Consider the reaction
NO2 + O3 NO3 + O2

Use the data shown to determine the rate law for the reaction.

Experiment

[NO2] M

[O3] M

Rate M s−1

1

0.21

0.70

6.30 x 10−3

2

0.63

0.70

1.89 x 10−2

3

0.21

0.35

3.15 x 10−3

4

0.42

0.35

6.30 x 10−3

Question

Consider the reaction
NO2 + O3  NO3 + O2

Use the data shown to determine the rate law for the reaction.

Experiment

[NO2] M

[O3] M

Rate M s−1

1

0.21

0.70

6.30 x 10−3

2

0.63

0.70

1.89 x 10−2

3

0.21

0.35

3.15 x 10−3

4

0.42

0.35

6.30 x  10−3

BlackTom AI · Accepted Answer

The question presents a kinetic data set for the reaction NO2 + O3 → NO3 + O2 and asks us to determine the rate law from the experiments.
First, I’ll compare how changes in [NO2] while keeping [O3] constant affect the rate. In experiments 1 and 2, [O3] is held at 0.70 M, while [NO2] changes from 0.21 to 0.63 M. If the rate roughly scales with the square of [NO2], we would expect the rate to increase by a factor of about 9 when [NO2] triples; however, the rate increases from 6.30×10−3 to 1.89×10−2, which is exactly a factor of 3, suggesting a first-order dependence on NO2 with respect to rate when [O3] is constant.
Next, I’ll examine how changes in [O3] while keeping [NO2] constant affect the rate. In experiments 1 and 3, [NO2] is kept at 0.21 M, while [O3] changes from 0.70 to 0.35 M. The rate changes from 6.30×10−3 to 3.15×10−3, which is a factor of 0.5 when [O3] is halved, consistent with first-order dependence on O3.
To confirm, I’ll compute an apparent rate constant k for each experiment using the proposed form rate = k [NO2][O3]. For experiment 1: k = (6.30×10−3) / (0.21 × 0.70) ≈ 0.0429. For experiment 2: k = (1.89×10−2) / (0.63 × 0.70) ≈ 0.0429. For experiment 3: k = (3.15×10−3) / (0.21 × 0.35) ≈ 0.0429. For experiment 4: k = (6.30×10−3) / (0.42 × 0.35) ≈ 0.0429. In every case, the calculated k is essentially the same, within rounding error, supporting a rate law where the rate is proportional to the product of [NO2] and [O3].
Now, consider whether alternative rate laws could fit the data. If the reaction were zero-order in NO2 or zero-order in O3, changing these concentrations would not produce the observed changes in rate; that is inconsistent with the data. If the reaction were first-order in NO2 but zero-order in O3, or vice versa, the rate would not scale correctly with simultaneous variations in both reactants as seen here. If either species had a fractional or non-integer order not matching the observed linear dependence on both concentrations, the computed k values would not stay constant across experiments. Therefore, the data strongly support a rate law where the rate is proportional to both reactants.
In summary, by analyzing how the rate changes with each reactant while computing k from each experiment, we observe consistent first-order behavior in both NO2 and O3, yielding a rate law of the form rate = k [NO2] [O3].

Consider the reaction NO2 + O3 NO3 + O2 Use the data shown to determine the rate law for the reaction. Experiment [NO2] M [O3] M Rate M s−1 1 0.21 0.70 6.30 x 10−3 2 0.63 0.70 1.89 x 10−2 3 0.21 0.35 3.15 x 10−3 4 0.42 0.35 6.30 x 10−3

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