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: A first-order reaction proceeds with a rate constant of 1.5x10  -5 s-1 at 25°C and 2.7x10-3 s-1 at 75°C. According to the Arrhenius equation, what is the activation energy of this reaction in kJ/mol?A

Solution: A first-order reaction proceeds with a rate constant of 1.5x10  -5 s-1 at 25°C and 2.7x10-3 s-1 at 75°C. According to the Arrhenius equation, what is the activation energy of this reaction in kJ/mol?A

Problem

A first-order reaction proceeds with a rate constant of 1.5x10  -5 s-1 at 25°C and 2.7x10-3 s-1 at 75°C. According to the Arrhenius equation, what is the activation energy of this reaction in kJ/mol?

A. 89.5 kJ/mol

B. 1.30 kJ/mol

C. 16.2 kJ/mol

D. 95.8 kJ/mol

Solution

We’re being asked to determine the activation energy (Ea) of a reaction. We’re given the rate constants at two different temperatures.


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



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)


View the complete written solution...