# Why is the oxidation of sulfur dioxide not termed as heat of formation of sulfur trioxide?

According to me, the following reaction's heat should be termed as heat of formation of sulfur trioxide:

$$\ce{SO2 + 1/2 O2 -> SO3}$$

However, in my textbook, they have that given it is not the formation reaction. Why is it so? And which is the formation reaction then?

• There are quite a few different reactions that end up in SO3. Which of them should be called formation? – Ivan Neretin Jun 24 '18 at 9:09
• Thanks for editing my question. After reading the edited question, I have come up with an answer(not sure if I am correct or not). The heat of formation of sulfur trioxide, instead of being the heat of reaction of the above reaction, is the sum of the heat of reaction of a two-step reaction(because the formation of sulfur trioxide is a two-step reaction). – Gordon Jun 24 '18 at 9:16
• @Gordon I don't think the number of steps is relevant. In fact, as Karl has mentioned in their answer below, it is entirely irrelevant whether it is possible to even produce the compound in that single reaction. – Gaurang Tandon Jun 24 '18 at 11:27
• @GaurangTandon Yes and no. If the actual synthesis of the product takes multiple steps then you can still calculate the heat of formation by adding up the details at each step. Yes the umber of steps doesn't matter, but real reactions of multiple steps can still get you to the answer. – matt_black Jun 24 '18 at 21:39

Heat of formation is the enthalpy difference to the elements in standard state as defined by IUPAC, which is in most cases their most thermodynamically stable form at STP.

It is completely irrelevant if the respective chemical equation makes any sense, i.e. it would work out in a laboratory. However high the number of intermediate steps to reach the desired product is, energy is always conserved.

• Always seems to be an exception. Black phosphorus is more thermodynamically stable than white phosphorus which was chosen as the standard state. – MaxW Jun 25 '18 at 2:40
• @MaxW Yes, i think white phosphorus is the only modification at STP which can be readily produced and has a well defined crystal structure and density. The polymeric forms (black, red, violett) are notorious for containing a huge amount of packing defects. – Karl Jun 25 '18 at 9:46

The heat of a reaction $\ce{A + B + ... -> X}$ is called the standard heat of formation of the product subject to a number of strict restrictions. These restrictions are:

1. Each of the reactants must be in their thermodynamically most stable state at the room temperature. For example, if you take bromine as a reactant, it must be in liquid state. If you take carbon, it must be graphite (s), etc. Also, polyatomic molecules must be written in complete molecules and not atoms (hence, $\ce{Br2}$ and not $\ce{Br}$)
2. Each of the reactants must themselves be elements and not compounds. For example, $\ce{C (s, graphite) + O2 (g) -> CO2 (g)}$ is correct, but $\ce{CO (g) + 1/2 O2 (g) -> CO2 (g)}$ is incorrect as $\ce{CO}$ is not an element.
3. There should be only one product and that too the one we're concerned about.
4. There should only be one mole of the product. Hence, $\ce{S (s) + 3/2 O2 (g) -> SO3 (g)}$ is correct, but $\ce{2S (s) + 3 O2 (g) -> 2 SO3 (g)}$ is incorrect (actually, the latter's heat of reaction is just double of the standard enthalpy of formation).

Can you now tell by which rule your given reaction fails to be a standard formation reaction?

Also, here's a fun little exercise: can you tell which of the following reactions are standard formation reactions?

1. $\ce{PCl5}$
• $\ce{P (white, s) + 5/2 Cl2 (g) -> PCl5 (s)}$
• $\ce{P (black, s) + 5/2 Cl2 (g) -> PCl5 (s)}$
• $\ce{P (white, s) + 5 Cl (g) -> PCl5 (s)}$
2. $\ce{KCl}$
• $\ce{K (s) + 1/2 Cl2 (g) -> KCl (s)}$
• $\ce{K (g) + 1/2 Cl2 (g) -> KCl (g)}$ (a homogenous mixture!)
• $\ce{2K (s) + Cl2 (g) -> 2KCl (s)}$
3. $\ce{O2}$
• $\ce{KClO3 (s) ->[\Delta] KClO (s) + O2 (g)}$
• $\ce{2Na2O (s) ->[\Delta] 4Na (s) + O2 (g)}$
• $\ce{H2O (g) + CO2 (g) -> 1/6 C6H12O6 (l) + O2 (g)}$