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Ch.17 - Chemical ThermodynamicsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch.17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds
Spontaneous vs Nonspontaneous Reactions
Entropy Calculations
Entropy Calculations: Phase Changes
Third Law of Thermodynamics
Gibbs Free Energy
Gibbs Free Energy Calculations
Gibbs Free Energy And Equilibrium

Calculations dealing with entropy take into account the effects of our chemical reaction and its surroundings. 

Entropy Calculations & Theory

Concept #1: The Total Entropy change of the Universe takes into account the contributions of the chemical reaction and its surroundings. 

Example #1: Calculate the total entropy change for a reaction with ∆Ssurr = 2.7 J/K and ∆Sºrxn = - 450.0 kJ/K.

Is this reaction spontaneous?

Concept #2: The entropy of the surroundings is based on the enthalpy of your reaction and its temperature

Example #2: Determine change in entropy of the universe for the following reaction at 32ºC.

2 A (g) + 5 B (s) → AB (g) + 2 C (g) ∆Hrxn = -140 kJ , ∆Srxn = 3.6 J/K

Concept #3: Like enthalpy, the entropy of a chemical reaction can be calculated from standard molar values.

Example #3: Calculate ∆Sºrxn for the following reaction at 25ºC.

2 NO (g) + O2 (g) → 2 NO2 (g) ∆Hrxn = -114.14 kJ

Practice: For the following reaction at 27 °C, calculate ∆S°rxn, ∆Ssurr, and ∆Stot. Determine if reaction is favorable.

Fe2O3 (s) + 3 H2 (g) → 2 Fe (s) + 3 H2O (g) ∆Hrxn = 98.8 kJ