Conservation law(s) of chemical reaction $\rm A \longrightarrow B + 2C$

Question

I am trying to find the conservation law for the following chemical reaction:

$$\rm A \longrightarrow B + 2C$$

where:

• A converts into $B$ and $C$ at a rate $k (k>0)$.
• $[A]_0$, $[B]_0$, $[C]_0$ are initial concentrations of $A$, $B$ and $C$
• $[B]_0$ and $[C]_0$ are initially $\rm 0\ M$.

I am thinking the conservation laws are:

• $[B] + [A] = [A]_0$
• $[C] + 2[A] = 2[A]_0$

Am I correct? Or am I getting it wrong?

Answer 1

Let $a, b, c : \mathbb R_0^+ \to \mathbb R_0^+$ be the time-varying amounts of chemical species $A, B, C$, respectively. The chemical reaction $A \to B + 2 C$ is modeled by the following system of $3$ coupled linear ODEs

$$\frac{\mathrm d}{\mathrm d t} \begin{bmatrix} a (t)\\ b (t)\\ c (t) \end{bmatrix} = \kappa \begin{bmatrix} -1 & 0 & 0\\ 1 & 0 & 0\\ 2 & 0 & 0\end{bmatrix} \begin{bmatrix} a (t)\\ b (t)\\ c (t) \end{bmatrix}$$

where $\kappa > 0$ is the rate at which species $A$ decays. The matrix above, which we henceforth denote by $\mathrm M$, is rank-$1$. Thus, the left and right null spaces of $\rm M$ are $2$-dimensional. Via Gaussian elimination, we quickly conclude that the left null space of $\rm M$ is parametrized as follows

$$\left\{ \gamma_1 \begin{bmatrix} 1\\ 1\\ 0\end{bmatrix} + \gamma_2 \begin{bmatrix} 2\\ 0\\ 1\end{bmatrix} : \gamma_1, \gamma_2 \in \mathbb R \right\}$$

Let the initial conditions be $a (0) =: a_0$ and $b (0) = c (0) = 0$. Choosing different values for $\gamma_1$ and $\gamma_2$, we obtain different conservation laws. For example,

• $(\gamma_1, \gamma_2) = (1, 0)$ produces $\dot a + \dot b = 0$. Integrating, we obtain $a (t) + b (t) = a_0$.

• $(\gamma_1, \gamma_2) = (0, 1)$ produces $2 \dot a + \dot c = 0$. Integrating, we obtain $2 a (t) + c (t) = 2 a_0$.

• $(\gamma_1, \gamma_2) = (2,-1)$ produces $2 \dot b - \dot c = 0$. Integrating, we obtain $2 b (t) - c (t) = 0$.

• $(\gamma_1, \gamma_2) = (1, 1)$ produces $3 \dot a + \dot b + \dot c = 0$. Integrating, we obtain $3 a (t) + b (t) + c (t) = 3 a_0$.

The general conservation law is, thus,

$$(\gamma_1 + 2 \gamma_2) \, a (t) + \gamma_1 b (t) + \gamma_2 c (t) = (\gamma_1 + 2 \gamma_2) \, a_0$$

and the solution of the system of ODEs is

$$\begin{bmatrix} a (t)\\ b (t)\\ c (t) \end{bmatrix} = \left( \mathrm I_3 + \left( 1 - \exp( -\kappa t) \right) \mathrm M \right) \begin{bmatrix} a_0\\ 0\\ 0\end{bmatrix} = a_0 \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix} + a_0 \left( 1 - \exp( -\kappa t) \right) \begin{bmatrix} -1\\ 1\\ 2\end{bmatrix}$$