Inorganic Chemistry/Catalysis

Introduction
Catalysis is the ability of some species to rapidly speed up the rate at which a chemical reaction proceeds. For historical reasons, the discipline is normally split into two sub-categories; homogeneous (homo = same, geneous = phase) and heterogeneous (hetero = different). Homogeneous catalysis is concerned with catalysts that are in the same phase as the chemical reactions they are speeding up. These reactions are normally in the liquid phase and include all of biology's enzymes. While the majority of homogeneous catalysis is in the liquid phase there are gas phase and solid phase homogeneous catalytic reactions. Heterogeneous catalysts have a catalyst that is in a different phase. This type of catalysis is responsible for the vast majority of 'bulk' chemicals that are produced each year that go into making all the things we take for granted around us such as plastics, and are also extensively used for refining oil in gasoline. This chapter focuses on heterogeneous catalysis and specifically how and why they work at all.

The key concept in catalysis, and indeed all of chemistry, is the tension between the thermodynamics of a particular reaction which tells you whether something should happen, and the kinetics which tells you how fast something will happen. This tension in the world around you is everywhere i.e. thermodynamics says the diamond on an engagement ring should turn into graphite, a more stable allotrope of carbon but the kinetics is such that you would be waiting a very, very long time before it happened! Catalysis works by giving chemical reactions that should happen a way to actually happen.

Conceptual Analogy
The standard way to think of a chemical reaction happening is to imagine the energy of a molecule as being represented by a z co-ordinate while the reaction 'proceeds' in the x-y plane. If this sounds confusing, just think of the energy as being like the height of land features like mountains relative to their position. A molecule in this landscape would be a place where you can't go downhill without having to go uphill first, like a crater lake. Even though on a global scale, a crater lake should flow down to the sea, the local conditions around it mean that it would have to first flow uphill, which we could postulate doesn't happen on a regular basis. This analogy helps to visualise what is going on in a chemical reaction: the atoms involved when they're joined in one fashion are in an energy 'lake' and aren't able to change, but given the right conditions, such as temperature or photons, they are able to get over the energy barrier and flow down to another energy 'lake' which has the atoms joined in another fashion. This energy barrier is called the activation energy. In our analogy, imagine drilling a hole from one lake to another lake, since the water doesn't have to go 'up and over', it can flow easily to another position. This is exactly what a catalyst does, it doesn't change the total energy of the system or where the system will eventually end up (the thermodynamics dictate this), it just provides an easier way for the system to get there and hence speeds it up (it increases the rate). This analogy paints a picture that helps you remember the text-book definition of a catalyst: a substance which increases the rate at which a thermodynamic equilibrium is obtained by lowering the activation energy of the reaction pathway.

That is what a catalyst does but it doesn't give us any idea of how. It is only useful in deciding whether you are observing catalysis in action, the mechanism of how heterogeneous catalysts work is far more subtle.

Mechanism
According to Surface absoprtion theory heterogeneous catalysis has five stages:

The ability for an atom or molecule to stick to the surface is known, brilliantly, as the Sticking Co-efficient. This is just the ratio or percentage of molecules that end up sticking on the surface.

Examples
In the contact process vanadium [V] oxide (V2O2) is solid whereas the reactants SO2 and O2 are gaseous.

More detailed method:

2V2O5(s) + 2SO2(g)⇔ 2SO3(g) + 2V2O4(s)

2V2O4(s) + O2 (g) ⇔ 2V2O5 (s)

Therefore: 2SO2(g) + O2(g)⇔ 2SO3(g)