David A. Tanzer, August 17, 2020, in unit Epidemic Models 2.
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”.
Next time we will return to the SIR model.
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