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Absolute Entropy and Entropy Change

Sophie Anderson

Sophie Anderson

7 min read

Next Topic - Gibbs Free Energy and Thermodynamic Favorability

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Study Guide Overview

This study guide covers entropy and spontaneity, focusing on absolute entropy (S°) and change in entropy (ΔS°). It explains how to calculate ΔS° using standard entropy values and predict its sign based on phase changes, number of moles, and molecular complexity. The guide also includes practice problems and emphasizes the concept of state functions. Key exam topics and tips for predicting and calculating entropy changes are provided.

Thermodynamics: Entropy and Spontaneity

Hey there, future AP Chem master! Let's dive into entropy, a concept that's all about disorder and spontaneity. This guide will help you nail those tricky questions on the exam. Let's get started!

Entropy (S) - The Measure of Disorder

Absolute Entropy (S°) vs. Change in Entropy (ΔS°)

  • Entropy (S) is a measure of the disorder or randomness of a system. Think of it as how many ways a system can be arranged.
Key Concept

Absolute Entropy (S°): This is the entropy of a substance at standard conditions (1 atm and 273 K). It's a measure of the number of possible states a molecule can take. You'll usually find these values in tables (like the one below) and you'll be given these values on the exam.

- **Change in Entropy (ΔS°)**: This is the change in disorder during a reaction. It's the difference between the entropy of the products and the reactants.

Standard Entropies


Caption: A table showing standard entropies (S°) for various substances. Note that these values are always positive, unlike ΔH° values.


State Functions: Pathway Independence

  • A state function is a property that depends only on the current state of the system, not on how it got there. It's pathway-independent.
  • Examples of state functions: Altitude change when climbing a mountain, enthalpy (H), and, you guessed it, entropy (S).
  • Non-state function example: Distance traveled. The distance you travel depends on the path you take.

State Function


Caption: Illustrating a state function. The change in altitude is the same regardless of the path taken.


Calculating ΔS°

  • Since entropy is a state function, we can calculate ΔS° using the following equation, which is very similar to how we calculated ΔH°:

ΔS°_rxn=ΣnS°_products−ΣnS°_reactantsΔS°\_{rxn} = ΣnS°\_{products} - ΣnS°\_{reactants}ΔS°_rxn=ΣnS°_products−ΣnS°_reactants

Where: - n = stoichiometric coefficient - S° = standard molar entropy

Exam Tip

Remember those coefficients! Don't forget to multiply the standard entropy of each substance by its stoichiometric coefficient from the balanced chemical equation.

Interpreting the Sign of ΔS°

  • ΔS° > 0 (Positive): The reaction increases disorder. The products are more disordered than the reactants. Think of it as things becoming more chaotic. 🤪
  • ΔS° < 0 (Negative): The reaction decreases disorder. The products are more ordered than the reactants. Things are becoming more organized. 🤓

Predicting the Sign of ΔS°

  • Phase Changes: Gases have higher entropy than liquids, and liquids have higher entropy than solids. (S_solid < S_liquid < S_gas)
  • Number of Moles: Reactions that increase the number of moles of gas usually have a positive ΔS°.
  • Complexity: Larger, more complex molecules tend to have higher entropy than smaller, simpler ones.

Practice Question

Practice Problems

1. Calculating ΔS°

Given the reaction:

KI(aq)+KNO3(aq)→KNO3(aq)+KI(aq)KI(aq) + KNO_3(aq) \rightarrow KNO_3(aq) + KI(aq)KI(aq)+KNO3​(aq)→KNO3​(aq)+KI(aq)

and the following standard molar entropies:

  • S°(KNO3(aq)) = 175 J/mol·K
  • S°(KI(aq)) = 150 J/mol·K
  • S°(KNO3(s)) = 250 J/mol·K
  • S°(KI(s)) = 125 J/mol·K

Calculate ΔS° for the reaction.

Solution:

ΔS°rxn=[1∗S°(KNO3(aq))+2∗S°(KI(aq))]−[1∗S°(KNO3(s))+2∗S°(KI(s))]ΔS°_{rxn} = [1 * S°(KNO_3(aq)) + 2 * S°(KI(aq))] - [1 * S°(KNO_3(s)) + 2 * S°(KI(s))]ΔS°rxn​=[1∗S°(KNO3​(aq))+2∗S°(KI(aq))]−[1∗S°(KNO3​(s))+2∗S°(KI(s))] ΔS°rxn=[1∗175+2∗150]−[1∗250+2∗125]=−25J/mol⋅KΔS°_{rxn} = [1 * 175 + 2 * 150] - [1 * 250 + 2 * 125] = -25 J/mol·KΔS°rxn​=[1∗175+2∗150]−[1∗250+2∗125]=−25J/mol⋅K

Answer: A. -25 J/mol·K

2. Predicting the Sign of ΔS°

Consider the reaction:

2Na(s)+Cl2(g)→2NaCl(s)2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)2Na(s)+Cl2​(g)→2NaCl(s)

Predict the sign of ΔS° for this reaction and explain.

Solution:

  • Reactants: 2 moles of solid and 1 mole of gas (3 moles total, including a gas).
  • Products: 2 moles of solid.
  • The reaction goes from a more disordered (gas present) state to a more ordered (only solid) state. Therefore, ΔS° is negative.

Answer: ΔS° is negative because the reaction decreases disorder.

Multiple Choice Questions

  1. Which of the following processes is expected to have a positive change in entropy (ΔS > 0)? (A) Freezing of water (B) Condensation of steam (C) Dissolving sugar in water (D) Formation of a crystal from a supersaturated solution

  2. For the reaction: N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)N2​(g)+3H2​(g)→2NH3​(g), the change in entropy (ΔS) is expected to be: (A) Positive because the number of moles of gas increases. (B) Negative because the number of moles of gas decreases. (C) Zero because the number of moles of gas remains the same. (D) Impossible to determine without more information.

Free Response Question

Consider the following reaction:

C2H4(g)+H2(g)→C2H6(g)C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)C2​H4​(g)+H2​(g)→C2​H6​(g)

(a) Predict the sign of ΔS° for the reaction. Justify your answer. (b) Given the following standard molar entropies:

  • S°(C2H4(g)) = 219.3 J/mol·K
  • S°(H2(g)) = 130.7 J/mol·K
  • S°(C2H6(g)) = 229.2 J/mol·K

Calculate ΔS° for the reaction. (c) If the reaction is exothermic, what does this indicate about the spontaneity of the reaction at low temperatures?

Scoring Breakdown:

(a) (2 points) - 1 point for predicting ΔS° is negative - 1 point for justification: The number of moles of gas decreases from 2 to 1, leading to a decrease in disorder.

(b) (2 points) - 1 point for correct setup: ΔS° = S°(C2H6) - [S°(C2H4) + S°(H2)] - 1 point for correct calculation: ΔS° = 229.2 - (219.3 + 130.7) = -120.8 J/mol·K

(c) (2 points) - 1 point for indicating that an exothermic reaction is favorable at low temperatures - 1 point for connecting to spontaneity: Since ΔH is negative and ΔS is negative, the reaction is spontaneous at low temperatures where the -TΔS term is small.


Final Exam Focus

  • Key Topics: Absolute vs. change in entropy, state functions, calculating ΔS°, predicting the sign of ΔS° based on phase changes and number of moles.
  • Common Question Types: MCQs on predicting the sign of ΔS°, FRQs involving calculation of ΔS° and connecting it to spontaneity.
  • Time Management: Quickly identify the number of moles of gas and phase changes to predict ΔS° signs. Use standard entropy values from the table to calculate ΔS° accurately.
  • Common Mistakes: Forgetting stoichiometric coefficients, mixing up signs, not understanding state function concept.

Memory Aid

Memory Aid: Remember "GAS" for predicting entropy changes: Gas > Aqueous > Solid. More moles of gas = higher entropy.


Exam Tip

Exam Tip: Always double-check your calculations and units. Pay close attention to the state symbols (s, l, g, aq) when predicting entropy changes.


You've got this! Keep practicing, and you'll ace the AP Chemistry exam. Good luck! 🚀

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Question 1 of 11

What does entropy primarily measure in a system? 🤔

The total energy of the system

The degree of disorder or randomness

The heat content of the system

The rate of a chemical reaction