# Implication of negative order on kinetics

In the reaction

$$\ce{2 O3 -> 3 O2}$$

the order of reaction that I have been taught is

$$\ce{O_3 \underset{k_{-1}}{\overset{k_{1}}{<=>}} O_2 + O}$$ $$\ce{O_3 + O \overset{k_{2}}{->} 2O_2}$$

$$\mathrm{rate} = k_{2}[\ce{O3}][\ce{O}]$$

$$k_{1}[\ce{O3}]= k_{-1}[\ce{O2}][\ce{O}]$$

So $$\mathrm{rate} = \frac{k_{1}k_{2}}{k_{-1}} [\ce{O3}]^2[\ce{O2}]^{-1}$$

What does this $$[\ce{O2}]^{-1}$$ part mean? For example, initially when there is no $$\ce{O2}$$, the rate should be $$\infty$$; how is that possible? Or is it normal and I should accept it?

• There is no "there is no". In reality, the concentration of anything anywhere is never exactly 0, so don't bother about the possible singularity. – Ivan Neretin May 21 '19 at 13:30
• It means that the reaction you quote is the overall reaction, a summary if you like, but in reality there are other reactions occurring, perhaps $\ce{O3 \to O2 + O^.}$, and when the full scheme is worked out the rate law is more complex. – porphyrin May 21 '19 at 13:30
• "The order of reaction that I have been taught" This is a statement that worries me a lot. The two ways I interpret this are (1) you were taught to derive the rate law in a certain way or (2) you were taught this specific rate law. The former statement has problems because the derivation would require knowledge of elementary steps, which are not provided here. The latter doesn't make sense because you can't teach a rate law for a reaction; you can only posit a mechanism consistent with experimental data and derive the associated rate law. – Zhe May 21 '19 at 13:56
• @Zhe The way we derived it was , the teacher gave us the mechanism. Told us what the Slowest step is and derives it from that – Mr.HiggsBoson May 21 '19 at 13:58
• Arrhenius-type rate equations are never valid at extremely low concentrations or extremely small samples, because they implicitly assume number of molecules is a continuous quantity. This is true even with idealised single-step processes. The discrete nature of matter requires statistical corrections when there aren't billions and billions of entities in your sample. This has the side-effect of removing any mathematical singularities, like the one you've encountered. – Nicolau Saker Neto May 21 '19 at 14:15

## 2 Answers

What does this $$[\ce{O2}]^{-1}$$ part mean?

It means the higher the concentration of oxygen, the lower the rate. The two-step mechanism you show relies on oxygen atoms (radicals) to react with ozone in the second step. Dioxygen acts as a radical scavenger (first reaction, reverse direction). So if dioxygen concentrations are high, the reaction will not proceed via this mechanism.

When ozone is produced as part of smog in the lower atmosphere (i.e. the air we breathe), this mechanism is not expected to help lower the ozone concentration. The equilibrium of the first reaction would be too far on the side of the reactants.

In the higher layers of the atmosphere, however, this sequence is more feasible. Concentrations of all species are extremely low, so bimolecular reactions are slow as well. In fact, in the image linked below, concentrations are given as molecule per mL (as apposed to mol / L) and rate a similarly expressed as molecules reacting per time unit.

https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry_files/image018.jpg

What actually happens in the ozone layer is a bit more complicated than the two step process you show, see for example https://www.ems.psu.edu/~brune/m532/meteo532_ch7_stratospheric_chemistry.htm

For example, initially when there is no O2, the rate should be ∞; how is that possible? Or is it normal and I should accept it?

See the excellent comment by Nicolau Saker Neto. Another way to say this is "the molecules don't know the concentration of species, but they still do their thing without worrying about division by zero".

• Thanks A lot, It was really helpful – Mr.HiggsBoson May 21 '19 at 16:06

The scheme gives $$rate=k_2\ce{[O3][O]}$$ and $$\displaystyle \frac{d\mathrm{[O]}}{dt} =k_1\mathrm{[O_3]}-k_{-1}\mathrm{[O_2][O] } -k_2\mathrm{[O_3][O] }=0$$ for steady state on the oxygen atom.

This gives the rate expression as: rate=$$\displaystyle \frac{k_1k_2\mathrm{[O_3]^2} }{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$$.

Thus the problem when the oxygen concentration is zero does not arise, which must be the case as initially pure ozone exists and then its rate of decomposition is $$k_1\mathrm{[O_3]}$$. At enormously high oxygen molecule concentration the dissociation is restricted because of the equilibrium step.