I'm curious about are the details of exothermic nature of 2H2 + O2 -> 2H2O reaction. The explanations I've seen just use a term "energy release" without specifying what type of energy it is. Assume that we have a perfect mixture of H2 and O2 in a closed vessel and produce a spark, which will initiate the reaction. What is the mechanism, which increases the velocity (temperature) of the molecules participating in the reaction?


3 Answers 3


You're asking two separate questions.

  1. What form of energy is produced when combining hydrogen and oxygen?
  2. How does that become thermal energy?

The answer to 1. is that energy can be released in different forms, such as electricity in a hydrogen-oxygen fuel cell, or light. An efficient fuel cell produces little heat.

Question 2. is best considered through thermodynamics. Theoretically, it might be possible to predict the position and velocity of a few unrestrained molecules of $\ce{H2}$ and $\ce{O2}$ combining to form $\ce{H2O}$ (ignoring quantum uncertainty). However, en masse, this is effectively random. Ah... and what is the random motion of particles called?

  • $\begingroup$ Answer 1 does not describe the environment in my post. Thermodynamics says nothing about how the energy of chemical bonds is converted to the kinetic energy of H2O produced by this reaction and in what form. It models the gas as a bunch of balls undergoing elastic collisions. How is the kinetic energy of 2 H2 and 1 O2 set of balls suddenly increased after they meet and magically transmute into 2 H2O balls? Can elastic ball model describe this reaction without using a sudden magic dv for the H2O balls? $\endgroup$ Dec 11, 2022 at 1:39
  • $\begingroup$ @PaulJurczak, what needs understanding? Electrons shift, the atoms to which they're attached move due to EMF. $\endgroup$ Dec 11, 2022 at 19:35
  • $\begingroup$ That's better. So are we talking about H2O molecule vibrating and transferring its newly acquired energy via elastic collisions with other molecules? $\endgroup$ Dec 12, 2022 at 5:47
  • $\begingroup$ @PaulJurczak Well, maybe instead of criticising, you should edit your question to actually ask what you want in a clear way? Using terminology like relaxation of intermediates instead of "form of energy" etc. $\endgroup$
    – Mithoron
    Dec 12, 2022 at 21:26
  • $\begingroup$ @Mithoron I have no idea what made you think I'm criticizing. My goal is to understand the physics underlying this reaction. I'm posting within the constraints of my knowledge. This is the first time I encountered the term relaxation of intermediates, so obviously I couldn't use it forming my question. I was hoping for a high school physics level explanation, Feynman style. $\endgroup$ Dec 13, 2022 at 1:16

The form of energy depends on the reaction conditions but the molecular mechanisms are complex

When water is formed from hydrogen reacting with oxygen, a large amount of energy must be released (that's thermodynamics). How it gets released depends on the conditions of the reaction.

When oxygen and hydrogen are reacted in a controlled way on a catalytic surface inside a fuel cell, much of the energy can be extracted as electricity. But the detailed mechanism involves complex reactions on a specific catalytic surface.

When hydrogen and oxygen gas is ignited, the reaction is fairly uncontrolled and the energy emerges as both light and heat. How this happens is not easy to describe in a simple, single reaction, not least because in a gas there are many molecular collisions happening all the time and it isn't just about two molecules colliding and yielding the product. Plus, the reaction will usually have multiple steps.

For example, a hydrogen molecule might collide with an oxygen molecule to yield an H• radical and an HOO• radical (I'm making up possibilities for illustrative purposes rather than trying to give a realistic description). Both species might have more kinetic energy than the original molecules so will move faster, but they are also, both, highly reactive so further collisions will often yield more reactions, some of which will release more energy. In some cases, the reactive species generated will decay spontaneously, emitting light. In many other cases they will bang into other molecules distributing the excess energy as kinetic energy throughout the remaining gas molecules. This will continue until only stable molecules are left.

The point of this is that excess energy from the formation of new molecules can be released as light or can be distributed very rapidly as kinetic energy to other molecules in the gas (and more kinetic energy in the gas is heat). At a molecular level, the excess energy from a specific chemical reaction can appear as rotational energy or vibrational energy in bonds or as kinetic energy from faster movement of the molecules. But the specific amounts of energy will distribute very rapidly across all the types of energy because molecular collisions tend to redistribute energy very, very rapidly to an equilibrium across all the types on energy in all the gas molecules.

Ultimately, at least in uncontrolled reactions, this means we see the excess as heat.

  • $\begingroup$ rotational energy or vibrational energy in bonds this makes sens to me, but kinetic energy from faster movement of the molecules must be secondary to it. A single event of molecules A and B colliding and merging into a resultant molecule AB does not increase total kinetic energy of the system A and B, modeled with elastic balls. Only collisions of AB with other molecules in the system can transfer some of that newly acquired rotational and vibrational energy into kinetic energy. Is that accurate? $\endgroup$ Dec 12, 2022 at 19:47
  • $\begingroup$ @PaulJurczak If two molecules collide but a bond breaks, releasing a lot of energy, then one of the resulting fragments may carry away a lot of kinetic energy (imagine a ball and spring model where the collision cause a highly compressed spring to unwind). But all the types of energy rapidly equilibrate anyway through further collisions. $\endgroup$
    – matt_black
    Dec 13, 2022 at 9:53
  • $\begingroup$ If two molecules collide but a bond breaks, releasing a lot of energy, then one of the resulting fragments may carry away a lot of kinetic energy - this scenario is easy to model, but what I'm curious about is colliding molecules fusing into a single one. This scenario is not compatible with the unwinding spring model, since there is only one molecule and the other end of the spring has nothing to "push" against. $\endgroup$ Dec 13, 2022 at 16:55
  • $\begingroup$ @PaulJurczak the "unwinding spring" is not a complete boding model but a very simple metaphor. It doesn't pretend to explain all possible reactions. BTW can you think of any strongly exothermic reactions involving two molecules just sticking together as the final step? $\endgroup$
    – matt_black
    Dec 14, 2022 at 23:16
  • $\begingroup$ can you think of any strongly exothermic reactions involving two molecules just sticking together as the final step? I'm not a chemist, so I will not volunteer with an example here. I realize that many (most) chemical reactions proceed in forward and reverse directions simultaneously, converging towards an equilibrium. I would just like to zoom in from a large scale thermodynamics model to interactions between individual molecules leading to their increased kinetic energy. Trying to understand it from the first principles. $\endgroup$ Dec 15, 2022 at 0:20

Thermal energy is released from the reaction $\ce{2H2 + O2 -> 2H2O}$. Before the reaction takes place the system of $\ce{H2}$ and $\ce{O2}$ molecules are in a metastable state: Physical state of O2 and H2

Given enough energy > activation energy the system gains enough energy to overcome the potential barrier and falls into a state of lower energy than the initial state.

  • $\begingroup$ Yes, but I'm asking about the physics of increasing the kinetic energy of H2, O2 and H2O molecules inside the vessel. What form of energy is being transformed into kinetic energy and how? $\endgroup$ Dec 10, 2022 at 23:11
  • $\begingroup$ You can try to put your sample in a heat bath,due to the temperature difference between the heat bath and the initial temperature of the sample, thermal energy will be transferred from the heat bath to your sample enabling the reaction. $\endgroup$
    – Volpina
    Dec 10, 2022 at 23:24

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