# Epidemic Models 2, Part 6

David A. Tanzer, August 17, 2020

# Digression: reaction rates in chemistry

The same kinetics occurs in chemical reaction networks. To illustrate, suppose there were two types of molecules, A and B, and that one A and one B molecule could collide to form a C molecule. So the reaction would be written $$A + B \rightarrow C$$.

Now imagine that a large number of A, B and C molecules are in a closed container, and that it is in the form of a gas, with A, B and C freely flying around, and bumping into each other.

Suppose, for specificity, that initially there are no C molecules, and an equal number of A and B molecules:

• Count(C) = 0
• Count(A) = Count(B)

Then, every time an A and a B collide to form a C, count(C) goes up by 1 and count(A) and count(B) go down by 1.

Now we have a stew of molecules.

So, what is the rate at which the reaction $A + B \rightarrow C$ occurs, at any moment in time?

First , it doesn’t depend on C at all – a reaction will occur whenever an A and a B collide (more realistically: when they collide in a sufficient way for the reaction to occur; at the required angles and speeds and orientations – but we’ll leave all this out for now, and simplify by saying the reaction occurs whenever A and B collide).

Again, the rule is rate(reaction) = RC * Count(A) * Count(B).

And for the same reasons: if we double the number of A’s while holding B fixed, collisions between A and B are twice as likely to occur, and so the rate of the reaction will double.

Question: what about if we had a reaction $A + A \rightarrow B$, what would be the rate equation. Well, it would be:

• RC * Count(A) * Count(A)

In this case, doubling the number of A’s will quadruple the chances of a reaction taking place – twice as likely to find an A as the “left participant” in the reaction, and twice as likely to find one as the “right participant”.