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: When N2O5(g) is heated it dissociates into N2O3(g) and O2(g) according to the following reaction:N2O5(g) ⇌ N2O3(g) + O2(g); Kc = 7.75 at a given temperatureThe N2O3(g) dissociates to give N2O(g) and O

Problem

When N2O5(g) is heated it dissociates into N2O3(g) and O2(g) according to the following reaction:

N2O5(g) ⇌ N2O3(g) + O2(g); Kc = 7.75 at a given temperature

The N2O3(g) dissociates to give N2O(g) and O2(g) according the following reaction:

N2O3(g) ⇌ N2O(g) + O2(g); Kc = 4.00 at the same temperature

When 4.00 mol of N2O5(g) is heated in a 1.00-L reaction vessel to this temperature, the concentration of O2(g) at equilibrium is 4.50 mol/L. Find the concentration of N2O5 in the equilibrium system.