Ch.14 - Chemical EquilibriumWorksheetSee 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: For the reaction2H2O(g) ⇌ 2H2(g) + O2(g)K = 2.4 x 10 -3 at a given temperature. At equilibrium in a 2.0-L container it is found that [H 2O(g)] = 1.1 x 10 -1 M and [H2(g)] = 1.9 x 10 -2 M. Calculate the moles of O2(g) present under these conditions.

Solution: For the reaction2H2O(g) ⇌ 2H2(g) + O2(g)K = 2.4 x 10 -3 at a given temperature. At equilibrium in a 2.0-L container it is found that [H 2O(g)] = 1.1 x 10 -1 M and [H2(g)] = 1.9 x 10 -2 M. Calculate th

Problem

For the reaction

2H2O(g) ⇌ 2H2(g) + O2(g)

K = 2.4 x 10 -3 at a given temperature. At equilibrium in a 2.0-L container it is found that [H 2O(g)] = 1.1 x 10 -1 M and [H2(g)] = 1.9 x 10 -2 M. Calculate the moles of O2(g) present under these conditions.