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: Consider this equilibrium reaction at 400 K. Br2 (g) + Cl2 (g) ⇌ 2BrCl (g)        Kc = 7.0 If the composition of the reaction mixture at 400 K is [BrCl] = 0.00415 M, [Br2] = 0.00366 M, and [Cl2] = 0.000672 M, what is the reaction quotient, Q? How is the reaction quotient related to the equilibrium constant, Kc, for this reaction?            (i) Q < K            (ii) Q > K           (iii) Q = K

Problem

Consider this equilibrium reaction at 400 K. 

Br2 (g) + Cl2 (g) ⇌ 2BrCl (g)        Kc = 7.0 

If the composition of the reaction mixture at 400 K is [BrCl] = 0.00415 M, [Br2] = 0.00366 M, and [Cl2] = 0.000672 M, what is the reaction quotient, Q? 

How is the reaction quotient related to the equilibrium constant, Kc, for this reaction? 

           (i) Q < K 

           (ii) Q > K

           (iii) Q = K