For a certain reaction at constant pressure, \(\Delta U=-95 \mathrm{~kJ}\) and \(56 \mathrm{~kJ}\) of expansion work is done by the system. What is \(\Delta H\) for this process?

Short Answer

Expert verified
delta H for the process is -39 kJ.

Step by step solution

01

Understanding the Relationship Between delta Udelta H, and Work

Recall the first law of thermodynamics, which states that the change in internal energy (delta U) of a system is equal to the heat added to the system (q) minus the work done by the system (w): delta U = q - w. At constant pressure, the heat added to the system is the change in enthalpy (delta H), so we have delta U = delta H - w.
02

Rearrange the Equation to Solve for delta H

To find delta H, we need to rearrange the first law of thermodynamics to solve for delta H. The rearranged equation is: delta H = delta U + w.
03

Substitute the Known Values

Substitute delta U = -95 kJ and w = 56 kJ (expansion work done by the system, which is positive since the system is doing work on the surroundings) into the equation delta H = delta U + w to find delta H.
04

Calculate delta H

Now we calculate delta H using the substituted values: delta H = (-95 kJ) + (56 kJ) = -39 kJ. This is the enthalpy change for the process.

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In essence, it studies the effects of changes in temperature, pressure, volume, and chemical structure that are associated with energy transitions in physical systems.

In the realm of thermodynamics, various laws describe the conservation and transfer of energy, and one of the central concepts is heat transfer and its transformation into work. This field is fundamental to understanding not only physics and chemistry but also vital in fields like engineering and meteorology, where energy transformations are crucial.
First Law of Thermodynamics
The first law of thermodynamics, which is also known as the law of energy conservation, states that the total amount of energy in an isolated system remains constant; energy cannot be created or destroyed, only transformed or transferred. For example, when we say that a system does work on its surroundings, we're observing energy transfer from the system to something else.

The mathematical expression of this law, \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy of the system, \( Q \) is the heat added to the system, and \( W \) is the work done by the system, elegantly summarizes the conservation of energy. This equation helps us understand how systems exchange energy, which is crucial for calculating quantities like the enthalpy change in a reaction.
Internal Energy
Internal energy, represented as \( \Delta U \) in thermodynamics, is the total energy contained within a system. It encompasses the energy held in chemical bonds, as well as kinetic and potential energy of particles in the system. In a chemical reaction, when bonds are broken and new bonds form, the internal energy of the system can either increase or decrease. This change in internal energy can be observed through temperature changes or work being done by or on the system.

In the context of our exercise, a negative \( \Delta U \) signifies that energy is being released by the system to the surroundings, typically in the form of heat or work. As a foundational concept, understanding internal energy is essential for interpreting and predicting the outcomes of thermodynamic processes.
Expansion Work
Expansion work occurs when a system changes in volume while exerting a force on its surroundings, such as when a gas expands against the pressure in its container. In thermodynamic terms, if a system does expansion work, it means the system is using its energy to push against an external pressure. This work, often referred to as 'P-V' (pressure-volume) work, can be calculated using the equation \( W = P \Delta V \) where \( P \) is the pressure and \( \Delta V \) is the change in volume.

In our textbook example, the system has done expansion work of 56 kJ. This indicates that the system expanded, performing work on its surroundings. The ability to calculate and understand expansion work is crucial for solving many thermodynamic problems, as it's a common way to release or absorb energy during reactions and processes.

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Most popular questions from this chapter

The internal energy of a system increased by \(400 \mathrm{~J}\) when it absorbed \(600 \mathrm{~J}\) of heat. (a) Was work done by or on the system? (b) How much work was done?

Identify the following systems as open, closed, or isolated: (a) gasoline burning in an automobile enginc; (b) mercury in a thermometer; (c) a living plant.

Two successive stages in the industrial manufacture of sulfuric acid are the combustion of sulfur and the oxidation of sulfur dioxide to sulfur trioxide. From the standard reaction enthalpies $$ \begin{gathered} \mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{~g}) \\ \Delta H^{\circ}=-296.83 \mathrm{~kJ} \\ 2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{SO}_{3}(\mathrm{~g}) \\ \Delta H^{\circ}=-791.44 \mathrm{~kJ} \end{gathered} $$ Calculate the reaction enthalpy for the oxidation of sulfur dioxide to sulfur trioxide in the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g})\).

Calculate the enthalpy of the reaction \(\mathrm{P}_{4}(\mathrm{~s})+10 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{PCl}_{5}(\mathrm{~s})\) from the reactions $$ \begin{gathered} \mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{PCl}_{3}(\mathrm{I}) \\ \Delta H^{\circ}=-1278.8 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{PCl}_{5}(\mathrm{~s}) \\ \Delta H^{m}=-124 \mathrm{~kJ} \end{gathered} $$

A calorimeter has a measured heat capacity of \(6.27 \mathrm{~kJ} \cdot\left({ }^{\circ} \mathrm{C}\right)^{-1}\). The combustion of \(1.84 \mathrm{~g}\) of magnesium led to a temperature change from \(21.30^{\circ} \mathrm{C}\) to \(28.56^{\circ} \mathrm{C}\). Calculare the enthalpy change of the reaction \(2 \mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(\mathrm{s})\).

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