I am working with the following kinetic scheme that represents pyrolysis of a biomass particle:

biomass pyrolysis

The paper that discusses this scheme provides the following rate constants:

  • K1 for the pathway of $\text{Biomass} \rightarrow (\text{Volatiles + Gases})_1$
  • K2 for the pathway of $\text{Biomass} \rightarrow (\text{Char})_1$
  • K3 for the pathway of $(\text{Volatiles + Gases})_1 \rightarrow (\text{Volatiles + Gases})_2 + (\text{Char})_2$

The rate equation provided in the paper for the conversion of biomass is $$\frac{\mathrm d\,B}{\mathrm dt} = -(K_1 + K_2)\,B$$

The rate equation for the char 1 component is given as $$\frac{\mathrm d\,C_1}{\mathrm dt} = K_2\,B - K_3\,C_1$$

And finally the rate equation provided for the char 2 component is $$\frac{\mathrm d\,C_2}{\mathrm dt} = \delta\,K_3\,C_1$$

where $\delta$ is a deposition coefficient reported as 1.45 in the article. Note that all components of the system are on a mass basis such as $\mathrm{kg/m^3}$.

It is fairly straightforward to calculate the concentrations of $B$, $C_1$, and $C_2$ at each time step given an initial concentration of $B$. However, I would like to determine the concentrations of the volatiles and gases in the system too. Since it doesn't seem possible to calculate the individual volatile and gas components from the given scheme, I would like to determine the concentration of the (Volatile + Gases) group which could be labeled as G. So how would I develop the overall rate equations for B, C, and G?

$$ \frac{\mathrm d\,B}{\mathrm dt} = -(K_1 + K_2)\,B \\ \frac{\mathrm d\,C}{\mathrm dt} = K_2\,B + ? \\ \frac{\mathrm d\,G}{\mathrm dt} = K_1\,B + ? $$

  • $\begingroup$ (V+G)2 is the same as C2 and (V+G)1 is "straightforward" as you said. $\endgroup$ Mar 21, 2016 at 20:18
  • $\begingroup$ @pH13 So the concentration of group (V+G)2 is the same as the concentration of C2; basically (V+G)2 = C2? $\endgroup$
    – wigging
    Mar 21, 2016 at 20:21
  • $\begingroup$ I'd say so, yes. As far as those subscripts are only indices. $\endgroup$ Mar 21, 2016 at 20:22
  • $\begingroup$ @pH13 That makes sense to me but what I'm actually trying to do is figure out the individual components. From (V+G)1 or (V+G)2 what is the concentration of V and G in each group? $\endgroup$
    – wigging
    Mar 21, 2016 at 20:42
  • $\begingroup$ As you don't know the amount of "Gases" I don't think that you can split it up further. $\endgroup$ Mar 21, 2016 at 20:46

1 Answer 1


The Biomass-equation is: $$\frac{\mathrm d B}{\mathrm d t} = -(k_1 + k_2) B$$ The (volatile + gases)$_1$ equation is: $$\frac{\mathrm d VG_1}{\mathrm d t} = k_1 B$$ The (volatile + gases)$_2$ equation is equal to equation for (Char)$_2$: $$\frac{\mathrm d VG_2}{\mathrm d t} = \frac{\mathrm d C_2}{\mathrm d t} = k_3 C_1$$ (It might be that it needs to be multiplied by 1/2.)

And the equation for (char)$_1$ is this: $$\frac{\mathrm d C_1}{\mathrm d t} = k_2 B-k_3 C_1$$

The equations for the complete set of (char)s and (volatile+gases)s are then the sums of the solutions for $C_1$ and $C_2$ resp. $VG_1$ and $VG_2$.

For initial concentration of $B(0)=1$ and everything else $f(0)=0$, as well as $k_1=2$, $k_2=10$ and $k_3=3$, you should get something similar to this: enter image description here

  • $\begingroup$ Thank you, I'll try this in my model and let you know if I need more help. Also, what did you use to create the plot? And what books would you recommend for learning about systems of kinetic reactions and how to compute their yields, i.e. something that discusses modern techniques or numerical methods for chemistry? $\endgroup$
    – wigging
    Mar 22, 2016 at 14:35
  • $\begingroup$ The calculation and the plot is made with Mathematica 10. Unfortunately I don't have any books to recommend as I'm not doing this very often. $\endgroup$ Mar 22, 2016 at 15:28

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