My friends and I were doing some problems from this year's IChO Preparatory Problems (PDF from the 49th International Chemistry Olympiad (2017)) when we stumbled upon a question which we had some confusion with.

Task 8. Decomposition of Nitrous Oxide
Nitrous oxide decomposes exothermically into nitrogen and oxygen, at a temperature of approximately $\pu{565 ^\circ C}$. $$\ce{2N2O (g) -> 2N2 (g) + O2 (g)}$$ This reaction follows the second-order kinetics when carried out entirely in the gas phase.

8.1) If the reaction is initiated with $[\ce {N2O}]$ equal to $\pu{0.108 mol dm-3}$, what will its concentration be after $\pu{1250 s}$ have elapsed at $\pu{565 ^\circ C}$? The rate constant for the second order decomposition of $\ce{N2O}$ is $\pu{1.10\times10^-3 dm3 mol-1 s-1}$ at this temperature.

Task 8 details the kinetics of the decomposition of nitrous oxide. We were confused with part 8.1 of the task which required us to find the concentration of the reactant after a particular duration of time has elapsed, given the temperature, initial concentration and rate constant of the reaction. We approached the question using two different methods:

I approached it using the method, as suggested in the solutions manual, of using the integrated rate law for 2nd order kinetics and substituting the values provided into that equation. My answer was exactly that stated in the solutions manual.

However, they approached it by first finding the half-life using the equation for the half-life for 2nd order kinetics and then, finding the concentration after knowing how many half-lives have passed. Their answer was slightly off.

After substituting random values into the equations for both methods, I realised that the "half-life method" which my friends used only gives a reasonable approximation when the time elapsed < half-life, gives the exact value when the time elapsed = half-life and goes completely off when the time elapsed > half-life. Why is that so? Since the half-life equation for 2nd order kinetics is derived from the integrated rate law (as shown on Chemistry LibreTexts), shouldn't they give the same answer?


1 Answer 1


Roughly speaking, half-life is not a thing at all for the second order (or any order other than first, for that matter).

What would they do after finding the number of half-lives? Raise 2 to that power? Too bad, because that only works for the exponential decay, which is the solution of the first-order kinetics and no other order. Really, what if the number of half-lives equals 2? After spending the first half-life, we now have a different half-life ahead of us, because it is concentration dependent, and the concentration has changed.

Following your link, we see the same statement, except they don't put enough emphasis on it, as to my taste:

For this reason, the concept of half-life for a second-order reaction is far less useful.

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    $\begingroup$ I understand what you mean. But how were they able to mathematically derive the half-life equation t1/2 = 1/(kCo) from the integrated rate law? Since the equation has been proven mathematically, it should work? $\endgroup$ Dec 14, 2017 at 5:47
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    $\begingroup$ They got it by plugging the halved concentration into the exact integrated rate law and solving for time. As such, this thing is indeed proven mathematically and surely does work. It does tell you when the concentration reaches ${1\over2}c_0$. But it shouldn't and doesn't tell you when it reaches ${1\over4}c_0$. Nor does it tell what the concentration will be in $2t_{1/2}$. $\endgroup$ Dec 14, 2017 at 6:15
  • $\begingroup$ So you are saying that it will only work for the 1st half-life, as I have also observed previously when trying to randomly substitute values, because the derivation has only been for the case of the 1st half-life? $\endgroup$ Dec 14, 2017 at 6:17
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    $\begingroup$ Yes, exactly.$\;$ $\endgroup$ Dec 14, 2017 at 6:18

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