AP Chemistry : Kinetics and Energy
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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?
253 second
44.4 second
88.8 seconds
22.2 seconds
11.1 seconds
22.2 seconds
Example Question #1 : Integrated Rate Laws
Based on the figure above, what is the order of reaction?
Third Order
First order
Squared Order
Zero Order
Second Order
First order
For a first order reaction, the ln [A]t is linear with t.
Example Question #1 : 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?
198 minutes
120 minutes
180 minutes
65 minutes
145 minutes
145 minutes
Example Question #4 : 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?
0.132 seconds
0.0729 second
0.0582 seconds
0.0314 seconds
0.0614 seconds
0.0729 second
Example Question #5 : Integrated Rate Laws
The half life of a first order reaction is 1.5 hours. What is the rate constant of this reaction?
1.5
0.46
0.21
0.52
0.75
0.46
Example Question #111 : Thermochemistry And Kinetics
Which of the following is true?
All of the above
If we know that a reaction is an elementary reaction, then we know its rate law.
The rate-determining step of a reaction is the rate of the slowest elementary step of its mechanism
In a reaction mechanism an intermediate is identical to an activated complex
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.
All of the above
All of the above describe elementary reactions and how they give an overall mechanism.
Example Question #112 : Thermochemistry And Kinetics
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 #113 : Thermochemistry And Kinetics
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 2 is limiting because the NO3 intermediate has to be formed before the reaction can occur.
Step 1 is limiting because the reaction can not go faster than its slowest step.
Not enough information
Step 2 is limiting because the fast step determines how quickly the reaction can occur.
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 #114 : Thermochemistry And Kinetics
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?
Endothermic
Exothermic
Endothermic
Since the products are higher in energy than the reactions, the reaction is endothermic.
Example Question #115 : Thermochemistry And Kinetics
Consider the following mechanism:
A + B -> R + C (slow)
A + R -> C (fast)
C
B
A
There are no intermediates
R
R
R is the intermediate. It is formed in Step 1 and consumed in Step 2.
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