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: At a particular temperature a 2.00-L flask at equilibrium contains 2.80 x 10 -4 mole of N2, 2.50 x 10 -5 mole of O2, and 2.00 x 10 -2 mole of N2O. Calculate K at this temperature for the reaction2N2(g

Problem

At a particular temperature a 2.00-L flask at equilibrium contains 2.80 x 10 -4 mole of N2, 2.50 x 10 -5 mole of O2, and 2.00 x 10 -2 mole of N2O. Calculate K at this temperature for the reaction

2N2(g) + O2(g) ⇌ 2N2O(g)

If [N2] = 2.00 x 10 -4 M, [N2O] = 0.200 M, and [O2] = 0.00245 M, does this represent a system at equilibrium?