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: Consider the hypothetical reaction A 2 (g) + B2 (g) → 2AB (g), where the rate law is:The value of the rate constant at 302°C is 2.45 x 10 -4 L/mol•s, and at 508°C the rate constant is 0.891 L/mol•s. W

Solution: Consider the hypothetical reaction A 2 (g) + B2 (g) → 2AB (g), where the rate law is:The value of the rate constant at 302°C is 2.45 x 10 -4 L/mol•s, and at 508°C the rate constant is 0.891 L/mol•s. W

Problem

Consider the hypothetical reaction A 2 (g) + B2 (g) → 2AB (g), where the rate law is:

The value of the rate constant at 302°C is 2.45 x 10 -4 L/mol•s, and at 508°C the rate constant is 0.891 L/mol•s. What is the activation energy for this reaction? What is the value of the rate constant for this reaction at 375°C?

Solution

Recall: The Arrhenius equation gives us the relationship of the rate constant k with temperature and activation energy Ea. The two-point form of the Arrhenius equation is:

We're given:

k1 = 2.45 × 10–4 L/mol•s          T1 = 302 ˚C + 273.15 = 575.15 K

k2 = 0.891 L/mol•s                   T2 = 508°C + 273.15 = 781.15 K

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