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: The desorption (leaving of the surface) of a single molecular layer of n-butane from a single crystal of aluminum oxide was found to be first order with a rate constant of 0.128/s at 150 K.If the surf

Solution: The desorption (leaving of the surface) of a single molecular layer of n-butane from a single crystal of aluminum oxide was found to be first order with a rate constant of 0.128/s at 150 K.If the surf

Problem

The desorption (leaving of the surface) of a single molecular layer of n-butane from a single crystal of aluminum oxide was found to be first order with a rate constant of 0.128/s at 150 K.

If the surface is initially completely covered with n-butane at 150 K, how long will it take for 25% of the molecules to desorb (leave the surface)?

Solution

We’re being asked to determine the time it would take for 25% of the molecules to desorb (leave the surface)

The integrated rate law for a first order reaction is as follows:


ln At = -kt + ln Ao


where:

[A]t = concentration at time t, 

[A]0 = initial

k = rate constant, 

t = time


Since no concentration is given, we can use the given percentages in place of concentration.

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