Question

The rate law for the decomposition of ozone to molecular oxygen is rate = k[O3]2 [O2] The mechanism proposed for this process is O3 0001 O + O2 O + O3 2O2 What is the rate law in terms of k1, k2, and k−1? Be sure to simplify the rate law before inputting your answer.

Answer 1

The rate law for the decomposition of ozone in terms of the constants k1, k2, and k−1 is rate = k1k2[O3]^2 / (k−1 + k2[O3]). This is derived using the proposed two-step mechanism and making an assumption that the backward reaction is much slower than the forward reaction.

Step-by-step explanation:

The proposed mechanism for the decomposition of ozone to molecular oxygen involves two elementary steps:

O3 O + O2 (rate constant k1)

O + O3 2O2 (rate constant k2)

There is also a reverse reaction to step 1 that should be considered, where O and O2 combine to reform O3 (rate constant k⁻1). According to the steady-state approximation, the concentration of intermediate species O remains constant because it is produced and consumed at the same rate. Therefore, the rate of its formation in step 1 is equal to its rate of consumption in step 2 and any reverse reaction.

The rate law for the given mechanism can be expressed as:

rate = k2[O][O3]

However, [O] is not directly measurable, so we need to link it to other reactants. We can express [O] from the equilibrium of step 1, assuming that the backward reaction is much slower than the forward reaction:

[O] = k1[O3] / (k⁻1 + k2[O3])

Then by substituting [O] back into the rate law:

rate = k2(k1[O3] / (k⁻1 + k2[O3]))[O3]

This simplifies to:

rate = k1k2[O3]^2 / (k⁻1 + k2[O3])

This rate law now relates the overall rate to the experimental rate constants k1, k2, and k⁻1.

Answer 2

The rate law is k =
`k([O3]^(2) )/([O2])\\`

Step-by-step explanation:

From the mechanism is necessary to derive the rate law from the elementary steps and explain the effects of [O2] on the rate

The first step is a reversible reaction. Assuming dynamic equilibrium is achieved, the rate of the forward reaction is equal to the rate of the backward reaction

rate(forward) = rate(backward)

k1 [O3] = k-1 [O] [O2]

[O] is not part of the rate law, so we need to express [O] in terms of other species

[O] =
`(k1 [O3])/(k-1[O2])`

from the second step

rate = k2[O] [O3]

substituting [O] from the first step

rate =
`k2 (k1 [O3] [O3])/(k-1[O2]) = (k2k1 [O3]^(2) )/(k-1[O2])\\`

k =
`(k2k1)/(k-1)\\`

The final rate law is then

k =
`k([O3]^(2) )/([O2])\\`

So, as the concentration os O2 increase the rate decrease. Also from the first step of the mechanism we can se that O2 can react to O to form back the reactant O3 resulting in decreased reaction rate.