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210

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: $$\ce{... 104 This is the reaction that occurs when hydrogen combusts:$$ \ce{2H2 + O2 -> 2H2O} $$Similarly, this is the combustion reaction for methane, a representative fuel:$$ \ce{CH4 + 2O2 -> CO2 + 2H2O} $$Water is a product in both of these reactions. Thus, water represents something that has already been oxidized by oxygen, and as such there is little ... 79 Combustion of small materials, such as a match or birthday candle, actually involve the release of volatile vapours, which themselves burn. It is not the solid material that burns. There needs to be a minimum amount of volatile material present in this combustion zone (just above the burning match) for the ignition to occur. As the combustion process ... 48 Combustion is a gas phase reaction. The heat of the flame vapourises the substrate and it's the vapour that reacts with the air. That's why heat is needed to get combustion started. Anyhow, wood contains lots of relatively volatile compounds so it's not too hard to get combustion started. Once combustion has started the heat of the flame keeps the reaction ... 39 \ce{H2O2} exists, and could be what you expect by burning water (since burning is really oxidizing, or adding oxygen.) However, in stoichiometric proportions, here is what would happen: Either further burning with outside oxygen$$ \ce{2H2O + O2 <--> 2H2O2} $$or decompose water to get oxygen, and reject hydrogen:$$ \ce{2H2O <--> H2O2 + ...

38

Actually... yes! Iron(II) oxide is thermodynamically unstable below $848~\mathrm K$. As it cools down to room temperature (it has to do it slowly) it disproportionates to iron(II,III) oxide and iron: $$\ce{4FeO -> Fe + Fe3O4}.$$ The iron is in a form of a fine powder, which is pyrophoric (it may catch a fire when exposed to air). You can see it in ...

32

A fire has no way to directly "draw in" oxygen for it to burn. It consumes what oxygen is in the immediate vicinity of the flame, depleting the air in oxygen compared to the concentration of oxygen in air further away. This sets up a concentration gradient such that oxygen will diffuse into the depleted region near the fire. An even more important process ...

32

The wick temperature does not have to be the same as the flame temperature.The flame is hottest at the bottom, but the wick is hottest at the top. For a candle, the wick burning isn't the intended purpose of the wick; light comes from burning wax (more generally: fuel), you want to burn the wax not the wick. Rather the purpose of a wick is to help fuel ...

28

The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a dissertation that presents a nice summary of the process here (this is a 1MB PDF). A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some ...

27

Since I will deal with all of the alkali metals in this answer, I think the question should also be broadened. There is no point in covering one single metal (sodium) without touching the others since it is the trend going down the group that we are interested in. All thermodynamic data is taken from Prof. M. Hayward's lecture notes at Oxford. So, firstly, ...

27

Because fire is not the same thing as light. Michael Faraday did a wonderful job of explaining how the candle works, and I direct you to look at it (there are also Youtube videos giving a modern take on this work) if you're interested. In short, the candle produces light, not because it is hot, but because it is sooty. The particles of soot glow when they ...

24

In normal candles, as you blow them out, you will see burning embers in the candles, which vaporises the wax and thus causes a ribbon of paraffin wax to rise up (this looks like a wisp of smoke). These embers are often not hot enough to light the ribbon of wax, so the candle goes out. However, in trick candles there is magnesium put in the wick. When ...

21

Black powder is a mixture of solids. As solids, they are not particularly inclined toward fast reactions between each other. Elevated temperature generally liven things up, because it allows diffusion of solids into each other. However, both carbon (charcoal) and potassium nitrate are relatively stable compounds, so even when potassium nitrate is melted, ...

21

This video shows how a black flame is achieved. If you illuminate the fire with a monochromatic light source (sodium vapor lamp) and introduce a species in the fire that absorbs that wavelength (sodium ions) then the fire will in fact appear black under the illumination. A screen shot is shown below: Click image for video.

19

The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Then not only methane is burnt, but also the arised hydrogen. So in "first" step: $\ce{CH_4 + O_2 -> CO_2 + 2H_2}$ but then the hydrogen will be also ...

18

Many volatile liquids are not combustible Dichloromethane (DCM) is a widely used solvent by chemists. It boils at around 40°C (the same as diethyl ether) but is not remotely combustible or flammable. Ether is both very volatile and very flammable, so much so that most labs would prefer not to have it used anywhere where flames or sparks could be present. ...

16

Most matches these days are safety matches: they're designed to need something more than ordinary levels of friction to ignite, by splitting the combustion materials between the match-head and the striking strip, i.e. the brownish paint thing down the side of the matchbox or across the front of the matchbook. Most "regular" matches now are safety matches. ...

16

Combustion is a gaseous phase phenomenon. Oil and gasoline have a high enough vapor pressure at ambient temperatures to produce a gaseous phase of fuel above the liquid. In contrast, hold a lighter up to a piece of wood and try to get it to light. It won't—at least not for quite some time. This is because solid fuels must first undergo endothermic pyrolysis ...

16

Yes, diamond will combust in air. Regardless of the ambient air temperature, e.g. your example of $21\ \mathrm{^\circ C}$, you of course have to heat it to it's ignition temperature somehow, whether in a furnace, a flame, etc. The autoignition temperature for diamond is around $900\ \mathrm{^\circ C}$ (source 1, source 2), compared to about $730\ \mathrm{^\... 16 Higher$\mathrm{RON}$seems possible. Yet the boiling point rises accordingly while heat of combustion remains roughly the same. Here are two compounds that fit all posed criteria. \begin{array}{|c|c|c|c|c|} \hline \mathbf{Molecule} & \mathrm{mp\ \mathrm{(^\circ C)}} & \mathrm{bp\ \mathrm{(^\circ C)}} & \Delta_\mathrm cH_\mathrm m^\circ\ (\... 15 Well first off, pure ethanol is hygroscopic; it attracts water, to the point that it will pull it out of the air. Ethanol and gasoline will mix, but ethanol, gasoline and water will not; the ethanol-water mixture will come out of solution and settle on the bottom of your tank. Add a little oxygen to the mix, and you get rust. However, the more common side ... 15 No, the paper will not burn without oxygen being present. Paper is made primarily of cellulose which is a polymer of glucose. If you heat paper in a vacuum the cellulose simply decomposes to$\ce{H2O}$,$\ce{CO2}$,$\ce{CO}$and carbon. As the paper decomposes it will "char" or turn brown to black as the cellulose polymer degrades. Here is a link to an ... 15 The fuel for the chemical reaction is the CH4 (in this case) and the oxygen. Assuming the CH4 isn't going to run out any time soon, the chemical reaction will continue as long as there is enough oxygen accessible. The force behind fire extinguishers can replace the oxygen available to the fire in its immediate area with CO2, and without one of the reactants ... 14 Compounds A-D all have the same molecular formula,$\ce{C6H12}. We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation $$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$ Any differences in the heat given off can be used to say that a compound is more stable (it ... 12 This is actually just one part of a much more general phenomena, which is the release of large, polycyclic aromatic compounds due to pyrolysis or combustion of organic matter. It is not specific to wood. So, to understand why these compounds form in a wood fire, or any other kind of fire, we must first understand what wood is, and then we must understand how ... 12 The butane and oxygen in the room are in a metastable state. An example energy diagram of a metastable state is shown in the image below. Before throwing in the cigarette, the butane and oxygen molecules are in the state labeled as 1. This state is metastable because there is a lower energy state (which makes it statistically more probable) available if ... 12 During pyrolysis, organic compounds are thermally decomposed in the absence of oxygen. The pyrolysis products are classified into categories based on their physical state of existence: char (solid), bio-oil (liquid) and non-condensable gases (gas). The relative proportions of these three product fractions significantly vary depending upon the process ... 12 Yes, absolutely. Burning refers to oxidation accompanied by a flame and there are numerous reactions in which oxygen takes no part. Most of these are either with elementary halogens, or compounds thereof. Here are some examples. \begin{align} \ce{2K + F2 &-> 2KF}\\ \ce{2Fe_\text{wool} + 3Cl2 &-> 2FeCl3}\\ \ce{H2O + ClF3 &-> HF + HCl ... 12 Curt F. is probably correct in his estimate that heptafluoropropane extinguishes fire primarily by physical means. According to an overview by Choy and Fong, 'An Introduction to Clean Agents: Heptafluoropropane' (Int. J. on Eng. Performance-Based Fire Codes, vol. 5, nr 4, p. 181-$184,$2003\$), For heptafluoropropane, the contribution of physical ...

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