True or false? A reaction mechanism is well-known for any reaction whose reactants and products are known.

If you know what a mechanism involves, then of course it would be false. If you know the stoichiometry of the reaction, then all you know are the starting reactants and the end products.

In other words, you know how it begins, and you know how it ends. But what happens in between is a complete mystery.

Let's take, for example, the bimolecular ozone destruction from McQuarrie's Physical Chemistry: A Molecular Approach. We can see that the overall reaction is:

#\mathbf(O_3(g) + O(g) -> 2O_2(g))#

If you only knew what was written above, then you would assume that the mechanism is for a one-step reaction, precisely like above:

1. #O(g)# collides with ozone in a particular orientation and knocks off an oxygen atom, creating two #O_2(g)# molecules as a result of the recombinations.

However, if we start involving #Cl(g)# in the mechanism, then we get two different mechanisms for the same reaction...

#O_3(g) + color(green)(Cl(g))=> color(green)(ClO(g)) + O_2(g)#
#color(green)(ClO(g)) + O(g) => O_2(g) + color(green)(Cl(g))#

1. Chlorine collides with ozone in a particular orientation to initiate the reaction as the catalyst, generating #ClO(g)# and #O_2(g)#.
2. The intermediate #ClO(g)# collides with a recently-knocked-off #O(g)# (that forms from a collision where the #Cl(g)# did not bond to the #O(g)#) to produce #O_2(g)# and regenerate the catalyst #Cl(g)#.
3. The reaction continues to repeat until the ozone is all gone.

Then, if we consider the overall reaction with chlorine catalysis and cancel out the intermediates and catalysts, we get:

Overall:

#O_3(g) + cancel(Cl(g)) => cancel(ClO(g)) + O_2(g)#
#cancel(ClO(g)) + O(g) => O_2(g) + cancel(Cl(g))#
#"---------------------------------------------"#
#O_3(g) + O(g) => 2O_2(g)#

which is the same reaction in general, but a different mechanism. Without #Cl(g)#, the reaction would not proceed like this.