As a personal problem, I am trying to find a rate law for the reaction $\ce{3 O2 -> 2 O3}$ 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 '17 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 '17 at 21:02
  • 1
    $\begingroup$ The rate of change of dioxygen also has three terms. $\endgroup$
    – porphyrin
    Mar 22 '17 at 10:12

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