Ch.13 - Chemical KineticsSee 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: Given the following rate law, how does the rate of reaction change if the concentration of Y is doubled? A. Rate = k[X]2[Y]3 A) The rate of reaction will increase by a factor of 9. B) The rate of reaction will increase by a factor of 2. C) The rate of reaction will increase by a factor of 8. D) The rate of reaction will increase by a factor of 4. E) The rate of reaction will remain unchanged.

Problem

Given the following rate law, how does the rate of reaction change if the concentration of Y is doubled?

A. Rate = k[X]2[Y]3

A) The rate of reaction will increase by a factor of 9.

B) The rate of reaction will increase by a factor of 2.

C) The rate of reaction will increase by a factor of 8.

D) The rate of reaction will increase by a factor of 4.

E) The rate of reaction will remain unchanged.