Ch.13 - Chemical KineticsWorksheetSee 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: For each of the following cases, identify the order with respect to the reactant, A. Case (A → products) Orderi) The half-life of A is independent of the initial concentration of [A].   ii) A two fold

Solution: For each of the following cases, identify the order with respect to the reactant, A. Case (A → products) Orderi) The half-life of A is independent of the initial concentration of [A].   ii) A two fold

Problem

For each of the following cases, identify the order with respect to the reactant, A. 

Case (A → products) Order

i) The half-life of A is independent of the initial concentration of [A].   

ii) A two fold increase in the initial concentration of A leads to a four fold increase in the initial rate.   

iii) A two fold increase in the initial concentration of A leads to a 1.41-fold increase in the initial rate.  

iv) The time required for [A] to decrease from [A]0 to [A]0/2 is equal to the time required for [A] to decrease from [A]0/2 to [A]0/4

v) The rate of decrease of [A] is a constant.

Solution

i) t1/2=ln2/k  This is the only half life equation which is not dependent on the initial concentration of a reactant, and it corresponds to 1st order reaction.

ii) Rate = k[A]x where x is the rate of [A] and overall reaction.
 If we increase [A] by 2, the only way for rate to increase by 4 would be if rate or x= 2.

Before: Rate = k[1]2
After: (Rate)4 = k[2]2= k[4]

So this would be 2nd order reaction

iii) Rate = k[A]x

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