Question

The reaction 2 NO₂(g) + O₃(g) N₂O₅(g) + O₂(g) was studied at a certain temperature with the following results:

Experiment [NO₂(g)] (M) [O₃(g)] (M) Rate (M/s)

1 0.643 0.643 31100

2 0.643 1.29 62300

3 1.29 0.643 62300

4 1.29 1.29 1.25e+05

What is the rate law for this reaction?

Answer 1

The rate law for the reaction 2 NO₂(g) + O₃(g) → N₂O₅(g) + O₂(g) is Rate = k [NO₂(g)] [O₃(g)].

Step-by-step explanation:

The rate law for the reaction 2 NO₂(g) + O₃(g) → N₂O₅(g) + O₂(g) can be determined by analyzing the initial rates of reaction at different concentrations of the reactants. In this case, the initial rates are given as follows:

Experiment 1: [NO₂(g)] = 0.643 M, [O₃(g)] = 0.643 M, Rate = 31100 M/s

Experiment 2: [NO₂(g)] = 0.643 M, [O₃(g)] = 1.29 M, Rate = 62300 M/s

Experiment 3: [NO₂(g)] = 1.29 M, [O₃(g)] = 0.643 M, Rate = 62300 M/s

Experiment 4: [NO₂(g)] = 1.29 M, [O₃(g)] = 1.29 M, Rate = 1.25e+05 M/s

From these experiments, we can see that the rate of reaction doubles when the concentration of either NO₂ or O₃ is doubled. Therefore, the rate law for this reaction is:

Rate = k [NO₂(g)] [O₃(g)]