04

Summarize the difference

In summary, the difference between ΔH and ΔE lies in the way energy is exchanged with the surroundings. ΔH accounts for both heat and pressure-volume work, while ΔE focuses on the internal energy of the system. Both quantities are essential in understanding the energetics of chemical reactions and thermodynamic processes.

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Key Concepts

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

ΔH (Enthalpy Change)
Enthalpy, represented by the symbol 'H', is a measure of the total energy of a thermodynamic system. It includes the internal energy, which is the energy required to create a system, and the energy required to make room for it by displacing its environment and establishing its volume and pressure.

The change in enthalpy (\textbf{\(\Delta H\)}) during a process is termed as the enthalpy change. It's a crucial concept in chemistry and physics, especially when dealing with heat exchange in chemical reactions or phase changes. In a reaction, \textbf{\(\Delta H\)} is positive in endothermic processes, where heat is absorbed from the surroundings, and negative in exothermic processes, where heat is released to the surroundings. Understanding enthalpy change is necessary for fields such as chemical engineering, material science, and thermodynamics, as it helps predict the energy exchange involved in various processes.
ΔE (Internal Energy Change)
Internal energy is a bit less intuitive than enthalpy because it's not something you feel directly like you might feel heat coming off an object. It's the sum of all the kinetic and potential energy of the particles in a system. This includes everything from the electrons moving around nuclei to the vibrations of atoms in a piece of metal. The symbol 'E' denotes internal energy.

The change in internal energy (\textbf{\(\Delta E\)}) is the difference in a system's internal energy from the initial state to the final state. It accounts for the energy changes happening inside the system due to chemical reactions or physical transformations, not involving the external work done by the system or heat transfer. For instance, when a chemical bond breaks, the system absorbs energy, increasing the \textbf{\(\Delta E\)}, and when a bond forms, it releases energy, decreasing the \textbf{\(\Delta E\)}. Variations in \textbf{\(\Delta E\)} are central to fields such as statistical mechanics and quantum chemistry.
Thermodynamics
Thermodynamics is essentially the science of energy, heat, work, and how they interplay within the physical world. It deals with the concepts of energy transfer and transformations and provides a framework for understanding phenomena such as why ice melts when left out at room temperature or why your car engine gets hot when running.

At the core of thermodynamics lie several laws that dictate how these energy transformations can occur, and they tell us things like energy cannot be created or destroyed (the First Law) and that the disorder, or entropy, of the universe tends to increase over time (the Second Law). Thermodynamics is not just theoretical—it's immensely practical. It helps us design engines, refrigerators, and even predict the spontaneous direction of chemical reactions—among a myriad of other practical applications.
Pressure-Volume Work
Pressure-volume work (\textbf{\(pV\)} work) is a concept in thermodynamics that occurs when a force is applied to change the volume of a system against an external pressure. Imagine pushing down on a syringe: here, you're doing work against the air pressure to decrease the volume. In a chemical system, pressure-volume work happens when gases expand or contract within a reaction.

Mathematically, it can be expressed as \textbf{\(w = -P\Delta V\)}, where \textbf{\(w\)} is the work done, \textbf{\(P\)} is the external pressure, and \textbf{\(\Delta V\)} is the change in volume. It is considered in situations where a change in the system's volume results in energy transfer, such as when a gas expands in a cylinder with a movable piston. It's an essential parameter in calculating the enthalpy change (\textbf{\(\Delta H\)}) and internal energy change (\textbf{\(\Delta E\)}) of a system under constant pressure.

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Most popular questions from this chapter

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \text { (unbalanced) } $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{~kJ}\). If \(2.50 \times 10^{2} \mathrm{~mL} \mathrm{~N}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50 \mathrm{~atm}\) and \(4.50 \times\) \(10^{2} \mathrm{~mL} \mathrm{O}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50\) atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

For the process \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) at \(298 \mathrm{~K}\) and \(1.0 \mathrm{~atm}\), \(\Delta H\) is more positive than \(\Delta E\) by \(2.5 \mathrm{~kJ} / \mathrm{mol}\). What does the \(2.5 \mathrm{~kJ} / \mathrm{mol}\) quantity represent?

Consider the dissolution of \(\mathrm{CaCl}_{2}\) : \(\mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{~kJ}\) An \(11.0-\mathrm{g}\) sample of \(\mathrm{CaCl}_{2}\) is dissolved in \(125 \mathrm{~g}\) water, with both substances at \(25.0^{\circ} \mathrm{C}\). Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\)

Given the following data $$ \begin{aligned} 2 \mathrm{O}_{3}(g) & \longrightarrow 3 \mathrm{O}_{2}(g) & \Delta H &=-427 \mathrm{~kJ} \\ \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{O}(g) & \Delta H &=495 \mathrm{~kJ} \\ \mathrm{NO}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-199 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{NO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}_{2}(g) $$

A gas absorbs \(45 \mathrm{~kJ}\) of heat and does \(29 \mathrm{~kJ}\) of work. Calculate \(\Delta E\).

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