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: Elemental boron can be forrmed by reaction of boron trichloride with hydrogen.  BCl3(g) + 1.5H2(g) → B(s) + 3HCl(g) Substance:    BCl3(g)       H2(g)          B(s)         HCl(g) (J/K•mol):          ?            130.6         5.87      186.8   If ΔS°rxn = 80.3 J/K, what is S° for BCl3(g)? A) -18.2   J/K•mol B) 18.2    J/K•mol C) 290.1  J/K•mol D) 355.4  J/K•mol E) 450.6  J/K•mol

Problem

Elemental boron can be forrmed by reaction of boron trichloride with hydrogen. 

BCl3(g) + 1.5H2(g) → B(s) + 3HCl(g)

Substance:    BCl3(g)       H2(g)          B(s)         HCl(g)

(J/K•mol):          ?            130.6         5.87      186.8

 

If ΔS°rxn = 80.3 J/K, what is S° for BCl3(g)?

A) -18.2   J/K•mol

B) 18.2    J/K•mol

C) 290.1  J/K•mol

D) 355.4  J/K•mol

E) 450.6  J/K•mol