[GeLe5000]

Hello.

 I've found this on a website:

 "The combustion reactions are therefore clearly exothermic. The temperatures reached during combustion reactions depend on this exothermicity but also on the complexity of the reaction. Thus, although the combustion of ethane
 C2H6 +  7/2 O2 ----> 2 CO2 +  3 H2O
 is about 15% more exothermic  than the combustion of acetylene
 C2H2 + 5/2 O2  ----> 2 CO2 + H2O
for the same volume of burned gas, the flame of acetylene is much hotter than that of ethane. This is explained by the fact that the heat produced by the combustion of ethane is divided into five molecules (2 CO2 et 3 H2O) against only three molecules (2 CO2 and 1 H2O) in the case of acetylene. "

Indeed dH = - 341 kcal / mole for ethane, - 300 kcal / mole for acetylene. (Calculated using enthalpies of formation)

 However, the heat produced by combustion, which is equal to the enthalpy change of the reaction, so the difference between the binding energy of the reactants and products, the difference between the energy required to break the bonds of the reactants and the energy released by the formation of the products bonds, this heat Q = ΔH, is what is released into the reaction environment, so that one can write ΔS_ext = - ΔH / T, that is to say, all the heat liberated by the reaction is responsible for the increase in entropy of the environment, heat of the flame, heat of the surrounding air, the molting metal ...

This story of  heat produced by the combustion which is spread out over different numbers of products molecules (CO2 and H2O) seems to me absurd. The heat of combustion is not distributed in molecules, it diffuses into the environment. What has to do with it the complexity of the reaction?

 Or would there be a loss of heat in the form of kinetic energy of a greater number of molecules and this would not influence a thermometer?

 So how is it that the flame of acetylene is warmer than that of ethane?

 Any idea?

[Borek]

Combustion is quite fast, before the heat gets spread it must be put somewhere - and initially it can be stored only directly in the reaction products. Heat transport is just too slow for the heat to move away instantly.

[GeLe5000]

Thus heat would consist in the kinetic energies of the products, amongst other things (photons) ?

But how comes that these high speed molecules don't directly contribute to the measured heat of the flame ? Probably because they don't measure it with a thermometer. They measure the light emitted by the flame ?

[Borek]

Thus heat would consist in the kinetic energies of the products, amongst other things (photons) ?

E~kT

But how comes that these high speed molecules don't directly contribute to the measured heat of the flame ? Probably because they don't measure it with a thermometer. They measure the light emitted by the flame ?

They do contribute. Kinetic energy of molecules is directly proportional to kT, plus the substance emits the radiation (think black body radiation). You can't separate these things, they occur at the same time. If you observe the light (more precisely: its color) you can measure the temperature AND you can be sure the kinetic energy of the molecules is proportional to this temperature.