Consider some general parallel first order process, where $\ce{A -> B}$ and $\ce{A -> C}$ (rate constants $k_1$ and $k_2$ respectively)

The overall rate equation of this process is said to be $-\frac{\mathrm d[A]}{\mathrm dt} = k_1[\mathrm A] + k_2[\mathrm A]$

I get the impression that this is derived from the addition of each individual rate equation, i.e. the addition of $-\frac{\mathrm d[\mathrm A]}{\mathrm dt} = k_1[\mathrm A]$ and $-\frac{\mathrm d[\mathrm A]}{\mathrm dt} = k_2[\mathrm A]$ would result in the overall rate equation shown above.

However, if this is the case, then why isn’t the rate equation actually $-\frac{2\,\mathrm d[\mathrm A]}{\mathrm dt} = k_1[\mathrm A] + k_2[\mathrm A]$? I mean, you’re adding two $-\frac{\mathrm d[\mathrm A]}{\mathrm dt}$ terms together; and so considering basic algebra, surely this would result in $-\frac{2\,\mathrm d[\mathrm A]}{\mathrm dt}$?

Or do differentials just not work in this way, and I’m simply confusing myself?

  • $\begingroup$ You just have two independent parallel decay pathways, so the faster one, with larger k, dominates. Just add the two k values: you do not have two separate pots of “A” to drain away. $\endgroup$
    – Ed V
    Feb 24, 2020 at 20:33
  • $\begingroup$ @Ed V I suppose the overall rate can therefore be seen as the rate constants of both individual pathways complementing each other? In other words, it’s as though the rate constants for both pathways combine to make one whole, ‘big’ rate constant and this is what dictates the overall rate? $\endgroup$ Feb 24, 2020 at 20:42
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    $\begingroup$ That is what I think. Imagine a silo full of water and two holes shot into it right at ground level. The water will drain from both holes, with faster drainage through the larger one (if one hole is larger). But this is equivalent to just having one hole that is somewhat larger than either of the other two. $\endgroup$
    – Ed V
    Feb 24, 2020 at 20:47
  • $\begingroup$ @EdV Yes, very aptly put: my frame of thought was something along those lines too, but more in the sense that the overall rate is simply the combined rates of each individual process, i.e. one ‘big’ rate... I understand it now, though! $\endgroup$ Feb 24, 2020 at 21:11
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    $\begingroup$ As you have worked out, species A can only decay as the sum of rate constants $k_1+k_2$. (It cannot sometimes decay with rate constant $k_1$ and at other times $k_2$). B appears with the total rate constant ($k_1+k_2$)as does C, however, the fraction of B produced is $k_1/(k_1+k_2)$ and the fraction of C $k_2/(k_1+k_2)$. $\endgroup$
    – porphyrin
    Feb 24, 2020 at 22:07

1 Answer 1


There is no need adding the number $2$ in front of the definition of the rate. So the rate equation is : $$-d[A]/dt = k_1 [A] + k_2[A] = (k_1 + k_2)[A]$$

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    $\begingroup$ Some of the comments posted already are more elaborate than this answer, which does not even address the main question. $\endgroup$ Feb 25, 2020 at 17:48

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