Book cover for Chemistry: The Molecular Nature of Matter

Chemistry: The Molecular Nature of Matter

Neil D. Jespersen, James E. Brady, Alison Hyslop

ISBN #9781118413920

7th Edition

3,064 Questions

Group icon
53,557 Students Helped

Homework Questions

Right arrow
Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This chapter on thermodynamics covers the fundamental laws governing energy transformations in chemical reactions. It begins with the conservation of energy through the first law, emphasizing the relationship between heat, work, and internal energy. The second and third laws introduce the concept of entropy and the conditions for spontaneous change. The introduction of the Gibbs free energy (?G) further integrates enthalpy and entropy considerations, enabling the prediction of reaction spontaneity and the calculation of equilibrium constants. Practical applications of these principles extend from combustion engines to sustainable energy systems, supported by calculations involving standard free energy changes and bond energies.

Learning Objectives

1

Describe the fundamental laws of thermodynamics and their implications for energy transformations.

2

Explain how the first law relates internal energy changes to heat and work in chemical reactions.

3

Analyze how the second and third laws introduce concepts of entropy and dictate the spontaneity of processes.

4

Apply the Gibbs free energy function (?G) to predict reaction spontaneity and maximum work output.

5

Calculate standard free energy changes, equilibrium constants, and bond energies to quantify chemical processes.

Key Concepts

CONCEPT

DEFINITION

First Law of Thermodynamics

A principle stating that energy is conserved in any process; the change in internal energy (ΔU) of a system equals the heat (Q) added to the system minus the work (W) done by the system (ΔU = Q - W).

Second Law of Thermodynamics

A principle that dictates spontaneous processes lead to an increase in the total entropy of the universe. It introduces the concept that energy transformations are not completely reversible.

Third Law of Thermodynamics

A principle stating that as a system approaches absolute zero, the entropy of a perfect crystal approaches zero, providing a reference point for determining absolute entropies.

Entropy

A measure of the disorder or randomness in a system, which increases in spontaneous processes as dictated by the second law.

Gibbs Free Energy (ΔG)

A thermodynamic function defined as ΔG = ΔH - TΔS; it predicts reaction spontaneity, where a negative ΔG indicates a reaction that can perform useful work.

Equilibrium Constant

A parameter that quantifies the ratio of product concentrations to reactant concentrations at equilibrium and is related to the standard free energy change (ΔG°).

Bond Energies

The amounts of energy required to break chemical bonds, which are used to estimate reaction enthalpies and, indirectly, free energy changes.

Example Problems

Example 1

What is the origin of the name thermodynamics?

Example 2

State the first law of thermodynamics in your own words. What equation defines the change in the internal energy in terms of heat and work? Define the meaning of the symbols, including the significance of their algebraic signs.

Example 3

How is a change in the internal energy defined in terms of the initial and final internal energies?

Example 4

What is the algebraic sign of $\Delta E$ for an endothermic change? Why?

Example 5

Which quantities in the statement of the first law are state functions and which are not?
Scroll left
Scroll right

Step-by-Step Explanations

QUESTION

Given that a chemical system absorbs 150 J of heat and performs 50 J of work, calculate the change in internal energy (ΔU).

STEP-BY-STEP ANSWER:

Step 1: Recall the first law of thermodynamics: ΔU = Q - W.
Step 2: Insert the provided values: Q = 150 J and W = 50 J.
Step 3: Perform the calculation: ΔU = 150 J - 50 J.
Step 4: Conclude that ΔU = 100 J.
Final Answer: The change in internal energy is 100 J.

First Law of Thermodynamics

QUESTION

For a reaction with an enthalpy change (ΔH) of -200 kJ, an entropy change (ΔS) of -0.5 kJ/K, at a temperature of 300 K, calculate ΔG.

STEP-BY-STEP ANSWER:

Step 1: Write down the Gibbs free energy equation: ΔG = ΔH - TΔS.
Step 2: Substitute the given values: ΔG = (-200 kJ) - (300 K)(-0.5 kJ/K).
Step 3: Multiply the temperature and entropy change: 300 K * -0.5 kJ/K = -150 kJ.
Step 4: Compute ΔG: ΔG = -200 kJ - (-150 kJ) = -200 kJ + 150 kJ.
Step 5: Conclude that ΔG = -50 kJ.
Final Answer: The Gibbs free energy change is -50 kJ, indicating spontaneity.

Gibbs Free Energy (ΔG)

QUESTION

Explain how the standard free energy change (ΔG°) is related to the equilibrium constant (K) for a reaction.

STEP-BY-STEP ANSWER:

Step 1: Recall the fundamental relationship between ΔG° and the equilibrium constant: ΔG° = -RT ln K, where R is the gas constant and T is temperature in Kelvin.
Step 2: Understand that a negative ΔG° corresponds to a large equilibrium constant (favoring products), which signifies a spontaneous reaction.
Step 3: Recognize that if ΔG° is positive, the equilibrium constant is low (favoring reactants) and the reaction is non-spontaneous under standard conditions.
Final Answer: The standard free energy change is inversely related to the equilibrium constant; a more negative ΔG° implies a larger K and a more product-favored equilibrium.

Equilibrium Constant and ΔG°

Scroll left
Scroll right

Common Mistakes

  • Misinterpreting the sign conventions for heat (Q) and work (W) in the first law, leading to incorrect calculations of internal energy.
  • Confusing entropy with disorder, without understanding its quantitative role in dictating reaction spontaneity.
  • Assuming that a positive ?G automatically means no reaction occurs, rather than indicating non-spontaneity under standard conditions.
  • Neglecting the effect of temperature on ?G, which can influence reaction spontaneity in ways not immediately obvious.
  • Overlooking the relationship between equilibrium constants and ?G°, thereby missing important insights into reaction dynamics and reversibility.