# What is $Z$ in collision theory of chemical reactions?

According to collision theory given by Max Trautz and William Lewis the rate of chemical reaction is proportional to $$Z$$ (number of collision per second per unit volume of reaction mixture)

Suppose a,b and c react to form d.

Then would Z include every possible collision like:

Collision between two different molecules like a and b or band c or a and c.

Collision between two same molecule like between a and a etc.

Collision between all three reactants at same time.

And most importantly does $$Z$$ include collision between reactants and products like a collide with d?

Or does $$Z$$ only include all collision (effective or non effective) of all reactants a,b and c at same time?

Here is what I want to say:

• Would Chemistry be a better home for this question? Commented Sep 25, 2023 at 7:08
• @Qmechanic yes but collision theory is based Kinetic theory of gases so I thought physics SE is better place Commented Sep 25, 2023 at 7:19

## 2 Answers

Yes, normally Z would imply collision of all the three reactants in the same time, since thee molecules have to be close to each other in order for chemical transformation to occur. Consequently, the law of mass action would involve product of the densities of all the three reagents: $$K\sim [a][b][c].$$

However, if there is an intermediate product, then we can talk about pairwise collisions only, e.g. $$a + b + c = ab + c = d.$$ But in this case we are talking about two subsequent chemical reactions.

Note that the reactions with many intermediate products are quite common - e.g., the difference between combustion and conventional oxidation reaction is the former involves formation of many intermediate products, which makes it very fast. See Can we call rusting of iron a combustion reaction?

Such reactions are not usually elementary reactions, but there is typically some reaction schema, requiring particular time order of collisions.

Also, if an elementary reaction has just a single product, the reaction rate depends on collisions with other molecules to pass the reaction energy while honoring momentum conservation.

E.g. $$\require{mhchem} \ce{2 H(g) <=> H2^{*}(g) ->[collision]H2(g)}$$