Question

Why does a catalyst effect the rate of reaction?

Answer 1

A catalyst speeds up the rate of reaction, either simply making it faster, or such that the reaction can readily occur, because it is essentially a reactive substance that alters the mechanism of the reaction to promote collisions in the reaction (the more optimal collisions, the faster the reaction).

Visually, it lowers the activation energy (or "energy barrier") within the reaction coordinate diagram for the reaction.

where `E_a` is the activation energy.

For example, in the following reaction mechanism for bimolecular ozone destruction from McQuarrie's Physical Chemistry: A Molecular Approach:

`O_3(g) + Cl(g) => ClO(g) + O_2(g)`
`ClO(g) + O(g) => O_2(g) + 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.

Without `Cl(g)`, the reaction would not proceed like this. Instead, it would proceed like this one-step mechanism:

`O_3(g) + O(g) => 2O_2(g)`

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.

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))`

vs.

`O_3(g) + O(g) => 2O_2(g)`

which is the same reaction in general, but a different mechanism (and thus a different rate law, but that's outside the scope of this question).