Ch.6 - Thermochemistry WorksheetSee 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: Find ΔH° for BaCO3 (s) → BaO (s) + CO2 (g) given that 2Ba (s) + O2 (g) → 2BaO (s)     ΔH° = -1107.0 kJBa (s) + CO2 (g) + 1/2O2 (g)→ BaCO3 (s)     ΔH° = - 822.5 kJ a. -1929.5 kJ b. -1376.0 kJ c. -284.5

Solution: Find ΔH° for BaCO3 (s) → BaO (s) + CO2 (g) given that 2Ba (s) + O2 (g) → 2BaO (s)     ΔH° = -1107.0 kJBa (s) + CO2 (g) + 1/2O2 (g)→ BaCO3 (s)     ΔH° = - 822.5 kJ a. -1929.5 kJ b. -1376.0 kJ c. -284.5

Problem

Find ΔH° for BaCO3 (s) → BaO (s) + CO2 (g) given that 

2Ba (s) + O2 (g) → 2BaO (s)     ΔH° = -1107.0 kJ

Ba (s) + CO2 (g) + 1/2O2 (g)→ BaCO3 (s)     ΔH° = - 822.5 kJ 

a. -1929.5 kJ 

b. -1376.0 kJ 

c. -284.5 kJ 

d. 269.0 kJ 

e. 537 kJ