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

Integrated Rate Law

General Chemistry

15. Chemical Kinetics

Integrated Rate Law

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Multiple Choice

The decomposition of HI to H2 and I2 is a second-order reaction with a rate constant of 3.8 × 10⁻⁵ M⁻¹·s⁻¹ at a certain temperature. If the initial concentration of HI is 0.401 M, calculate the amount of time (in days) it will take to consume 70.8% of HI.

2.3 days

4.5 days

1.5 days

3.7 days

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Verified step by step guidance

Identify the type of reaction: The problem states that the decomposition of HI is a second-order reaction. For a second-order reaction, the rate law is given by \( \text{Rate} = k[A]^2 \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

Use the integrated rate law for a second-order reaction: The integrated rate law for a second-order reaction is \( \frac{1}{[A]} = kt + \frac{1}{[A_0]} \), where \( [A] \) is the concentration at time \( t \), \( [A_0] \) is the initial concentration, and \( k \) is the rate constant.

Determine the final concentration of HI: Since 70.8% of HI is consumed, 29.2% remains. Calculate the final concentration \( [HI] = 0.292 \times [HI_0] = 0.292 \times 0.401 \text{ M} \).

Substitute the known values into the integrated rate law: Plug in \( [A_0] = 0.401 \text{ M} \), \( [A] = 0.292 \times 0.401 \text{ M} \), and \( k = 3.8 \times 10^{-5} \text{ M}^{-1}\cdot\text{s}^{-1} \) into the equation \( \frac{1}{[A]} = kt + \frac{1}{[A_0]} \).

Solve for \( t \) and convert to days: Rearrange the equation to solve for \( t \) in seconds, then convert the time from seconds to days by dividing by the number of seconds in a day (86400 seconds/day).