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: A sample of pure carbon dioxide at 500 K undergoes spontaneous decomposition to produce the following reaction:2 CO2 (g) ⇌ 2 CO (g)  +  O2 (g)The equilibrium constant KP is 215. Through the reaction it is shown that the partial pressure of O2 is 0.30 at equilibrium. Calculate the partial pressures of CO and CO2 

Solution: A sample of pure carbon dioxide at 500 K undergoes spontaneous decomposition to produce the following reaction:2 CO2 (g) ⇌ 2 CO (g)  +  O2 (g)The equilibrium constant KP is 215. Through the reaction i

Problem

A sample of pure carbon dioxide at 500 K undergoes spontaneous decomposition to produce the following reaction:

2 CO2 (g) ⇌ 2 CO (g)  +  O2 (g)

The equilibrium constant KP is 215. Through the reaction it is shown that the partial pressure of O2 is 0.30 at equilibrium. Calculate the partial pressures of CO and CO2