Steady-state approximation for the destruction of ozone

Question

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 $\ce{[O3]}$.

Answer 1

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

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

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

Answer 2

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

Rate$=−\frac{d[\ce{O3}]}{dt}=k_2[\ce{O}][\ce{O3}]$.

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:

$$Rate=−\frac{d[\ce{O3}]}{dt}=k_1[\ce{O3}]-k_{-1}[\ce{O2}][\ce{O}]+k_2[\ce{O}][\ce{O3}]$$

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

$$\frac{d[\ce{O}]}{dt}=0=k_1[\ce{O3}]−k_{−1}[\ce{O2}][\ce{O}]−k_2[\ce{O}][\ce{O3}]$$

Rearranging terms, we get

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

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

$$Rate=−\frac{d[\ce{O3}]}{dt}=k_1[\ce{O3}]-k_{-1}[\ce{O2}][\ce{O}]+k_2[\ce{O}][\ce{O3}]$$$$=k_2[\ce{O}][\ce{O3}]+k_2[\ce{O}][\ce{O3}]$$$$=2k_2[\ce{O}][\ce{O3}]$$

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

$$Rate=\frac{2k_1k_2[\ce{O3}]^2}{k_{−1}[\ce{O2}]+k_2[\ce{O3}]}$$