AP Chemistry : Thermochemistry and Kinetics

Study concepts, example questions & explanations for AP Chemistry

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Example Questions

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?

Possible Answers:

180 minutes

145 minutes

120 minutes

198 minutes

65 minutes

Correct answer:

145 minutes

Explanation:

 

 

Example Question #3 : 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?

Possible Answers:

0.0314 seconds

0.132 seconds

0.0614 seconds

0.0582 seconds

0.0729 second

Correct answer:

0.0729 second

Explanation:

Example Question #4 : Integrated Rate Laws

The half life of a first order reaction is 1.5 hours.  What is the rate constant of this reaction?

Possible Answers:

1.5

0.52

0.21

0.75

0.46

Correct answer:

0.46

Explanation:

 

Example Question #1 : Reaction Mechanisms

Which of the following is true?

Possible Answers:

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

In a reaction mechanism an intermediate is identical to an activated complex

If we know that a reaction is an elementary reaction, then we know its rate law.

Correct answer:

All of the above

Explanation:

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 

02

The rate law for the formation of NOBr based on this mechanism is rate =           .

Possible Answers:

Correct answer:

Explanation:

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:

 

  and:  

 

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)

Possible Answers:

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 1 is limiting because the NO2 is a reactant.

Step 2 is limiting because the fast step determines how quickly the reaction can occur.

Correct answer:

Step 1 is limiting because the reaction can not go faster than its slowest step.

Explanation:

The reaction can never go faster than its slowest step. 

Example Question #4 : Reaction Mechanisms

04

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?

Possible Answers:

Exothermic

Endothermic

Correct answer:

Endothermic

Explanation:

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)

Possible Answers:

C

B

R

A

There are no intermediates

Correct answer:

R

Explanation:

R is the intermediate.  It is formed in Step 1 and consumed in Step 2.

Example Question #1 : Hess' Law

Using the following data to calculate the enthalpy change in:  3 NO2(g) + H2O (l) ->  2 HNO3 (aq) +  NO (g)

04q

Possible Answers:

 68.5 kJ/mol

-68.5 kJ/mol

264 kJ/mol

137 kJ/mol

-137 kJ/mol

Correct answer:

-137 kJ/mol

Explanation:

04a

Example Question #2 : Hess' Law

Using the following data to calculate the enthalpy change in:  2 CH4 (g) + 3 O2-> 2 CO (g) + 4 H2O (l)

05q

Possible Answers:

214.2 kJ/mol

-1214 kJ/mol

-418.4 kJ/mol

-214.2 kJ/mol

418.4 kJ/mol

Correct answer:

-1214 kJ/mol

Explanation:

05a

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