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: Food rots about 50 times more rapidly at 25 °C than when it is stored at 6 °C. Determine the overall activation energy for the processes responsible for its decomposition.A. 30 kJ/molB. 142 kJ/molC. -

Problem

Food rots about 50 times more rapidly at 25 °C than when it is stored at 6 °C. Determine the overall activation energy for the processes responsible for its decomposition.

A. 30 kJ/mol
B. 142 kJ/mol
C. -30 kJ/mol
D.  -142 kJ/mol
E. not enough information given

Solution

We’re being asked to determine the activation energy (Ea) for the process responsible when food rots about 50 times more rapidly at 25 °C than when it is stored at 6 °C. 


This means we need to use the two-point form of the Arrhenius Equation:


ln k2k1=-EaR[1T2-1T1]


where k1 = rate constant at T1

k2 = rate constant at T2

Ea = activation energy (in J/mol)

R = gas constant (8.314 J/mol•K)

T1 and T2 = temperature (in K)


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