I can imagine one aspect of flammability (the ability of a substance to ignite on fire) to include calculating the kinetics and thermodynamics of the reaction of the system with oxygen. This is not generalizable though, since in some systems (e.g. alkali metals) there is no need for oxygen in its pure form for the system to catch on fire. What is the "general" way to determine flammability?

  • 1
    $\begingroup$ Very interesting question, the rare bookmark for me. $\endgroup$ Oct 26, 2020 at 15:13
  • 2
    $\begingroup$ While not a full answer at all, I'd like to point out that computational generation of the burning of a specific molecule is an active field of research, a recent review can be found here: sciencedirect.com/science/article/abs/pii/S0360128520300964 . In short, we can generate a reasonable potential energy landscape with combinatorical methods, reduce that mechanism using some advanced tools from the theory of differential equations, solve the so-called master equation on the reduced mechanism, and plug the chemical rate constant into combustion modeling tools to get real properties $\endgroup$
    – user430
    Apr 26, 2021 at 15:25

1 Answer 1


Unlike other properties of matter such as electrical conductivity and heat capacity, which can be defined quite precisely, flammability is more like "hardness" which is more vaguely defined (for example using Mohs hardness scale).

The flammability of a substance may seem to be categorized rather arbitrarily, for example the HMIS (Hazardous Materials Identification System) classifies flammability as follows:

Rating Description Examples
0 Liquids or non-aerosols that don't burn under "normal" conditions. $\ce{H2O}$, $\ce{CCl4}$,$\ce{CO2}$
1 Liquids or non-aerosols that will only burn above 94°C. Canola oil
2 Liquids or non-aerosols that will only burn between 38-94°C *OR* aerosols with a heat of combustion $\le 20$ kJ/g. Diesel fuel
3 Liquids or non-aerosols that will only burn between 23-38°C *OR* with a boiling point at or above 38°C and flash point below 23°C *OR* aerosols with a heat of combustion $\le 30$ kJ/g but $\gt 20$ kJ/g. $\ce{C3H6O}$ (acetone)
4 Liquids or non-aerosols with a boiling point below 38°C and flash point below 23°C *OR* aerosols with a heat of combustion $\gt 30$ kJ/g. $\ce{C4H10}$ (butane)

So the answer depends not only on the classification system you're using to define flammability, but can also depend on other things such as whether the substance is an aerosol, or a non-aerosol / liquid.

Let's start with aerosols since the HMIS classification of flammability is more precise in this case, since it's based on the heat of combustion (energy released during combustion, divided by the amount of it). We do need to define "combustion" though, and contrary to the aspect described in your question, which is the "ability to ignite fire": combustion does not always result in fire. It is just a high-temperature exothermic redox reaction involving fuel and an oxidant. In fact alkali metals are sometimes considered to be nonflammable (again contrary to what you wrote in your question, but also contrary to the HMIS flammability classification, so this is a good example of the lack of standardization in these definitions) but they are combustible.

You are correct though, that the oxidant does not have to be oxygen, for example it could be chlorine (hydrogen burns in chlorine), fluorine, chlorine trifluoride, nitrous oxide, nitric acid, etc.

Now we've learned that determining flammability of an aerosol may be practically impossible, because there is no way to check every single possible candidate for the oxidant.

However if we make the question more specific, for example "how can we computationally determine combustibility of an aerosol with in the presence of an oxygen, chlorine, nitrogen or fluorine oxidant?" then it becomes easier (but still difficult).

You would computationally study the reaction:

$$ \ce{Fuel + Oxidant → Products}\tag{1} $$

for example in the case of alkali metals, which you mentioned in your question, we could have:

$$ \ce{6Li + N2 → 2Li3N}\tag{2}. $$

Now all you would have to do would be to computationally calculate how much energy is released in that reaction, for which you can get a decent estimate with the help of an software, which will help you get estimates on the energies of each substance in the equation.

For liquids or non-aerosols, the situation is more complicated and your best route might be to follow a procedure for doing a simulation such as what I described in my answers to:


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .