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AP Chemistry 5.3 Concentration Change Over Time- Exam Style questions - FRQs- New Syllabus

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Question

At elevated temperatures, \( \mathrm{NO_2} \) undergoes decomposition in the gas phase, forming \( \mathrm{NO} \) and \( \mathrm{O_2} \), as represented by the following equation.
\( \mathrm{2\,NO_2 \rightarrow 2\,NO + O_2} \)
A scientist measures the change in \( [\mathrm{NO_2}] \) over the first \( 100.\ \mathrm{s} \) of the reaction at \( 546^\circ\mathrm{C} \). The scientist uses the data collected from the experiment to generate the following two graphs.
Based on these data, the scientist makes the claim that the rate law for the reaction is \( \mathrm{rate} = k[\mathrm{NO_2}]^2 \).
(a) Explain how the graphs indicate 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 \( 6.52 \times 10^{-7}\ \mathrm{M/s} \). Determine the rate of appearance, in \( \mathrm{M/s} \), of \( \mathrm{O_2} \) at this same point in the reaction.
(c) \( \mathrm{NO_2} \) is a molecule that contains an odd number of electrons and can be oxidized to form the \( \mathrm{NO_2^+} \) ion. In \( \mathrm{NO_2} \), the unpaired electron is presumed to be delocalized on the nitrogen atom, as shown in the Lewis diagram in the box on the left.
(i) In the box on the right, complete the Lewis diagram for \( \mathrm{NO_2^+} \). Be sure to show all bonding and nonbonding electrons.
(ii) A student makes the claim that the bond angles in \( \mathrm{NO_2} \) and \( \mathrm{NO_2^+} \) are different from each other. Do you agree or disagree with the student’s claim? Justify your answer.

Most-appropriate topic codes (AP Chemistry):

• Topic \( 5.3 \) — Concentration Changes Over Time (Part \( \mathrm{(a)} \))
• Topic \( 4.5 \) — Stoichiometry (Part \( \mathrm{(b)} \))
• Topic \( 2.5 \) — Lewis Diagrams (Part \( \mathrm{(c)(i)} \))
• Topic \( 2.7 \) — VSEPR and Hybridization (Part \( \mathrm{(c)(ii)} \))
▶️ Answer/Explanation

(a)
For a correct explanation:
The plot of \( \dfrac{1}{[\mathrm{NO_2}]} \) versus time is the most linear, indicating that the reaction is second order with respect to \( \mathrm{NO_2} \).
For a second-order reaction in one reactant, the integrated rate law predicts a linear relationship between \( \dfrac{1}{[\mathrm{A}]} \) and time, so the straight-line graph supports the claim.

(b)
For the correct calculated value:
From the balanced equation, \( \mathrm{2\,NO_2 \rightarrow 2\,NO + O_2} \)

This means: \( \dfrac{1\ \mathrm{mol\ O_2}}{2\ \mathrm{mol\ NO_2}} \)

Therefore, \( 6.52 \times 10^{-7}\ \mathrm{M/s} \times \dfrac{1\ \mathrm{mol\ O_2}}{2\ \mathrm{mol\ NO_2}} = 3.26 \times 10^{-7}\ \mathrm{M/s} \)

So, the rate of appearance of \( \mathrm{O_2} \) is \( \boxed{3.26 \times 10^{-7}\ \mathrm{M/s}} \).

(c)(i)
For the correct Lewis diagram:

In \( \mathrm{NO_2^+} \), nitrogen forms two double bonds to oxygen and has no unpaired electron.
Each oxygen has two lone pairs, and the entire structure is enclosed in brackets with an overall \( + \) charge.

(c)(ii)
For the correct answer and a valid justification, consistent with part \( \mathrm{(c)(i)} \):
Agree. The angle of \( \mathrm{NO_2^+} \) is different from the angle in \( \mathrm{NO_2} \) because there would no longer be a nonbonding electron on the central atom in \( \mathrm{NO_2^+} \), and the oxygen atoms would spread farther apart, forming a linear structure with a \( 180^\circ \) bond angle.

Another valid way to justify this is by hybridization:
The hybridization of \( \mathrm{N} \) in \( \mathrm{NO_2} \) is \( \mathrm{sp^2} \), which gives a bond angle of approximately \( 120^\circ \). The hybridization of \( \mathrm{N} \) in \( \mathrm{NO_2^+} \) is \( \mathrm{sp} \), which gives a bond angle of \( 180^\circ \).

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