AP Chemistry Unit 5: Kinetics

Explore AP Chemistry Unit 5 questions on reaction rates, rate laws, integrated kinetics, collision theory, reaction mechanisms, catalysts, and activation-energy profiles.

Covered syllabus points

5.1 Reaction Rates 5.2 Introduction to Rate Law 5.3 Concentration Changes Over Time 5.4 Elementary Reactions 5.5 Collision Model 5.6 Reaction Energy Profile 5.7 Introduction to Reaction Mechanisms 5.8 Reaction Mechanism and Rate Law 5.9 Pre-Equilibrium Approximation 5.10 Multistep Reaction Energy Profile 5.11 Catalysis

[Maximum number: 2]

At elevated temperatures, $\mathrm{NO}_{2}$ undergoes decomposition in the gas phase, forming NO and $\mathrm{O}_{2}$ as represented by the following equation.

2 \mathrm{NO}_{2} \rightarrow 2 \mathrm{NO}+\mathrm{O}_{2}

A scientist measures the change in $\left[\mathrm{NO}_{2}\right]$ over the first 100. s of the reaction at 546°C. The scientist uses the data collected from the experiment to generate the following two graphs.

Figure for Question 5 Kinetics question 1 — AP Chemistry

Based on these data, the scientist makes the claim that the rate law for the reaction is rate $=k\left[\mathrm{NO}_{2}\right]^{2}$ .

(a)

Explain how the graphs indicate that the reaction is second order with respect to $\mathrm{NO}_{2}$ .

[ 1 ]

For a correct explanation: 1 point
The plot of $\frac{1}{\left[\mathrm{NO}_{2}\right]}$ versus time is the most linear, indicating that the reaction is second order with respect to $\mathrm{NO}_{2}$ .

(b)

At a certain point in the reaction, the rate of disappearance of $\mathrm{NO}_{2}$ is determined to be $6.52 \times 10^{-7} M / \mathrm{s}$ . Determine the rate of appearance, in M / s, of $\mathrm{O}_{2}$ at this same point in the reaction.

[ 1 ]

For the correct calculated value: 1 point

6.52 \times 10^{-7} M / \mathrm{s} \times \frac{1 \mathrm{~mol} \mathrm{O}_{2}}{2 \mathrm{~mol} \mathrm{NO}_{2}}=3.26 \times 10^{-7} M / \mathrm{s}

[Maximum number: 3]

Answer the following questions about ascorbic acid (vitamin C).

(a)

Dehydroascorbic acid (DHAsc) can be produced by reacting ascorbic acid with the triiodide

ion, $\mathrm{I}_{3}{ }^{-}$ , as represented by the following equation.

\mathrm{HAsc}+\mathrm{I}_{3}^{-} \rightarrow \mathrm{DHAsc}+3 \mathrm{I}^{-}+2 \mathrm{H}^{+}

The student runs three trials of the reaction with different initial concentrations of HAsc and

$\mathrm{I}_{3}{ }^{-}$ , producing the following data.

[ 3 ]

(i)

The rate law for the reaction is rate $=k[\mathrm{HAsc}]\left[\mathrm{I}_{3}{ }^{-}\right]$ . Explain how the data in the table

support the conclusion that the reaction is first order with respect to [HAsc].

[ 1 ]

C (i) For a correct explanation.

Examples of acceptable responses may include the following:

- $\frac{4.914 \times 10^{-4}}{2.457 \times 10^{-4}}=\left(\frac{0.900}{0.450}\right)^{a}$ , thus a=1

- Comparing trials 1 and 3, the rate doubles when the concentration of ascorbic acid is

doubled and the triiodide ion concentration is constant, indicating that the process is

first order with respect to ascorbic acid.

(ii)

Calculate the value of the rate constant, k, for the reaction. Include units with your

answer.

[ 2 ]

(ii) For the correct calculated value:

rate $=k[\mathrm{HAsc}]\left[\mathrm{I}_{3}{ }^{-}\right]$

Using trial 1 data:

$k=\frac{\mathrm{rate}}{[\mathrm{HAsc}]\left[\mathrm{I}_{3}^{-}\right]}=\frac{2.457 \times 10^{-4} \mathrm{M} / \mathrm{s}}{(0.450 \mathrm{M})(1.200 \mathrm{M})}=4.55 \times 10^{-4} \mathrm{M}^{-1} \mathrm{~s}^{-1}$

For the correct units:

$M^{-1} \mathrm{~s}^{-1}$

Point 07

Point 08

Point 09

[Maximum number: 1]

The rate law may be written using stoichiometric coefficients for which of the following?

Precipitation reactions

Acid-base reactions

Elementary processes

Solubility reactions

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Covered syllabus points

5 Kinetics question 1

Question (a)

Question (b)

5 Kinetics question 2

Question (a)

Question (i)

Question (ii)

5 Kinetics question 3