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: When a reaction is run at 150.0°C, data shows that k = 28.68 s  -1 and Ea = 36.25 kJ/mol. What temperature would the reaction have to be run in order for k to be 1521 s-1? a. 234.8 °Cb. 274.1 °Cc. 414

Solution: When a reaction is run at 150.0°C, data shows that k = 28.68 s  -1 and Ea = 36.25 kJ/mol. What temperature would the reaction have to be run in order for k to be 1521 s-1? a. 234.8 °Cb. 274.1 °Cc. 414

Problem

When a reaction is run at 150.0°C, data shows that k = 28.68 s  -1 and Ea = 36.25 kJ/mol. What temperature would the reaction have to be run in order for k to be 1521 s-1

a. 234.8 °C

b. 274.1 °C

c. 414.9 °C

d. 173.7 °C

e. -1.11 °C

Solution

We’re being asked to determine the temperature at which the rate constant of a reaction is 1521 s–1


We’re given the rate constant at another temperature and the activation energy of the reaction. This means we need to use the two-point form of the Arrhenius Equation:



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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