Does anyone know the analytical solutions to the following rate laws:

1) $\frac{\mathrm{d}\rho_\text{W}}{\mathrm{d}t} = -\rho_\text{W} \cdot (K_1+K_2+K_3)$

2) $\frac{\mathrm{d}\rho_\text{T}}{\mathrm{d}t} = \rho_\text{W} \cdot K_2 - (K_4+K_5) \cdot \rho_\text{T}$

3) $\frac{\mathrm{d}\rho_\text{C}}{\mathrm{d}t} = \rho_\text{W} \cdot K_3 + \rho_\text{T} \cdot K_5$

As an example, I have determined the following analytical solution for this rate law:

$\frac{\mathrm{d}\rho_\text{W}}{\mathrm{d}t} = -\rho_\text{W} \cdot (K_1+K_2+K_3) \qquad \Rightarrow \qquad \rho_\text{W} = \rho_{\text{W}0} \cdot \operatorname{e}^{-(K_1+K_2+K_3) \cdot t}$

where $\rho_{\text{W}0} = $ initial density, $K = $ reaction rate constant, $t =$ time

The scheme for the reactions are:

  1. $\ce{wood} \xrightarrow{K_{1}}{} \ce{gas}$

  2. $\ce{wood} \xrightarrow{K_{2}}{} \ce{tar}$

  3. $\ce{wood} \xrightarrow{K_{3}}{} \ce{char}$

  4. $\ce{tar} \xrightarrow{K_{4}}{} \ce{gas}$

  5. $\ce{tar} \xrightarrow{K_{5}}{} \ce{char}$

Any help with the three rate laws above would be greatly appreciated.


2 Answers 2


This is the solution I get:

$$\begin{align}\rho _T(t) &= -\frac{K_2 \rho _{w0} \left(e^{\left(-K_1-K_2-K_3\right) t}-e^{t (-K4-K5)}\right)}{K_1+K_2+K_3-K4-K5}\\\rho _W(t) &= e^{\left(-K_1-K_2-K_3\right) t} \rho _{w0}\\\rho_C(t) &=\tiny{-\frac{\rho _{w0} e^{-t \left(K_1+K_2+K_3+K4+K5\right)} \left(K_2^2 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)+K_2 \left(K5 (K4+K5) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)+K_1 K5 e^{\left(K_1+K_2+K_3\right) t} \left(e^{t (K4+K5)}-1\right)-K_3 \left(-(K4+2 K5) e^{t \left(K_1+K_2+K_3+K4+K5\right)}+K5 e^{\left(K_1+K_2+K_3\right) t}+(K4+K5) e^{t (K4+K5)}\right)\right)+K_3 (K4+K5) \left(-K_1-K_3+K4+K5\right) \left(e^{t (K4+K5)}-e^{t \left(K_1+K_2+K_3+K4+K5\right)}\right)\right)}{\left(K_1+K_2+K_3\right) (K4+K5) \left(-K_1-K_2-K_3+K4+K5\right)}}\end{align}$$

Solving this by hand isn't that hard, as the coupling is unidrectional. So, we can solve for $\rho_W$ first, and substitute it in the equation for $\frac{d\rho_T}{dt}$. Now, we have an equation of the form $\frac{dy}{dt}=Ay+Be^{kt}$, which has a solution of the form $y=Ce^{At}+De^{kt}$, where C,D can be found by back-substitution. Now, we subtitute the values for $\rho_T, \rho_W$ in the last equation, and we get an equation of the form $\frac{d\rho_C}{dt}=f(t)$ ($f(t)$ here is some combination of exponentials), which can easily be solved by integrating $f(t)$


For what it's worth, problems like this can be solved very nicely with Mathematica and I find that it provides access to some really cool chemistry even when the math is a bit overwhelming for some students. One can obtain the same solution as @ManishEarth with the following Mathematica code:

soln = DSolve[
   wood'[t] == -wood[t] (k1 + k2 + k3),
   tar'[t] == wood[t]*k2 - (k4 + k5)*tar[t],
   char'[t] == wood[t]*k3 + tar[t]*k5,
   wood[0] == wood0,
   tar[0] == 0,
   char[0] == 0},
  {wood[t], tar[t], char[t]}, t]

The fun part is how easy it is to visualize the results:

Mathematica graphics

And there are ways to dynamically manipulate the plot such that one can explore the effect of the rate constants on the time-dependent concentrations.

enter image description here

The mathematical rigor of my solution is naturally a bit weak, but oftentimes my emphasis is on teaching the science.

  • 2
    $\begingroup$ Wow, thanks for pointing this out in Mathematica. I use Matlab for everything at work. Do you know if you can dynamically adjust parameters in Matlab like you show in Mathematica? $\endgroup$
    – wigging
    Aug 18, 2013 at 15:32
  • 2
    $\begingroup$ I haven't used Matlab for a number of years now. There is this code which provides some type of manipulative features, but I don't think Matlab is as advanced in this type of visualization as Mathematica is (at the moment, at least). $\endgroup$ Aug 18, 2013 at 15:50

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