Hydrogen is flammable, and for any fire to burn it needs oxygen. Why does a compound made of hydrogen and oxygen put out fires instead of catalyzing them? I understand that hydrogen and water are chemically different compounds, but what causes water to be non-flammable?

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    $\begingroup$ So if wood burns, why don't the ashes burn? $\endgroup$ Commented Dec 19, 2016 at 6:46

7 Answers 7


You can think of water as the ash from burning hydrogen: it's already given off as much energy as possible from reacting hydrogen with oxygen.

You can, however, still burn it. You just need an even stronger oxidizer than oxygen. There aren't many of them, but fluorine will work,

$$ \ce{2F2 + 2H2O -> 4HF + O2} $$

as will chlorine trifluoride:

$$ \ce{ClF3 + 2H2O -> 3HF + HCl + O2} $$

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    $\begingroup$ I wonder what burning water looks like. $\endgroup$
    – knoftrix
    Commented Apr 27, 2019 at 3:44
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    $\begingroup$ @SauravSingh, given the nature of the chemicals I've listed, I wouldn't want to be anywhere near a fire that's burning water. $\endgroup$
    – Mark
    Commented Apr 27, 2019 at 4:22
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    $\begingroup$ Burning water with an excess of ClF3 with "complete combustion" will not leave HCl it will leave Cl2. In fact, 2HCl + 1/2O2 => H2O + Cl2 can happen in the gas phase with a suitable catalyst. Cl2 and O2 trade off as stronger oxidizer depending on what is being oxidized, but F2 and ClF3 are much stronger. $\endgroup$ Commented Jul 31, 2021 at 23:25

This is the reaction that occurs when hydrogen combusts:

$$ \ce{2H2 + O2 -> 2H2O} $$

Similarly, this is the combustion reaction for methane, a representative fuel:

$$ \ce{CH4 + 2O2 -> CO2 + 2H2O} $$

Water is a product in both of these reactions. Thus, water represents something that has already been oxidized by oxygen, and as such there is little further energetic gain to be made by trying to react it again with oxygen.

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    $\begingroup$ I maybe reading to much into your wording - "little further energetic gain". Should it be "no further energetic gain"? I am not a chemist, so this is just a point of curiosity. $\endgroup$ Commented Dec 12, 2016 at 21:17
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    $\begingroup$ @DigitalTrauma Ah, yes, sorry. That phrasing was rhetorical, not scientific. I can't imagine any products of a water/oxygen reaction that would have lower energy. $\endgroup$
    – hBy2Py
    Commented Dec 12, 2016 at 21:23
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    $\begingroup$ Sodium and Magnesium come to mind as well, @Mark $\endgroup$ Commented Dec 12, 2016 at 23:11
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    $\begingroup$ The OP asked why water puts out fire. Many fuels need to be hotter than 100 C to burn. Water cools the fuel and combustion stops. Don't try this with hydrocarbon fires, though. $\endgroup$ Commented Dec 12, 2016 at 23:16
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    $\begingroup$ @NateLockwood Re "Sodium and Magnesium": Those do not "burn water" (by capturing the hydrogen's electrons better than oxygen does), but on the contrary reduce the oxygen by providing electrons even better than the hydrogen does. They get burned by the oxygen in the water, but they do not burn the water's hydrogen. Quite the contrary. $\endgroup$ Commented Dec 13, 2016 at 12:09

$\ce{H2O2}$ exists, and could be what you expect by burning water (since burning is really oxidizing, or adding oxygen.)

However, in stoichiometric proportions, here is what would happen:

Either further burning with outside oxygen

$$ \ce{2H2O + O2 <--> 2H2O2} $$

or decompose water to get oxygen, and reject hydrogen:

$$ \ce{2H2O <--> H2O2 + H2} $$

To determine if that can work, we can simply compute the energy result:

  • $\ce{H-H}$ bond energy is $432$,
  • $\ce{H-O}$ is $459$,
  • $\ce{O-O}$ is $142$,
  • $\ce{O=O}$ is $494$

(all in $\mathrm{kJ/mol}$)

Also, $\ce{H2O}$ is $\ce{H-O-H}$ and $\ce{H2O2}$ is $\ce{H-O-O-H}$.

For the first equation, we have:

  • $2 \times 2 \times \ce{H-O} + \ce{O=O} = 2 \times 2 \times 459 + 494 = 2330$
  • $2 \times (\ce{H-O} + \ce{O-O} + \ce{O-H}) = 2 \times (459 + 142 + 459) = 2120$
  • Result is $-210\ \mathrm{kJ/mol}$

For the second equation:

  • $2 \times 2 \times \ce{H-O} = 2 \times 2 \times 459 = 1836$
  • $(\ce{H-O} + \ce{O-O} + \ce{O-H}) + \ce{H-H} = (459 + 142 + 459) + 432 = 1492$
  • Result is $-344\ \mathrm{kJ/mol}$

So we see that both reactions are endothermic, so what is more likely to happen is:

$$ \ce{H2O2 + H2 -> 2H2O} $$ $$ \ce{2H2O2 -> 2H2O + O2} $$

As a matter of comparison:

$$ \ce{2H2 + O2 -> 2H2O} $$

Has the following energy:

$$2\times 432 + 494 \ce{->} 4\times 459$$

This is an exothermic reaction with an energy of $+478\ \mathrm{kJ/mol}$

  • $\begingroup$ It's really great to see the numbers, here, however, as a non-chemist, I am confused. Why does the reaction move towards the compound with higher energy? Doesn't entropy cause systems to tend towards low-energy states? $\endgroup$
    – fuzzyTew
    Commented Dec 18, 2016 at 12:32
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    $\begingroup$ @fuzzyTew the energy in the bound is the energy released during the reaction when the bound is formed, which is also the amount of energy you need to provide to break it. So the lower energy bounds are the first to break, and the system tends to go toward high-energy bounds, because they are hard to break. $\endgroup$
    – njzk2
    Commented Dec 18, 2016 at 17:07
  • $\begingroup$ I see. Comparing with @Paraquat's answer, I think the energy of the bond or molecule system itself would be the negative of the number you state in this answer. D_e = 432 kJ / mol. $\endgroup$
    – fuzzyTew
    Commented Dec 18, 2016 at 17:32

This becomes intuitive once you accept that any chemical compound will 'want' to be in the state with the lowest energy, the ground state for a given temperature and pressure. For example, if you plot the energy of the system of two hydrogen atoms against the separation of the atoms, it looks like this:

Potential energy surface for hydrogen

So when you bring together two hydrogen atoms, they form a molecule and release an amount of energy $D_e$. The $\ce{H2}$ molecule is the ground state for hydrogen, so at room temperature and pressure, you would only expect to see hydrogen gas in the form of molecules. It's difficult to draw an analogous graph for water because the molecule has more degrees of freedom (in fact, you would need to draw a seven-dimensional graph), but you would find $\ce{H2O}$ as the ground state of a system with two parts hydrogen to one part oxygen.

So back to the question: why does water put out fire instead of catalyzing it? It really depends on what you are burning. Let's assume it's wood, which is mostly carbon and hydrogen, but for the sake of simplicity, we'll just use methane, which is similar:

$$\ce{CH4 + O2 -> CO2 + 2H2O + energy}$$

The energy released on the right-hand side of the equation is the 'burning'. The ground state of most systems involving carbon, oxygen and hydrogen is some combination of carbon dioxide gas and water. This combination has the lowest energy, and to get anything else, you would have to pump energy into the system.

However, water isn't the ground state of all chemical systems, for example, sodium, hydrogen, and oxygen. This can be seen if you Google a video of sodium reacting with water. In this case, the ground state of the system is sodium hydroxide and hydrogen gas, so the water does 'burn' in this case:

$$\ce{2Na + 2H2O -> 2NaOH + H2}$$


Because the hydrogen in water is already oxidized. When magnesium burns in water, the magnesium is being oxidized because the magnesium takes (binds with) the oxygen from the water molecules. This causes the release of hydrogen gas, which would/could burn in air (air is about 1/5 oxygen, 4/5 nitrogen) to form water again.


When different chemicals react, they form products that have completely different properties than what went into them.

In this case, hydrogen and oxygen are both very reactive, which is why they react together (or 'combust') to form water. The water that is formed is much less reactive, and does not easily react with oxygen, and so will not burn.

Because water is so non-reactive, it's interaction with the fire is entirely physical. It absorbs a lot of heat very quickly and prevents oxygen from getting to the fuel. Between those two effects, it's the simplest way to put out a wood fire.


Water is already burnt hydrogen: $$\ce{H2 + 1/2O2 -> H2O}$$

The reaction of oxidizing water ever more (to peroxide, for example) is non favorable and the peroxide will decompose as:

$$\ce{H2O2 -> H2O + 1/2O2}$$

with ΔH of −98.2 kJ/mol and a ΔS of 70.5 J/(mol·K)

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    $\begingroup$ This is true but doesn't really answer the question directly. $\endgroup$
    – bon
    Commented Oct 28, 2017 at 11:04

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