As a personal problem, I am trying to find a rate law for the reaction $\ce{2 O3 -> 3 O2}$ using the steady-state approximation.

The elementary reactions are, \eqref{eq:equilib} (forward rate constant $k_1$, reverse rate constant $k_{-1}$), and \eqref{eq:destroy} (forward rate constant $k_2$): \begin{align} \ce{O3 &<=> O2 + O} \tag1\label{eq:equilib}\\ \ce{O + O3 &-> 2 O2} \tag2\label{eq:destroy} \end{align}

I go about by setting the total rate of disappearance of $\ce{O}$ equal to that of its formation. Thus,

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

and rearranging gives $$[\ce{O}] = \frac{k_1[\ce{O3}]}{k_{-1}[\ce{O2}] + k_2[\ce{O3}]}$$

Next, I express the overall rate in terms of the disappearance of $\ce{O3}$. This means that

$$\text{Rate} = -\frac{\mathrm{d}[\ce{O3}]}{\mathrm{d}t} = k_2[\ce{O}][\ce{O3}].$$

Then I can substitute the expression for $\ce{[O]}$ to get an expression for the overall rate, and I get

$$\text{Rate} = \frac{k_1 k_2[\ce{O3}]^2}{k_{-1}[\ce{O2}] + k_2[\ce{O3}]}$$

With the step above lies my question. Is $-\frac{\mathrm{d}[\ce{O3}]}{\mathrm{d}t}$ equal to what I have above, or does it also include the rate of disappearance of $\ce{[O3]}$ from the first elementary reaction?

$$\text{Rate} = -\frac{\mathrm{d}[\ce{O3}]}{\mathrm{d}t} = k_1[\ce{O3}] + k_2[\ce{O}][\ce{O3}],$$ as opposed to just $$\text{Rate} = k_2[\ce{O}][\ce{O3}]?$$

I suppose the former is correct, because when $k_2$ is small, the expression reduces to one obtained by simpler methods ($\text{Rate} = \frac{k_1k_2[\ce{O3}]^2}{k_{-1}[\ce{O2}]}$). However, it seems that I do need to include the first elementary step, since that is involved in consuming $\ce{[O3]}$.

  • 1
    $\begingroup$ The rate of change of ozone has three terms, two from the equilibrium reaction and one from the reaction with O atoms, $\frac{d\ce{[O_3]}}{dt} = -\ce{k_1[O_3] +k_{-1}[O][O2] - k_2[O][O_3] } $ (To get steady state the general way would be to put $d[O]/dt = 0$ which gives the terms you quote) $\endgroup$
    – porphyrin
    Mar 21, 2017 at 14:35
  • $\begingroup$ Yet if I defined the rate of reaction based on the rate of appearance of $\ce{O_2}$, as $\ce{\frac{1}{2}\frac{d[O_2]}{dt}}$, then I get a different expression: Rate = $\ce{k_2[O][O_3]}$, which differs from that given by using -$\ce{\frac{d[O_3]}{dt}}$ $\endgroup$
    – Yunfei Ma
    Mar 21, 2017 at 21:02
  • 1
    $\begingroup$ The rate of change of dioxygen also has three terms. $\endgroup$
    – porphyrin
    Mar 22, 2017 at 10:12

2 Answers 2


Applying the steady-state and rate-determinign-step aproximations, we have: $$\ce{2 O3 -> 3 O2}$$

$$\ce{O3 + M <=> O2 + O + M}$$ $$\ce{O + O3 -> 2 O2}$$ $$R=\frac{1}{2}\frac{\mathrm{d}[\ce{O2}]}{\mathrm{d}t} = k_2[\ce{O}][\ce{O3}]$$ $$\frac{\mathrm{d}[\ce{O}]}{\mathrm{d}t} =0= k_1[\ce{O3}][\ce{M}] -k_{-1}[\ce{O2}][\ce{O}][\ce{M}] - k_2[\ce{O}][\ce{O3}]$$ $$[\ce{O}] = \frac{k_1[\ce{O3}][\ce{M}]}{k_{-1}[\ce{O2}][\ce{M}] + k_2[\ce{O3}]}$$ $$R=\frac{1}{2}\frac{\mathrm{d}[\ce{O2}]}{\mathrm{d}t} =\frac{k_1k_2[\ce{O3}]^2[\ce{M}]}{k_{-1}[\ce{O2}][\ce{M}] + k_2[\ce{O3}]}$$ $$k_{-1}[\ce{O2}][\ce{M}] >> k_2[\ce{O3}]$$

$${Rate} = \frac{k_1k_2[\ce{O3}]^2}{k_{-1}[\ce{O2}]}=k_{eff}\frac{[\ce{O3}]^2}{[\ce{O2}]}$$

  • $\begingroup$ why are you assuming that k−1[O2][M]>>k2[O3]? That is not a given here. If you make that assumption, you don't even need to do the steady-state approximation, as you can just use the equilibrium expression for the first step. $\endgroup$
    – Andrew
    Sep 12, 2022 at 12:15
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    $\begingroup$ Due to the fact that k-1>>k2, because the second elementary reaction is the rate determining step. This means that its rate is much slower, so the k2 is also tiny $\endgroup$ Sep 13, 2022 at 13:39
  • $\begingroup$ Nowhere in the question does it say that the second elementary reaction is the rate determining step $\endgroup$
    – Andrew
    Sep 13, 2022 at 15:05
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    $\begingroup$ Well, it's just how it is in real life this specific reaction. $\endgroup$ Sep 13, 2022 at 21:50
  • $\begingroup$ And that's the way reaction rates are deduced. $\endgroup$ Sep 13, 2022 at 21:51

The error in your logic is when you write:

Next, I express the overall rate in terms of the disappearance of $\ce{O3}$. This means that


For any reaction $\ce{aA + bB -> cC}$, the overall rate is given by

Rate $=-\frac{1}{a}\frac{d[\ce{A}]}{dt}=-\frac{1}{b}\frac{d[\ce{B}]}{dt}=\frac{1}{c}\frac{d[\ce{C}]}{dt}$.

So in your case, the overall stoichiometry is $\ce{3O2 -> 2 O3}$, so your overall rate is equal to $-\frac{1}{2}\frac{d[\ce{O3}]}{dt}$.

Going back to your quoted expression, we instead have

Rate$=−\frac{1}{2}\frac{d[\ce{O3}]}{dt}=k_2[\ce{O}][\ce{O3}]$, so $=\frac{d[\ce{O3}]}{dt}=-2k_2[\ce{O}][\ce{O3}]$.

From there, substitute just as you did to get the final answer.

UPDATE: Apparently this was problem 7.3 of the International Chemistry Olympiad exam in 2005. Their derivation of the answer is different from mine (though giving the same result), so I'm adding it here for completeness, with some explanatory text for the steps.

Start with the expression for the net rate of decomposition of ozone:


Make a steady-state assumption for the intermediate O (as shown in the question above) to get the equation


Rearranging terms, we get


The left side represents two of the terms in our original equation, so we can substitute the right side term for them:


To finish, we substitute in the expression for [O] that can also be derived from the steady-state equation (as shown in the question) and get a final result of


  • $\begingroup$ I'm sorry, but I think you're wrong. $\endgroup$ Sep 13, 2022 at 21:54
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    $\begingroup$ First, if you consider the global reaction rate as the decreasing of the ozone, you must also consider the 3 terms involved with it. $\endgroup$ Sep 13, 2022 at 21:57
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    $\begingroup$ Second, is a mistake to say that the stoichiometric coefficient of the ozone is 2, because the definition is only valid for elementary steps (you are considering the global reaction), so it would be 1. $\endgroup$ Sep 13, 2022 at 21:58
  • $\begingroup$ The reason you do not need to consider all three terms is the steady-state assumption, which requires that step 1 happens only once (net) per reaction 2. If forward step 1 is faster than step 2, it is offset by the reverse step 1, so that O that is produced in step 1 is consumed in step 2 at the same rate. That means that the overall rate of O3 consumption is intuitively just 2 times the rate of step 2. $\endgroup$
    – Andrew
    Sep 13, 2022 at 22:33
  • $\begingroup$ Well, that's another way to solve it, but less practical, due to its complexity. $\endgroup$ Sep 15, 2022 at 9:23

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