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For a reaction at constant pressure, the enthalpy change is \(+30 \mathrm{kJ} .\) During the reaction, the system expands and does \(25 \mathrm{kJ}\) of work. What is the change in internal energy for the reaction? (Section 13.5 )

Short Answer

Expert verified
The change in internal energy is \(5 \text{kJ}\).

Step by step solution

01

Understand the Problem

We are given that the enthalpy change \( \Delta H \) for a reaction is \(+30 \text{kJ}\), and the system does \(25 \text{kJ}\) of work by expanding against external pressure. We need to find the change in internal energy \( \Delta U \).
02

Recall the First Law of Thermodynamics

The First Law of Thermodynamics relates enthalpy \( \Delta H \), internal energy \( \Delta U \), and work \( w \) done as: \[ \Delta U = \Delta H - w \] where "work done by the system" is considered as \(w\).
03

Substitute Given Values

Using the equation \( \Delta U = \Delta H - w \), substitute \( \Delta H = +30 \text{kJ} \) and \( w = 25 \text{kJ} \): \[ \Delta U = 30 \text{kJ} - 25 \text{kJ} \]
04

Calculate the Change in Internal Energy

Perform the calculation to find \( \Delta U \): \[ \Delta U = 30 \text{kJ} - 25 \text{kJ} = 5 \text{kJ} \]
05

Interpret the Result

The result \( \Delta U = 5 \text{kJ} \) means that the internal energy of the system has increased by \( 5 \text{kJ} \) as a result of the reaction when considering both the enthalpy change and the work done.

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Key Concepts

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

Enthalpy
Enthalpy is an important concept in thermodynamics, particularly when we deal with reactions at constant pressure. It is a measure of the total energy of a thermodynamic system, commonly symbolized as \( H \). Enthalpy encompasses both the internal energy \( U \) of the system and the energy required to "make room" for it by displacing its environment. For reactions, the change in enthalpy \( \Delta H \) is crucial because it reflects the amount of heat absorbed or released, assuming the pressure remains constant.
In the exercise provided, the enthalpy change was given as \(+30 \text{kJ}\). This positive value indicates that the reaction absorbs heat from its surroundings, categorizing it as endothermic. Endothermic reactions require energy input, which increases the energy content of the system.
When you see a reference to enthalpy change in a problem, you’re often asked to quantify the heat exchange at constant pressure. Thus, understanding enthalpy helps you gauge how thermodynamic processes affect the energy stored within a system.
First Law of Thermodynamics
The First Law of Thermodynamics is a core principle in physics and chemistry, describing the conservation of energy. It states that the total energy of an isolated system remains constant and can neither be created nor destroyed, only transformed or transferred. When applied to thermodynamic processes, this law can be expressed as:
\[ \Delta U = q - w \]
Where:
  • \( \Delta U \) is the change in internal energy
  • \( q \) is the heat added to the system
  • \( w \) is the work done by the system
For cases at constant pressure, like our exercise, the equation is often adapted to encompass enthalpy changes:
\[ \Delta U = \Delta H - w \]
This formula highlights the relationships between different forms of energy within a system. It's essential for solving problems where you need to find internal energy changes when given enthalpy changes and work values. By using the First Law of Thermodynamics, you can clearly track energy flow in chemical reactions, thereby providing insights into these dynamic processes.
Work Done
Work done in the context of thermodynamics refers to the energy transfer due to a force acting over a distance, like when a system expands or contracts. It is typically represented by the symbol \( w \) and is a pivotal component in energy conservation equations. In our exercise, the system does \( 25 \text{kJ} \) of work by expanding, which is energy the system uses to "push back" the surroundings.
When discussing thermodynamic problems, 'work done by the system' is considered positive, because the system is using energy. However, in the equation relating internal energy to enthalpy, the convention of subtracting the work term (\( w \)) reflects that this energy is leaving the system.
Understanding work in thermodynamics helps explain how systems exchange energy with their environment. Whenever a system performs work, it can result in a change in internal energy, as we observed with \( \Delta U \) in the exercise. Being comfortable with these concepts supports grasping broader thermodynamic principles.

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Most popular questions from this chapter

Calculate the energy transferred as work, \(w\), when a gas is comprossed from \(250 \mathrm{cm}^{3}\) to \(125 \mathrm{cm}^{3}\) by an external prossure of \(10 \mathrm{kPa}\). (Section 13.5 )

A slice of banana weighing \(2.7 \mathrm{g}\) was burned in oxygen in a bomb calorimeter and produced a temperature rise of \(3.05 \mathrm{K}\) In the same calorimeter, the combustion of \(0.316 \mathrm{g}\) of benzoic acid produced a temperature rise of \(3.24 \mathrm{K}\). Ac U for benzoic acid is \(-3251 \mathrm{kJmol}^{-1}\). If the average mass of a banana is \(125 \mathrm{g}\), how much energy in (a) \(\mathrm{kJ}\) and (b) calories (kcal) can be obtained on average from a banana? (1 calorie \(=4.18 \mathrm{kJ}\) ) (Section 13.5 )

Calculate the energy needed to melt \(750 \mathrm{kg}\) of sodium metal at \(371 \mathrm{K}\). (The standard enthalpy change of fusion of sodium is \(+2.60 \mathrm{kJmol}^{-1} .\) ) (Section 13.2

The decomposition at constant volume of \(1 \mathrm{mol}\) of gaseous krypton difluoride \(\left(\mathrm{KrF}_{2}\right)\) to its elements at \(298 \mathrm{K}\) gives out \(59.4 \mathrm{kJ}\) of heat. Calculate the standard enthalpy change of formation, \(\Delta_{t} H_{298}^{\circ}\), of solid \(\mathrm{KrF}_{2}\). (The enthalpy change of sublimation of solid \(\mathrm{KrF}_{2}\) is \(+41 \mathrm{kJmol}^{-1}\). ) (Section \(13.3)\).

Calcite and aragonite are two forms of calcium carbonate. Calculate the enthalpy change for the transition from calcite to aragonite. (AH(calcite) \(=-1206.9 \mathrm{kJ} \mathrm{mol}^{-1}\) and \(\Delta \mathrm{H}\) (aragonite) \(=\) \(-1207.1 \mathrm{kJ} \mathrm{mol}^{-1}\), j (Section 13.3)

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