I have a question regarding the rates of parallel reactions. I have gathered experimental data for the overall reaction (i.e. conversions and selectivities and specific times) but due to the two reactions running parallel to each other for this reaction means I can't use a simple integrated rate plot.

Say for

$$ \ce{A ->[$\large k_1$] B} \\ \ce{A ->[$\large k_2$] C} $$

By substitution and integration, I have found that:

$$ [A](t) = [A](0) \mathrm{exp}(-(k_1+k_2)t) $$

and then the following equations can be derived in terms of the products formed:

$$ \frac{d[B]}{dt} = k_1 [A](0) \mathrm{exp}(-(k_1+k_2)t) $$

which gives

$$ [B](t) = \frac{k_1[A](0)}{k_1+k_2}(1-\mathrm{exp}(-(k_1+k_2)t) $$

when integrated. Also,

$$ [C](t) = \frac{k_2[A](0)}{k_1+k_2}(1-\mathrm{exp}(-(k_1+k_2)t) $$

is the equation for the other parallel reaction.

My initial thoughts were to plot the graphs (as I know the values for $[C]$, $[B]$, and $[A]$) and form some simultaneous equations, but this is an exponential plot and so i'm not sure how this would work.. or if I could even plot it, as I don't know $k_1$ and $k_2$ - that is what I need to determine.

  • $\begingroup$ To clarify, you know the values of [A](t) and either [B](t) or [C](t) for some value of $t$ other than zero? $\endgroup$
    – Tyberius
    Mar 25, 2017 at 22:51

1 Answer 1


One way you could get $k_1$ and $k_2$ is to first find their sum using your equation for $[A](t)$ at a known $t$.

Once you have done that, you can solve for $k_1$ using your equation for $[B](t)$ at a known $t$, since you now have the denominator of your first factor. You could alternatively find $k_2$ using $[C](t)$.

Once you have $k_1$, then $k_2$ is easily found from the sum.

  • 1
    $\begingroup$ Yes that is how I have done this.. well sort of - I found the total k from equation [A]t by a first order rate plot, and then I used [B]t/[C]t = k1/k2, and used an average of the concentration ratios (as I had a few values for different t values) and then solved the two equations to get the values! Thanks though :D $\endgroup$ Mar 26, 2017 at 15:44
  • $\begingroup$ As B and C always appear with the same rate constant ($k_1+k_2) as A disappears. The only way to get individual rate constants is to look at the relative proportions (i.e. yields) of B and C produced. $\endgroup$
    – porphyrin
    Mar 26, 2017 at 21:16

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