AP Chemistry : Kinetics and Energy
Study concepts, example questions & explanations for AP Chemistry
All AP Chemistry Resources
Example Questions
Example Question #1 : Integrated Rate Laws
The rate constant for a second-order reaction is 0.15 M-1s-1. If the initial concentration of the reactant is 0.30 M, how long does it take for the concentration to decrease to 0.15 M?
11.1 seconds
44.4 second
253 second
88.8 seconds
22.2 seconds
22.2 seconds
Example Question #1 : Integrated Rate Laws
Based on the figure above, what is the order of reaction?
Zero Order
Squared Order
Second Order
Third Order
First order
First order
For a first order reaction, the ln [A]t is linear with t.
Example Question #2 : Integrated Rate Laws
A compound decomposes by a first-order process. If 25.0% of the compound decomposes in 60 minutes, the half-life of the compound is?
145 minutes
65 minutes
198 minutes
120 minutes
180 minutes
145 minutes
Example Question #1 : Integrated Rate Laws
Cyclopentane is unstable and decomposes by a first order reaction. The rate constant for this reaction is 9.5 s-1. What is the half life of the reaction?
0.0729 second
0.0582 seconds
0.0614 seconds
0.0314 seconds
0.132 seconds
0.0729 second
Example Question #1 : Integrated Rate Laws
The half life of a first order reaction is 1.5 hours. What is the rate constant of this reaction?
0.21
1.5
0.46
0.52
0.75
0.46
Example Question #1 : Reaction Mechanisms
Which of the following is true?
In a reaction mechanism an intermediate is identical to an activated complex
All of the above
If we know that a reaction is an elementary reaction, then we know its rate law.
Since intermediate compounds can be formed, the chemical equations for the elementary reaction in a multistep mechanism do not always have to add to give the chemical equation of the overall process.
The rate-determining step of a reaction is the rate of the slowest elementary step of its mechanism
All of the above
All of the above describe elementary reactions and how they give an overall mechanism.
Example Question #2 : Reaction Mechanisms
A possible mechanism for the overall reaction Br2 (g) + 2 NO (g) -> 2 NOBr(g) is
The rate law for the formation of NOBr based on this mechanism is rate = .
Based on the slowest step the rate law would be: Rate = k2 [NOBr2] [NO], but one cannot have a rate law in terms of an intermediate (NOBr2).
Because the first reaction is at equilibrium the rate in the forward direction is equal to that in the reverse, thus:
Substitution yields:
Example Question #1 : Reaction Mechanisms
For the reaction NO2 (g) + CO (g) -> NO (g) + CO2 (g), the reaction was experimentally determined to be Rate = k[NO2]2. If the reaction has the following mechanism, what is the rate limiting step, and why?
Step 1: 2 NO2 -> NO3 + NO (slow)
Step 2: NO3 + CO -> NO2 + CO2 (fast)
Step 1 is limiting because the NO2 is a reactant.
Step 1 is limiting because the reaction can not go faster than its slowest step.
Step 2 is limiting because the NO3 intermediate has to be formed before the reaction can occur.
Step 2 is limiting because the fast step determines how quickly the reaction can occur.
Not enough information
Step 1 is limiting because the reaction can not go faster than its slowest step.
The reaction can never go faster than its slowest step.
Example Question #4 : Reaction Mechanisms
Based on the figure above, what arrows corresponds to the activation energy of the rate limiting step and the energy of reaction? Is the reaction endo- or exothermic?
Exothermic
Endothermic
Endothermic
Since the products are higher in energy than the reactions, the reaction is endothermic.
Example Question #5 : Reaction Mechanisms
Consider the following mechanism:
A + B -> R + C (slow)
A + R -> C (fast)
There are no intermediates
C
R
A
B
R
R is the intermediate. It is formed in Step 1 and consumed in Step 2.
All AP Chemistry Resources
