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
Problem

A reaction has a rate constant k = 8.54 x 10-4 s-1 at 45 °C and an activation energy, Ea = 90.8 kJ. What is the value of k at 25 °C?

A. 8.52 x 10-4 s-1 

B. 4.46 x 10-3 s-1

C. 8.54 x 10-5 s-1

D. 4.46 x 10-4 s-1

E. 8.54 x 10-3 s-1

Solution

We’re being asked to determine the rate constant at 25°C of the reaction when k = 8.54 x 10-4 s -1 at 45°C with a EA = 90.8 kJ


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


lnk2k1 = -EaR1T2-1T1


where:

k1 = rate constant at T1 

k2 = rate constant at T

Ea = activation energy (in J/mol) 

R = gas constant (8.314 J/mol•K) 

T1 and T2 = temperature (in K).

Solution BlurView Complete Written Solution