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

Dinitrogen tetraoxide, N 2O 4, decomposes to nitrogen dioxide, NO 2, in a first-order process. If k = 2.5 x 10 3 s -1 at 5 oC and k = 7.0 x 10 4 s -1 at 25°C, what is the activation energy for the decomposition?

a. 214 kJ/mol

b. 361 kJ/mol

c. 323 kJ/mol

d. 115 kJ/mol

e. 151 kJ/mol

Solution

We’re being asked to determine the activation energy of the reaction when k = 2.5 x 10 3 s -1 at 5°C and k = 7.0 x 10 4 s -1 at 25°C


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).

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