Conservation rule in closed systems

Question

In a closed system the reacting species interconvert but do not leave the system. Therefore the amounts of different species in the system would be bound by a conservation rule.

A simple example: $$\ce{A -> B}$$

Here the conservation rule would be: $\ce{A + B = constant}$, which says that at any point of time during the reaction the sum of the amount of both the species would be constant.

A more complex example $$\ce{2A -> B}$$

Here, the conservation rule would be: $\ce{A + 2B = constant}$.

Let's take a more complex system:

$$\begin{align} \ce{A &-> B} \\ \ce{2B &-> C} \\ \ce{C &-> D + E} \\ \ce{2A + C &-> F} \end{align}$$

From empirical analysis I can deduce that the conservation rule would be: $$\ce{A + B + 2C + D + E + 4F = constant}$$

However, I am not able to deduce a mathematical equation that gives these co-efficients. I assume that the stoichiometry matrix may give a clue but I am not sure. The stoichiometry matrix for the above system is:

$$\mathbf{S}=\begin{bmatrix} -1 & 0 & 0 & -2 \\ 1 & -2 & 0 & 0 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

This is relatively easier to figure out empirically as there is only one starting reactant (A). For reaction systems with multiple starting reactants (as shown below), the deduction of conservation rule(s) becomes even more difficult.

$$\ce{aA + bB -> cC + dD}$$ The problem seems quite elementary but I am not able to solve it. Am I missing something obvious?

Answer 1

With 6 components and 4 reactions, you must have two conservation rules (that is, if the reactions are independent, which they are).

Mathematically, we find ourselves in a 6D space (as there are 6 components), we move in a 4D subspace defined by the 4 reactions, and we need a basis for the 2D subspace normal to it, which is precisely where we can't go. The basis vectors will be our conservation laws.

You may use any CAS or your own hands to solve the problem. In Wolfram Mathematica, this is done as follows:

NullSpace[{
{1, -1, 0, 0, 0, 0},
{0, 2, -1, 0, 0, 0},
{0, 0, 1, -1, -1, 0},
{2, 0, 1, 0, 0, -1}
}]

As expected, the answer includes two vectors. One is $\{1, 1, 2, 2, 0, 4\}$. Another is $\{0, 0, 0, -1, 1, 0\}$, which translates to $\rm [D] - [E] = const$ ($0$ if there were no initial amounts), which is obviously true, since both are produced in the same reaction and never spent. That's your second conservation law.

BTW, the sum of the two vectors corresponds to the conservation law you already know. A basis can be chosen in a multitude of ways, after all.

That's what you do in any problem of this sort. That's linear algebra; it is about as simple and robust as a cleaving axe. It will never run out of gas, or freeze because its usage license has expired, or refuse to chop the incompatible firewood. It just works.

So you prepare the stoichiometry matrix (which you already did; I just transposed it), find the normal subspace, translate the basis back to your compounds. That would be it. The notion of time is irrelevant. The notion of differential equations is irrelevant. The notion of limiting reactants is irrelevant. It is just vectors and spaces. They do not even know you apply them to chemistry, nor do they need to.

So it goes.