AP Chemistry 5.2 Introduction to Rate Law - Exam Style questions - FRQs- New Syllabus

(a)
For the correct calculated value:
Accept one of the following:

• \( k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{1.67\ \mathrm{hr}} = 0.415\ \mathrm{hr^{-1}} \)

• \( k = \dfrac{\ln[A]_0 – \ln[A]_t}{t} = \dfrac{\ln(0.160) – \ln(0.0800)}{1.67\ \mathrm{hr}} = 0.415\ \mathrm{hr^{-1}} \)

• \( k = \dfrac{\ln[A]_0 – \ln[A]_t}{t} = \dfrac{\ln(0.160) – \ln(0.0400)}{3.33\ \mathrm{hr}} = 0.416\ \mathrm{hr^{-1}} \)

• \( k = \dfrac{\ln[A]_0 – \ln[A]_t}{t} = \dfrac{\ln(0.160) – \ln(0.0200)}{5.00\ \mathrm{hr}} = 0.416\ \mathrm{hr^{-1}} \)

For the correct units, consistent with the calculated value:
\( \mathrm{hr^{-1}} \)

A quick check: the concentration halves from \( 0.160 \rightarrow 0.0800 \rightarrow 0.0400 \rightarrow 0.0200 \) in nearly equal time intervals, so the reaction is behaving like a first-order process, which matches the given rate law.

Therefore, \( \boxed{k \approx 0.415\ \mathrm{hr^{-1}}} \).

(b)
For the correct answer and a valid justification:
Step \( 1 \) is the rate-determining step. The rate law of elementary step \( 1 \) is \( \mathrm{rate} = k[\mathrm{N_2O_5}] \), which is consistent with the first-order kinetics of the overall rate law.

Step \( 2 \) would involve both \( \mathrm{NO_2} \) and \( \mathrm{NO_3} \), and step \( 3 \) would involve \( \mathrm{N_2O_5} \) and \( \mathrm{NO} \). Since the observed rate law depends only on \( [\mathrm{N_2O_5}] \), step \( 1 \) must be the slow step.

(c)
For the correct answer:
Remain the same. The rate constant, \( k \), is independent of concentration and will remain the same at constant temperature.

Doubling the initial concentration would change the initial rate, but not the value of \( k \). For a given reaction, \( k \) changes only when conditions such as temperature change.