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Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: (a) \(q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J} .\) (b) A system releases \(66.1 \mathrm{~kJ}\) of heat to its surroundings while the surroundings do \(44.0 \mathrm{~kJ}\) of work on the system.

Short Answer

Expert verified
(a): \( \Delta E = -0.077 \) kJ; exothermic. (b): \( \Delta E = -22.1 \) kJ; exothermic.

Step by step solution

01

Understanding the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy \( \Delta E \) of a system is given by the sum of the heat \( q \) added to the system and the work \( w \) done on the system: \[ \Delta E = q + w \].
02

Calculating \( \Delta E \) for Case (a)

Given: \( q = 0.763 \) kJ and \( w = -840 \) J. We need to convert everything to the same unit, so convert \( w \) to kJ: \( w = -840 \) J = \(-0.840 \) kJ. Now substitute into the formula: \[ \Delta E = 0.763 \text{kJ} + (-0.840 \text{kJ}) = -0.077 \text{kJ} \].
03

Determining the Nature of the Process for Case (a)

Since \( \Delta E = -0.077 \text{kJ} \), which is negative, the process is exothermic (the system loses energy to the surroundings).
04

Calculating \( \Delta E \) for Case (b)

In this case, the system releases heat \( q = -66.1 \) kJ and the surroundings do work on the system \( w = 44.0 \) kJ. Therefore: \[ \Delta E = -66.1 \text{kJ} + 44.0 \text{kJ} = -22.1 \text{kJ} \].
05

Determining the Nature of the Process for Case (b)

Since \( \Delta E = -22.1 \text{kJ} \), which is negative, the process is exothermic (the system loses energy to the surroundings).

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Key Concepts

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

First Law of Thermodynamics
The first law of thermodynamics serves as a fundamental principle in understanding energy changes in a system. This law is sometimes referred to as the law of energy conservation. Essentially, it states that the total energy of an isolated system remains constant. In simpler terms, energy can neither be created nor destroyed; it can only be transformed from one form to another. This is mathematically represented by the equation: \[ \Delta E = q + w \]where:
  • \( \Delta E \) is the change in internal energy of the system
  • \( q \) is the heat added to the system
  • \( w \) is the work done on the system
When you apply this understanding to a problem, the challenge is to determine how much energy is exchanged in the form of heat and work. Internal energy change \( \Delta E \) can be positive or negative, indicating whether the system gains or loses energy respectively.
This equation is vital for solving problems in thermodynamics, and knowing how to apply it can help in determining if a process is endothermic or exothermic effectively.
Endothermic Process
An endothermic process is one where energy is absorbed from the surroundings into the system. During such processes, the overall energy of the system increases. An easy way to remember this is by relating it to the term 'endo-', which means 'inside'.
In practical terms, if \( \Delta E \) is positive, this indicates an endothermic process. This means that more energy is absorbed into the system than is released through work or other means. You may observe endothermic processes in daily life, such as:
  • Melting ice - heat is absorbed from the environment to convert solid ice into liquid water.
  • Photosynthesis - plants absorb sunlight to convert carbon dioxide and water into glucose and oxygen.
These processes require energy input to occur. Understanding when a process is endothermic involves calculating \( \Delta E \) and observing a positive value, which shows that the system has gained energy.
Exothermic Process
In contrast to endothermic processes, an exothermic process is characterized by the release of energy from the system to its surroundings. As a result, the system loses internal energy, indicated by a negative \( \Delta E \). The term 'exo-' can be associated with 'outside', reflecting that the energy is being transferred out of the system.
If you calculate \( \Delta E \) and find it to be negative, this confirms that the process is exothermic. A few examples of exothermic processes include:
  • Combustion - burning of fuels releases heat energy into the surroundings.
  • Respiration - the metabolic process of breaking down glucose to provide energy for cells also releases heat.
Such processes are often spontaneous and may produce heat, light, or sometimes even a sound. In evaluating thermodynamic problems, a negative \( \Delta E \) will clearly indicate an exothermic reaction.

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Most popular questions from this chapter

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{I}_{2}(s)\) or \(1 \mathrm{~mol} \mathrm{l}_{2}(g)\) at the same temperature, (b) \(2 \mathrm{~mol}\) of iodine atoms or \(1 \mathrm{~mol}\) of \(\mathrm{I}_{2},(\mathbf{c}) 1 \mathrm{~mol} \mathrm{I}_{2}(g)\) and \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or \(2 \mathrm{~mol} \mathrm{HI}(g)\) at \(25^{\circ} \mathrm{C},\left(\right.\) d) \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g})\) at \(300{ }^{\circ} \mathrm{C} .\)

Consider the following hypothetical reactions: $$ \begin{array}{l} \mathrm{A} \longrightarrow \mathrm{B} \quad \Delta H_{I}=+60 \mathrm{k} \mathrm{J} \\ \mathrm{B} \longrightarrow \mathrm{C} \quad \Delta H_{I}=-90 \mathrm{k} \mathrm{J} \end{array} $$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(\mathrm{A} \longrightarrow \mathrm{C}\) (b) Construct an enthalpy diagram for substances \(A, B\), and \(\mathrm{C},\) and show how Hess's law applies.

A \(100-\mathrm{kg}\) man decides to add to his exercise routine by walking up six flights of stairs \((30 \mathrm{~m}) 10\) times per day. He figures that the work required to increase his potential energy in this way will permit him to eat an extra order of French fries, at 245 Cal, without adding to his weight. Is he correct in this assumption?

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is produced by plants as follows: \(12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(I) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g)\) $$ \Delta H=5645 \mathrm{~kJ} $$ About \(4.8 \mathrm{~g}\) of sucrose is produced per day per square meter of the earth's surface. The energy for this endothermic reaction is supplied by the sunlight. About \(0.1 \%\) of the sunlight that reaches the earth is used to produce sucrose. Calculate the total energy the sun supplies for each square meter of surface area. Give your answer in kilowatts per square meter \(\left(\mathrm{kW} / \mathrm{m}^{2}\right.\) where \(\left.1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\right)\)

(a) What is meant by the term state function? (b) Give an example of a quantity that is a state function and one that is not. \((\mathbf{c})\) Is the volume of a system a state function? Why or why not?

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