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 plot of k versus 1/T has a slope of -7,445. What is E a, for this reaction?A, 90.79 kJ/molB. 895.5 J/molC. 304.7 kJ/molD. 61.90 kJ/molE. 610.5 J/mol

Solution: A plot of k versus 1/T has a slope of -7,445. What is E a, for this reaction?A, 90.79 kJ/molB. 895.5 J/molC. 304.7 kJ/molD. 61.90 kJ/molE. 610.5 J/mol

Problem

A plot of k versus 1/T has a slope of -7,445. What is E a, for this reaction?

A, 90.79 kJ/mol

B. 895.5 J/mol

C. 304.7 kJ/mol

D. 61.90 kJ/mol

E. 610.5 J/mol

Solution

We’re being asked to determine the activation energy (Ea) of a reaction given that the plot of k vs. 1/T has a slope of –7445.


We’re given the plot of k (y) vs. 1/T (x)

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


ln k=-EaR 1T + ln A


where: 

k = rate constant

Ea = activation energy (in J/mol)

R = gas constant (8.314 J/mol • K)

T = temperature (in K)

A = Arrhenius constant or frequency factor


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