I am a sophomore. A friend and I ran across a kinetic chemistry exercise. We solved it in a slightly different way and we can't come to an agreement on who is right. We would appreciate some help:

Destruction of stratospheric ozone as determined by using the steady-state approximation.

The balance of ozone in the stratosphere is of critical concern because this molecule absorbs ultraviolet light that would be harmful to life at Earth's surface. The principal production mechanism for ozone is recombination of $\ce O$ atoms with $\ce{O_2}$. The principal destruction mechanism is that given below. There is increasing concern over alternative destruction mechanisms involving molecules introduced into the stratosphere by human activity.

Determine the destruction rate of ozone in the following mechanism. $$\begin{align}\ce{O3 + M &<=>[k_1][k_{-1}] O2 + O + M}\\ \\ \ce{O3 + O &->[k_2] 2 O2}\end{align}$$

Our answers:

We agree on the beginning, so we have :

$$\begin{align} -\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} &= k_1[\ce{O_3}][\ce{M}] - k_{-1}[\ce{O_2}][\ce O][\ce M] + k_2[\ce{O_3}][\ce O]\\ -\frac{\mathrm d[\ce{M}]}{\mathrm dt} &= k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M] \\ -\frac{\mathrm d[\ce O]}{\mathrm dt} &= -k_1[\ce{O_3}][\ce M] + k_{-1}[\ce{O_2}][\ce O][\ce M] +k_2[\ce{O_3}][\ce O] \end{align}$$

Now here comes the disagreement, we need to use the steady-state approximation to say that one of the above speed is null:

My friend argues that $-\frac{\mathrm d[\ce O]}{\mathrm dt}$ should be approximated to null because it is a transient compound.

Even if in most other exercise, it would be correct, I disagree with him because in the proposed mechanism, $\ce{O_3}$ and $\ce O$ play a symmetrical part. So if we approximate $-\frac{\mathrm d[\ce O]}{\mathrm dt}$ to $0$ we should also approximate $-\frac{\mathrm d[\ce{O_3}]}{\mathrm dt}$ to $0$. Since we want to calculate this specific value, we don't want to approximate it. In my demonstration, I approximate $-\frac{\mathrm d[\ce{M}]}{\mathrm dt}$ to $0$, arguing that the compound $\ce M$ is on both side of the equilibrium equation and therefore it's concentration will remain constant.

We both manage to find a result:
My friend’s calculation:

$$ - \frac{\mathrm d[\ce O]}{\mathrm dt} = -k_1[\ce{O_3}][\ce M] + k_{-1}[\ce{O_2}][\ce O][\ce M] + k_2[\ce{O_3}][\ce O] = 0 $$ $$ k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M] = k_2[\ce{O_3}][\ce O] $$ We substitute the left member in the speed of ozone: $$ -\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = 2\times k_2[\ce{O_3}][\ce O] $$ Because $[\ce O]$ is transient, it is difficult to mesure experimentally, so we expresse it in function of $[\ce{O_3}], [\ce{O_2}]$ and $[\ce M]$ : $$ k_1[\ce{O_3}][\ce M] = [\ce O] \times ( k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M] ) $$ $$ [\ce O] = \frac{k_1[\ce{O_3}][\ce M]}{k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M]} $$ And finally we have my friend’s result: $$ -\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = \frac{2k_1k_2[\ce{O_3}]^2[\ce M]}{k_2[\ce{O_3}] + k_{-1}[\ce{O_2}][\ce M]} $$

Now my calculations: $$ -\frac{\mathrm d[\ce M]}{\mathrm dt} = k_1[\ce{O_3}][\ce M] - k_{-1}[\ce{O_2}][\ce O][\ce M] = 0 $$ Which allow us to simplify the destruction rate of ozone: $$ -\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = k_2[\ce{O_3}][\ce O] $$ We already disagree by a factor 2 on the rate. But if we replace $[\ce O]$ like before, we obtain a much simpler result: $$\begin{align} k_1[\ce{O_3}][\ce M] &= k_{-1}[\ce{O_2}][\ce{O}][\ce M] \\ [\ce{O}] &= \frac{k_1[\ce{O_3}][\ce M]}{k_{-1}[\ce{O_2}][\ce M]} \\ [\ce O] &= \frac{k_1[\ce{O_3}]}{k_{-1}[\ce{O_2}]} \end{align}$$ Finally I have: $$ -\frac{\mathrm d[\ce{O_3}]}{\mathrm dt} = \frac{k_1k_2[\ce{O_3}]^2}{k_{-1}[\ce{O_2}]} $$

Which one of us is wrong and why?

  • 1
    $\begingroup$ There is no reason to set $\frac{d\ce{[O3]}}{dt}$ to zero. If that were true, the reaction would have zero rate. The steady state approximation allows you to say that the intermediate concentration is low and therefore the change in the concentration of intermediate is very close to zero. There's no reason why it can't be produced very quickly and consumed equally quickly. Therefore, your claim is incorrect. $\endgroup$
    – Zhe
    Oct 11, 2017 at 21:05
  • $\begingroup$ Thank you for your comment. What I don't understand is how $\frac{d[O]}{dt}$ can be considered null without considering $\frac{d[O_3]}{dt}$ to be null at the same time. Unless $k_1 >> k_{-1}$ I don't see how it's possible... The wording doesn't specify values or order of magnitude on $k_1$, $k_{-1}$ or $k_2$. $\endgroup$
    – Amon
    Oct 11, 2017 at 21:43
  • $\begingroup$ Ignore the reverse reaction for a minute. Consider buckets one on top of the other each with a hole in the bottom. The top bucket is full of water. The water drops into bucket 2 and immediately drops out. The rate of water leaving the top bucket is significant. The rate of water accumulating in the second lower bucket is effectively zero. $\endgroup$
    – Zhe
    Oct 11, 2017 at 21:55
  • 2
    $\begingroup$ I don't understand why you want to make the ozone concentration constant. That would imply no reaction. $\ce{O3}$ and $\ce{O}$ don't have the same role at all. One is a reactant and the other is an intermediate. Steady state approximation only applies to one of them. The change in concentration of $\ce{M}$ seems fine, but your intuition about the rate of change of ozone concentration is wrong. $\endgroup$
    – Zhe
    Oct 12, 2017 at 0:06
  • 1
    $\begingroup$ I think you misunderstood what M is. M is the concentration of all molecules in the atmosphere, including O2 and N2. It is so large that it doesn't change significantly with time. $\endgroup$ Oct 12, 2017 at 13:20

1 Answer 1


Dioxygen is split in the stratosphere by UV light into oxygen atoms. This is the main source of free oxygen atoms. Their production rate depends on the intensity of the UV light.

In steady-state we can assume that the concentration of oxygen atoms is constant.

$$\frac{d[\ce{O}]}{dt} = 0 \tag{1}$$

For the above destruction mechanism we can formulate the rate laws

$$\begin{align} \frac{d[\ce{O3}]}{dt} &= -k_1 [\ce{O3}] [\ce{M}] + k_{-1} [\ce{O}] [\ce{O2}] [\ce{M}] - k_2 [\ce{O}] [\ce{O3}] \tag{2}\\ \\ \frac{d[\ce{O}]}{dt} &= k_1 [\ce{O3}] [\ce{M}] - k_{-1} [\ce{O}] [\ce{O2}] [\ce{M}] - k_2 [\ce{O}] [\ce{O3}] \tag{3} \end{align}$$

Adding equation $(2)$ and $(3)$ which by $(1)$ is $0$ we get

$$\frac{d[\ce{O3}]}{dt} = -2~k_2 [\ce{O}] [\ce{O3}] \tag{4}$$

With equations $(1)$ and $(3)$ we can express the concentration of oxygen atoms as

$$[\ce{O}] = \frac{k_1 [\ce{O3}] [\ce{M}]}{k_{-1} [\ce{O2}] [\ce{M}] + k_2 [\ce{O3}]} \tag{5}$$

Substituting $(5)$ into equation $(4)$ we get

$$\frac{d[\ce{O3}]}{dt} = \frac{-2~k_1~k_2 [\ce{O3}]^2 [\ce{M}]}{k_{-1} [\ce{O2}] [\ce{M}] + k_2 [\ce{O3}]} \tag{6}$$

So your friend seems to be right.

  • $\begingroup$ Hi, thank you for you answer, however it doesn't really answer my question. What allow you to make the approximation $\frac{\mathrm d[\ce{O}]}{\mathrm dt} = 0$ ? Just saying "steady-state" feels like you're waving a magic wand and poof the problem is gone... That's not really satisfactory. $\endgroup$
    – Amon
    Oct 14, 2017 at 20:08
  • 2
    $\begingroup$ If we can assume that the primary source of oxygen atoms is the flow of UV light, the we have a constant rate of production and, according to the above reactions a variable rate of destruction. When these are equal we effectively have a transformation ov UV to heat (hot Ms) and a constant concentration of oxygen atoms. $\endgroup$
    – aventurin
    Oct 14, 2017 at 20:47

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