# Why do alkenes (olefins) burn hotter than alkanes (paraffins), even though alkanes are denser and more hydrogen-rich?

Burning hydrogen releases more total energy than combusting carbon, and all alkanes are denser as well, so why are flame temperatures for alkene molecules higher than their equivalent alkanes?

P.S.: Also, is there a place on the web listing specific energies and energy densities for alkenes (olefins)?

• The more water vapor in flame the lower temp. Aug 5 '21 at 17:35
• Actually this is not entirely correct or a complete picture. Flames have been studied for a long time, even intentional injection of steam can increase flame temperature. For example this study concluded that "The results indicate that water vapor does not act as an inert diluent but instead inhibits carbon formation and gives rise to a greater heat release than in its absence." Combustion and Flame, 27 (1976), pg. 205-215 Aug 5 '21 at 19:25
• @M.Farooq nice comment. I would have answered like Mithoron, without reading this. Aug 6 '21 at 7:35

Flame temperatures are not primarily determined by the thermodynamics of the reactions in the flame

Intuitively it might seem that the temperature of a flame is dependent on the thermodynamics of the reactions occurring in the flame (eg total energy released, as the question assumes).

But flames are far more complex than a simple point of view based on equilibrium thermodynamic properties might suggest. Apart from anything else, flames are dynamic. And they depend strongly on the detailed conditions in the burner where the flame occurs. This should be fairly obvious from the known behaviour of flames in a bunsen burner. A well mixed flame where the fuel and air have the right ratio is very hot but the same fuel will produce a much cooler flame when insufficient air is mixed into it. The detailed conditions creating the flame matter.

But, for any given fuel, even well-mixed gases give different temperatures. But even this is not primarily determined by the thermodynamics of the reaction. Kinetics matters. The speed with which the oxygen can react with the fuel depends a lot on the nature of the fuel, so we probably should not expect the temperature to be strongly dependent on the simple thermodynamics of the reaction. If one reaction happens much faster than another, the flame could easily be much hotter even if the other reaction ultimately releases more energy (flames are dynamic and the temperature depends both on the amount of energy released and how fast it can be released).

And observations of how hot flames can be certainly suggest that kinetics might well dominate. Alkenes are more reactive than alkanes (and the flames are hotter) and alkynes more reactive still. So the hottest widely used flame in welding in the oxy-acetylene flame. Many casual explanations online seem to get the nature of this effect wrong, claiming that the acetylene triple bond "has a lot of energy to release" which is just wrong. But it is far more reactive than an alkane so that energy release probably happens much faster leading to a much hotter flame.

The main point is that you have to think about kinetics, flame conditions as well as thermodynamics to understand flame temperatures. And this means the results are more complex than a simple perspective based on just one factor suggests.

Double-bonds are less stable, i.e., their formation is less exothermic, so it takes less energy to break one double bond than two singles.

Or one triple vs. three single bonds. Consider triple-bonded acetylene -- there are not many hotter flames than from oxy-acetylene combustion... well, there is dicyanoacetylene, $$\ce{N≡C−C≡C−C≡N}$$, with three triple bonds in one compact molecule. Not quite up to plasma torch level, though.

See Derek Lowe's Things I Won't Work With for another stressed-out substance.