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

Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 10 °C and k = 4.0 x 103 s-1 at 30 °C, what is the activation energy for the decomposition?

a. 34998 kJ/mol
b. -35.0 kJ/mol
c. 34998 kJ/mol
d. 35.0 kJ/mol
e. 14.7 kJ/mol


We’re being asked to determine the activation energy (Ea) when nitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2

We’re given the rates at two different temperatures.

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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