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

Arrhenius Equation

General Chemistry

15. Chemical Kinetics

Arrhenius Equation

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Multiple Choice

Using the Arrhenius Equation, if the activation energy for a reaction is 80 kJ/mol, at what temperature will the rate be ten times faster than at 0 degrees Celsius?

Approximately 350 K

Approximately 298 K

Approximately 400 K

Approximately 500 K

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Verified step by step guidance

Start by understanding 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 (8.314 J/mol·K), and \( T \) is the temperature in Kelvin.

Since the rate is ten times faster, the new rate constant \( k_2 \) is ten times the original rate constant \( k_1 \). Therefore, \( \frac{k_2}{k_1} = 10 \).

Use the ratio of the Arrhenius equations for the two temperatures: \( \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{R T_2}}}{A e^{-\frac{E_a}{R T_1}}} = e^{-\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)} \).

Substitute \( \frac{k_2}{k_1} = 10 \), \( E_a = 80000 \) J/mol, \( R = 8.314 \) J/mol·K, and \( T_1 = 273 \) K (0 degrees Celsius) into the equation: \( 10 = e^{-\frac{80000}{8.314} \left( \frac{1}{T_2} - \frac{1}{273} \right)} \).

Solve for \( T_2 \) by taking the natural logarithm of both sides and rearranging the equation: \( \ln(10) = -\frac{80000}{8.314} \left( \frac{1}{T_2} - \frac{1}{273} \right) \). Calculate \( T_2 \) to find the temperature at which the rate is ten times faster.