Ch.13 - Chemical KineticsWorksheetSee 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 mechanism for the oxidation of HBr by O2 to form 2 H2 O and Br2 4HBr(g)+O2(g)2H2O(g)+2Br2(g) is shown below. HBr(g)+O2(g)HOOBr(g) HOOBr(g)+HBr(g)2HOBr(g) HOBr(g)+HBr(g)H2O(g)+Br

Problem

The mechanism for the oxidation of HBr by O2 to form 2 H2 O and Br2
4HBr(g)+O2(g)2H2O(g)+2Br2(g)
is shown below.
HBr(g)+O2(g)HOOBr(g)
HOOBr(g)+HBr(g)2HOBr(g)
HOBr(g)+HBr(g)H2O(g)+Br2(g).

HBr does not react with O2 at a measurable rate at room temperature under ordinary conditions. What can you infer from this about the magnitude of the activation energy for the rate-determining step?