At high temperatures nitrogen dioxide decomposes into nitrogen oxide and oxygen gas according to $$\ce{2 NO_2 -> 2 NO + O_2}$$

I have a question now. Let's say at a certain temperature I'm given data for the formation of $\ce{O_2}$. For example: At $t = 0$ we have $[\ce{O_2}] = 0$ mol/L, at $t = 50$, $[\ce{O_2}] = 7.10 \cdot 10^{-4}$ mol/L, at $t = 100, [\ce{O_2}] = 1.205 \cdot 10^{-3}$ and so on until a certain time.

If I'm given that the initial concentration of $\ce{NO_2}$ is $0.006M$, how can I know what the concentration of $\ce{NO_2}$ is at certain times? I'm asked to determine the order of the reaction on the basis of the given data, and also to present graphically the rate constant.

To do that I need to know if $\ln([\ce{NO_2}])$ versus time $t$ is a straight line. But how to find $[\ce{NO_2}]$ in the first place?

  • $\begingroup$ If you had a reaction A -> B, with [A] = 1 M and no B present at the start, and you are told that after 60 seconds [B] = 0.5 M, how would you calculate [A] at t = 60 s? $\endgroup$ Jun 2, 2015 at 14:00
  • $\begingroup$ I would do (initial concentration of A) - (amount of formation of B). So 1M - 0.5M = 0.5M of A at 60 s. But can we apply this? This assumes theres a linear relationship between reactants and products. Is this the case also in my example? Since the stoichiometric coefficient is 2 there. $\endgroup$
    – Kamil
    Jun 2, 2015 at 14:05
  • 1
    $\begingroup$ Stoichiometry is always linear. If one molecule of $\ce{O2}$ is produced, two molecules of $\ce{NO2}$ have decomposed, so you just have a factor of 2: $[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$. $\endgroup$ Jun 2, 2015 at 14:08
  • $\begingroup$ What about the presence of the other product, NO. Won't that influence the amount that is taken away? $\endgroup$
    – Kamil
    Jun 2, 2015 at 14:11
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    $\begingroup$ That would be the case if you have two competing reactions: A -> B and A -> C. But here the production of NO, the production of $\ce{O2}$, and the decomposition of $\ce{NO2}$ are all linearly dependent on each other because they are part of the same reaction. Essentially, for every molecule of $\ce{O2}$ you produce, you also have to produce two molecules of NO, and two molecules of $\ce{NO2}$ have to be decomposed. So their concentrations are all linked: $[\ce{O2}]_t = 0.5 [\ce{NO}]_t = 0.5([\ce{NO2}]_i - [\ce{NO2}]_t)$. $\endgroup$ Jun 2, 2015 at 14:18

1 Answer 1


In this reaction, the production of $\ce{O2}$ is tied to the decomposition of $\ce{NO2}$; for each molecule of $\ce{O2}$ that is produced, there have to be 2 molecules of $\ce{NO2}$ that have decomposed. So:

$$2 \Delta n_{\ce{O2}} = -\Delta n_{\ce{NO2}}$$

where $\Delta n$ represents a change in the number of moles. The question does not specify, but I will assume that the volume of the reaction is kept constant; since concentration is amount divided by volume, this means that the same relation holds for concentrations:

$$2\Delta [\ce{O2}] = -\Delta [\ce{NO2}]$$

Since the initial concentration of $\ce{O2}$ is zero, $\Delta [\ce{O2}] = [\ce{O2}]$, and therefore

$$2[\ce{O2}]_t = [\ce{NO2}]_i - [\ce{NO2}]_t $$

$$[\ce{NO2}]_t = [\ce{NO2}]_i - 2[\ce{O2}]_t$$

To use the numbers in the question, at $t = 50$ s we have $[\ce{NO_2}] = (0.006 - 2(7.10\cdot 10^{-4}))$ M = $4.58\cdot 10^{-3}$ M.


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