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 by a factor of 1 million, how much lower must the activation energy barrier be when sucrose is involved?

Approximately 1 kcal/mol lower

Approximately 20 kcal/mol lower

Approximately 10 kcal/mol lower

Approximately 5.7 kcal/mol lower

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Verified step by step guidance

Understand the relationship between the rate of a reaction and the activation energy. The rate of a reaction is exponentially related to the activation energy through 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.

Recognize that the enzyme increases the rate of the reaction by a factor of 1 million. This means the rate constant \( k \) with the enzyme is 1 million times greater than the rate constant \( k_0 \) without the enzyme.

Set up the equation relating the rate constants with and without the enzyme: \( \frac{k}{k_0} = e^{-\frac{E_a - E_{a0}}{RT}} = 1,000,000 \), where \( E_a \) is the activation energy with the enzyme and \( E_{a0} \) is the activation energy without the enzyme.

Take the natural logarithm of both sides to solve for the change in activation energy: \( \ln(1,000,000) = -\frac{E_a - E_{a0}}{RT} \).

Rearrange the equation to find the difference in activation energy: \( E_{a0} - E_a = RT \ln(1,000,000) \). Use the value of \( R \) (approximately 1.987 cal/mol·K) and assume a typical temperature (e.g., 298 K) to calculate the approximate change in activation energy.