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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

Gibbs Free Energy and the Equilibrium Constant are the two most prominent variables used to determine spontaneity.

Gibbs Free Energy-Equilibrium Constant Calculations

Concept #1: If the equilibrium constant and temperature are known for a reaction then the standard Gibbs Free Energy can be calculated

Example #1: A certain reaction takes place at 25ºC and has an equilibrium constant of 2.8 x 104. Determine the Gibbs free energy of the reaction.

Practice: For reaction, Ag2CO3 (s) ⇌ Ag2O (s) + CO2 (g), the ∆Hº = 79.14 kJ/mol, ∆Sº = 167.2 J/K.

Determine the equilibrium constant at which the temperature is 365.1 K.

Concept #2: When standard conditions are not held then the nonstandard version of Gibbs Free Energy must be used. 

Practice: Consider a hypothetical reaction at 38 ºC, X2 (g) + 2 Y (s) ⇌ 3 Z (g), with a ∆G of −75.8 kJ.

Concentrations of reactants and products: [X2] = 1.4 M, [Y] = 0.34 M, [Z] = 2.6 M. Calculate Keq of the reaction under standard conditions.

Example #2: The given reaction has a ∆Gº of -374 kJ, and partial pressures of SF4, F2, SF6 are 0.63 atm, 0.95 atm, 1.7 atm respectively. Calculate the ∆Grxn for this reaction. 

SF4 (g) + F2 (g) → SF6 (g)