The steps below represent a proposed mechanism for the catalyzed oxidation of $\ce{CO}$ by $\ce{O3}$:

$$\begin{align} &\text{Step 1: } & \ce{NO2(g) + CO(g) &-> NO(g) + CO2(g)} \\ &\text{Step 2: } & \ce{NO(g) + O3(g) &-> NO2(g) + O2(g)} \\ \end{align}$$

Which of the following statements is true based on this mechanism?

  1. The overall products are $\ce{NO2}$ and $\ce{O2}$.
  2. The overall products are $\ce{NO}$ and $\ce{CO2}$.
  3. The overall products are $\ce{NO}$ and $\ce{O2}$.
  4. Either $\ce{NO2}$ or $\ce{NO}$ can be viewed as a catalyst. (correct answer)
  5. Both $\ce{NO2}$ and $\ce{NO}$ are intermediates.

Based on the given proposed mechanism for the catalyzed oxidation of $\ce{CO}$ by $\ce{O3}$, the answer for above question suggests that either $\ce{NO}$ or $\ce{NO2}$ can be considered as a catalyst.

Thus, I'm very confused as to why the answer is 4. I would think it is 5) because $\ce{NO}$ is produced first and than consumed later on. I understand $\ce{NO2}$ does not have that same pattern, however I thought that it cannot be a catalyst because $\ce{O3}$ is already a catalyst.

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    $\begingroup$ Ozone is what is oxidizing the carbon monoxide and the two nitrogen oxides are catalysts, just as answer 4 says. $\endgroup$ – Ed V Apr 26 '20 at 1:38
  • $\begingroup$ Why is NO not an intermediate though? I learned it that an intermediate shows up in the products first and then shows up in the reactants $\endgroup$ – Lat Apr 26 '20 at 1:43
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    $\begingroup$ I do not know what you learned about intermediates, but what matters is that ozone cannot (in this problem anyway) directly oxidize CO to $\ce{CO_2}$, so $\ce{NO_2}$ does the task and would end as NO, except that ozone then oxidizes NO back to $\ce{NO_2}$. The nitrogen oxides just cycle around. I do not see any harm in thinking of NO as an intermediate, but it is the cycling pair that comprise the catalyst. $\endgroup$ – Ed V Apr 26 '20 at 1:48
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    $\begingroup$ @MathewMahindaratne Ah, but suppose you start with a gas mixture of $\ce{O3}$, CO and NO. Then the $\ce{O3}$ will oxidize the NO to $\ce{NO2}$, then the first reaction takes place and the net result is the same. So, for the order in the problem, I tend to agree: $\ce{NO2}$ gets the ball rolling and NO is intermediate and essential in regeneration of $\ce{NO2}$. But if the second reaction goes first, then the nitrogen oxides change roles. Maybe this is a distinction without a difference! $\endgroup$ – Ed V Apr 26 '20 at 2:02
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    $\begingroup$ @MathewMahindaratne Done! I am still woefully slow at doing the formatting. $\endgroup$ – Ed V Apr 26 '20 at 2:31

My take on this problem is as follows. Suppose a flask contains an equimolar mixture of $\ce{CO}$ and $\ce{O_3}$. The net reaction between the two gases will be

$$\ce{CO (g) + O3 (g) -> CO2 (g) + O2 (g)} \tag{1}$$

But it needs a catalyst. If $\ce{NO_2}$ gas is introduced, then the reaction proceeds via the two reactions given in the OP's question:

$$\ce{NO_2 (g) + CO (g) -> CO2 (g) + NO (g)} \tag{2}$$ and $$\ce{NO (g) + O3 (g) -> NO2 (g) + O2 (g)} \tag{3}$$

So the nitrogen oxides cycle around. On the other hand, if $\ce{NO}$ gas is introduced, then the reaction proceeds via 'flipped' equations:

$$\ce{NO (g) + O3 (g) -> NO2 (g) + O2 (g)} \tag{4}$$ and $$\ce{NO_2 (g) + CO (g) -> CO2 (g) + NO (g)} \tag{5}$$

and again the nitrogen oxides cycle around. So it does not really matter whether $\ce{NO_2}$ gas or $\ce{NO}$ gas is introduced first: the two nitrogen oxides are the overall catalytst and either could be considered to be an intermediate. Basically, this is a distinction without a difference.

  • $\begingroup$ Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this! $\endgroup$ – Ed V May 26 '20 at 1:38

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