Book cover for General Chemistry: Principles and Modern Applications

General Chemistry: Principles and Modern Applications

Ralph H. Petrucci, F. Geoffrey Herring, Jeffry D. Madura, Carey Bissonnette

ISBN #9780132931281

11th Edition

3,230 Questions

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293,395 Students Helped

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This chapter integrates two classical views of entropy—Boltzmann's statistical approach and Clausius’s classical thermodynamics—to build a comprehensive understanding of spontaneous change in chemical systems. Through the exploration of entropy, solution properties, and Gibbs energy, students learn how intermolecular forces and concentration units affect solution behavior. The chapter emphasizes the criteria of the second law of thermodynamics for spontaneity and provides real-world applications that underscore the importance of colligative properties in everyday contexts.

Learning Objectives

1

Explain the contrasting perspectives of entropy from Boltzmann’s (microscopic) and Clausius’s (macroscopic) viewpoints.

2

Analyze how intermolecular forces and concentration units (molarity and molality) affect solution behavior and ideality.

3

Apply the second law of thermodynamics to determine the spontaneity of chemical processes using entropy and Gibbs energy.

4

Describe the significance of colligative properties in practical applications such as automotive antifreeze and medical IV solutions.

5

Evaluate how Gibbs energy changes with temperature and composition to predict spontaneous change in systems.

Key Concepts

CONCEPT

DEFINITION

Entropy (S)

A measure of the disorder or randomness in a system; in Boltzmann’s view, S = k ln(W), where W is the number of microscopic arrangements.

Solute

The substance that is dissolved in a solution.

Solvent

The medium in which the solute is dissolved to form a solution.

Molarity

A concentration unit representing the number of moles of solute per liter of solution, sensitive to temperature changes due to volume variation.

Molality

A concentration unit defined as moles of solute per kilogram of solvent, independent of temperature.

Colligative Properties

Properties of solutions that depend only on the number of solute particles, not on their identity. Examples include boiling point elevation and freezing point depression.

Gibbs Energy (G)

A thermodynamic function defined as G = H - TS, used to predict the spontaneity of a process; a negative ΔG indicates spontaneous change.

Second Law of Thermodynamics

A fundamental law stating that the entropy of an isolated system tends to increase over time, leading to spontaneous change.

Example Problems

Example 1

Consider a system of five distinguishable particles confined to a one-dimensional box of length $L .$ Describe how the following actions affect the number of accessible microstates and the entropy of the system: (a) increasing the length of the box to $2 L$ for fixed total energy (b) increasing the total energy for constant length $L$

Example 2

Consider a sample of ideal gas initially in a volume $V$ at temperature $T$ and pressure $P .$ Does the entropy of this system increase, decrease, or stay the same in the following processes? (a) The gas expands isothermally. (b) The pressure is increased at constant temperature. (c) The gas is heated at constant pressure.

Example 3

The standard molar entropy of $\mathrm{H}_{2}(\mathrm{g})$ is $S^{\circ}=$ $130.7 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ at $298 \mathrm{K} .$ Use this value, together with Boltzmann's equation, to determine the number of microstates, $W$, for one mole of $\mathrm{H}_{2}(\mathrm{g})$ at $298 \mathrm{K}$ and 1 bar. Reflect on the magnitude of your calculated value by describing how to write the number in decimal form.

Example 4

In a 1985 paper in the Journal of Chemical Thermodynamics, the standard molar entropy of chalcopyrite, $\mathrm{CuFeS}_{2},$ is given as $0.012 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}$ at $5 \mathrm{K}$. [See R. A. Robie et. al., J. Chem. Thermodyn., 17, 481 (1985).] Use this value to estimate the number of microstates for one picogram $\left(1 \times 10^{-12} \mathrm{g}\right)$ of $\mathrm{CuFeS}_{2}$ at $5 \mathrm{K}$ and 1 bar. Report your answer in the manner described in Exercise 3.

Example 5

Indicate whether each of the following changes represents an increase or a decrease in entropy in a system, and explain your reasoning: (a) the freezing of ethanol; (b) the sublimation of dry ice; (c) the burning of a rocket fuel.
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Step-by-Step Explanations

QUESTION

How do you calculate the change in entropy (ΔS) using Boltzmann's relation when the number of microstates increases?

STEP-BY-STEP ANSWER:

Step 1: Identify the number of microstates, W, before and after a process.
Step 2: Use the Boltzmann formula S = k ln(W) to determine the initial and final entropy.
Step 3: Compute the entropy change ΔS as ΔS = k ln(W_final) - k ln(W_initial).
Final Answer: ΔS = k ln(W_final/W_initial), representing the increase or decrease in system disorder.

Boltzmann’s Entropy Calculation

QUESTION

Given the enthalpy and entropy changes for a process, how can you determine if the process is spontaneous at a given temperature?

STEP-BY-STEP ANSWER:

Step 1: Write down the Gibbs energy equation: ΔG = ΔH - TΔS.
Step 2: Insert the known values of enthalpy change (ΔH) and entropy change (ΔS) into the equation.
Step 3: Multiply the absolute temperature (T) with ΔS.
Step 4: Subtract TΔS from ΔH to compute ΔG.
Step 5: Conclude that if ΔG is negative, the process is spontaneous; if positive, it is non-spontaneous.
Final Answer: The sign of ΔG based on ΔG = ΔH - TΔS determines the spontaneity of the process.

Determining Spontaneity via Gibbs Energy

QUESTION

How do colligative properties explain the practical use of antifreeze in automotive applications?

STEP-BY-STEP ANSWER:

Step 1: Recognize that colligative properties depend on the number of solute particles in a solvent.
Step 2: Understand that adding antifreeze increases the number of dissolved particles, lowering the freezing point of the solution.
Step 3: Apply this principle to explain that the lowered freezing point prevents the coolant from freezing in cold temperatures.
Final Answer: The addition of solute (antifreeze) leads to a depression in the freezing point due to colligative properties, ensuring the coolant remains liquid in low temperatures.

Colligative Properties Application

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Common Mistakes

  • Confusing entropy with enthalpy; remembering that entropy is related to disorder while enthalpy relates to heat content.
  • Overlooking the temperature dependence of molarity due to the volume sensitivity of solutions.
  • Assuming all solutions are ideal, thereby ignoring the role of intermolecular forces in non-ideal behavior.
  • Misinterpreting colligative properties by focusing on solute identity rather than the number of solute particles.
  • Neglecting the importance of the negative sign in ?G when evaluating whether a process is spontaneous.