I have parallel equations: \begin{align}\ce{A &->[k_b] B},&\ce{A &->[k_c] C}.\end{align} I understand how to determine formation of $\ce{A}$ (with the integration) but I really keep going in circles with integrating $\ce{B}$. If I can have some assistance in how to integrate $\ce{B}$, then I will be fine for integrating $\ce{C}$.

This is what I have: \begin{align} \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= k_b[\ce{A_0}] \cdot \mathrm{e}^{(-k_b+k_c)t}\\ \int_{[\ce{A_0}]}^{[\ce{A}]}\frac{\mathrm{d}[B]}{\mathrm{d}t} &= \int k_b\cdot[A] \mathrm{d}t\\ \ln\frac{[\ce{B}]}{[\ce{A_0}]} &= \int k_b\cdot[\ce{A_0}] \cdot \mathrm{e}^{-(k_b+k_c)\cdot t}\mathrm{d}t\\ \end{align}

To save the embarrassment, once I raise to the "$\mathrm{e}$" of both sides, I get \begin{align} \frac{[\ce{B}]}{[\ce{A_0}]} &= \mathrm{e}^{\frac{k_b\cdot[\ce{A_0}]}{(k_b+k_c)}t} \cdot \mathrm{e}^{\mathrm{e}^ {-(k_b+k_c)t}}\\ [\ce{B}] &=[\ce{A_0}] \mathrm{e}^{\frac{k_b\cdot[\ce{A_0}]}{k_b+k_c}t} \cdot \mathrm{e}^{\mathrm{e}^{-(k_b+k_c)t}}\\ \end{align}

Can someone show me the steps of integrating the $\int k_b\cdot[\ce{A}] \mathrm{d}t$? Very basic steps are available in solutions, but it still does not fully show the integration/algebra and I am just very lost.

  • $\begingroup$ I submitted an edit to improve the formatting of your post. Let me know if everything was rewritten as you intended. $\endgroup$
    – Tyberius
    Commented Apr 15, 2017 at 3:00
  • $\begingroup$ I have only corrected some of the bad MathJax, there might still be errors in the constants introduced by @Tyberius, please check again. $\endgroup$ Commented Apr 15, 2017 at 3:24

1 Answer 1


Start by solving the rate equation for $\ce{[A]}$: \begin{align} \frac{\mathrm{d}\ce{[A]}}{\mathrm{d}t} &= -k_\mathrm{B}\ce{[A]} - k_\mathrm{C}\ce{[A]}& \Longleftrightarrow&& \ce{[A]} &= \ce{[A]_0}\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t). \end{align} We can then solve the rate equation for $\ce{[B]}$: \begin{align} \frac{\mathrm{d}\ce{[B]}}{\mathrm{d}t} &= k_\mathrm{B}\ce{[A]} \\ &= k_\mathrm{B}\ce{[A]_0}\exp\left(-(k_\mathrm{B} + k_\mathrm{C}\right)t)\\ \ce{[B]}-\ce{[B]_0} &= \int_{\ce{[B]_0}}^{\ce{[B]}}\mathrm{d}\ce{[B]}' \\ &= k_\mathrm{B}\ce{[A]_0}\int_0^t\mathrm{d}t'\,\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t')\\ &= \frac{k_\mathrm{B}\ce{[A]_0}}{k_\mathrm{B} + k_\mathrm{C}}[1-\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t)]. \end{align} I use primes to distinguish the variable of integration from the limits of integration.

  • 1
    $\begingroup$ The expression given is not correct, one limit has been missed, I have edited please check $\endgroup$
    – porphyrin
    Commented Apr 15, 2017 at 9:46
  • $\begingroup$ Just to add, the term in front of the exponential ($kB/(k_B+k_C)$is the fraction going towards B. $\endgroup$
    – porphyrin
    Commented Apr 15, 2017 at 9:47
  • $\begingroup$ Thank you @a-cyclohexane-molecule. Can you just lastly explain the [B]}-[B]_o and where that comes from? I know the ln[B]/[A]0 = ln[b] - ln[A]o but if you "e" both sides, I don't see how you get your equation, and then I am back to the original problem so there must be a step I do not know. And in the integration you ended up with a one, but I dont understand how that comes about. $\endgroup$
    – juliodesa
    Commented Apr 15, 2017 at 13:46
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    $\begingroup$ Sorry about the mistake; thanks for fixing it. @juliodesa, $\ce{[B]}$ and $\ce{[B]_0}$ are the bounds of our integral -- we start with $\ce{[B]_0}$ concentration of species B, and have $\ce{[B]}$ concentration of species B at time $t$. The one comes from evaluating the integral at the lower bound of $t=0$. $\endgroup$ Commented Apr 15, 2017 at 14:01
  • $\begingroup$ and also note that the rate constant at which B appears is $k_b+k_c$ and similarly for appearance of C and disappearance of A, this is true even though the rate constant to B is $k_b$ and that to C, $k_c$. $\endgroup$
    – porphyrin
    Commented Apr 15, 2017 at 21:33

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