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

Solution: The standard free energy of formation of CS 2 (l) is 65.27 kJ⋅mol −1 at 298 K. This means that at 298 K a) CS2 (l) will not spontaneously form C (s) + 2 S (s). b) CS2 (l) is thermodynamically unstable. c) CS2 (l) is thermodynamically stable. d) No catalyst can be found to decompose CS 2 (l) into its elements. e) CS2 (l) has a negative entropy.

Problem

The standard free energy of formation of CS (l) is 65.27 kJ⋅mol −1 at 298 K. This means that at 298 K

a) CS(l) will not spontaneously form C (s) + 2 S (s).

b) CS(l) is thermodynamically unstable.

c) CS(l) is thermodynamically stable.

d) No catalyst can be found to decompose CS (l) into its elements.

e) CS(l) has a negative entropy.