Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

15. Chemical Kinetics

Energy Diagrams

General Chemistry

15. Chemical Kinetics

Energy Diagrams

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Multiple Choice

If an enzyme increases the rate of the hydrolysis reaction of sucrose into glucose and fructose by a factor of 7 million, how much lower must the activation barrier be?

Approximately 50 kJ mol⁻¹ lower

Approximately 70 kJ mol⁻¹ lower

Approximately 10 kJ mol⁻¹ lower

Approximately 30 kJ mol⁻¹ lower

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Verified step by step guidance

Understand that the rate of a reaction is related to the activation energy by the Arrhenius equation: \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

Recognize that the enzyme increases the reaction rate by a factor of 7 million, which means the new rate constant \( k' = 7,000,000 \times k \).

Set up the equation for the rate constants before and after the enzyme is added: \( k' = A e^{-\frac{E_a'}{RT}} \) and \( k = A e^{-\frac{E_a}{RT}} \).

Divide the equation for \( k' \) by the equation for \( k \) to eliminate \( A \): \( \frac{k'}{k} = e^{-\frac{E_a'}{RT}} / e^{-\frac{E_a}{RT}} = e^{-\frac{E_a' - E_a}{RT}} \).

Solve for the change in activation energy \( \Delta E_a = E_a - E_a' \) using the natural logarithm: \( \ln(7,000,000) = \frac{\Delta E_a}{RT} \). Rearrange to find \( \Delta E_a = RT \ln(7,000,000) \).