Ch.14 - Chemical EquilibriumWorksheetSee all chapters
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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 reversible reactionA(g) ⇌ B(g) Which K value would indicate that there is more B than A at equilibrium?A. K = 9x10-9B. K = 0.2C. K = 9000D. K = 9x109


For the reversible reaction

A(g) ⇌ B(g) 

Which K value would indicate that there is more B than A at equilibrium?

A. K = 9x10-9

B. K = 0.2

C. K = 9000

D. K = 9x109


The equilibrium constant expression is written as:


$K = \frac{[B]}{[A]}$

For this problem, the expression should look as follows:

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