I know this might be a silly question, but should I sustain the exact energy or more for the enthalpy of endothermic reaction to occur or can I give it in parts?

Let's say a certain reaction requires $100\ \mathrm{kJ}$ to occur, do I have to give it $100\ \mathrm{kJ}$ or more to occur or can I give it $50\ \mathrm{kJ}$, two times? Assuming that no energy is lost. If it is possible, what about real-life, where energy is in fact lost?

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    $\begingroup$ It's not a matter of silliness, but fundamental misunderstanding. You think you need to freeze coal to burn it? No? So why you'd think something similar about reaction that is endo- instead of exothermic? $\endgroup$
    – Mithoron
    Dec 25, 2022 at 20:57
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    $\begingroup$ It may depend on the phenomena. If you want to heat one or two liters water, you may provide $50$ kJ than $50$ kJ later on. But if you want to break a bond whose bond energy is $100$ kJ/mol, you must provide $100$ kJ/mol. Providing half of it is useless. This energy will be lost. $\endgroup$
    – Maurice
    Dec 25, 2022 at 21:01
  • $\begingroup$ @Maurice Does that mean that the bond will absorb the 50 kJ but do nothing with them? Where will the 50 kJ go? $\endgroup$ Dec 25, 2022 at 21:11
  • $\begingroup$ @BelalBahaa. The bond will simply not absorb $50$ kJ/mol $\endgroup$
    – Maurice
    Dec 26, 2022 at 12:23

2 Answers 2


No reaction has to be given the exact activation energy to proceed

There is a deep misunderstanding here about what activation energy means. It isn't a property of the bulk reaction that is happening, but more like a property of the individual molecules undergoing reactions.

The point is that all the molecules in a bulk compound or mixture do not have the same energy. Simplistically consider a reaction where two different molecules need to collide with more than a certain energy to cause a reaction (the net result of which might absorb more energy ("endothermic") or release more energy ("exothermic"). Many of the individual molecules will bang together without enough energy and nothing will happen. Some will have enough energy and the reaction will proceed. But the individual molecules will always have a wide range of individual energies (As Geoff says, in a Boltzmann distribution). What matters for whether the reaction happens is not how much total energy there is, but what proportion of the individual molecules have more than the activation energy threshold.

That proportion is what varies with the energy you add to the system. It isn't about getting the added energy exactly right: it is about getting enough to achieve some proportion of the molecules with the right level.

In an endothermic reaction this is still true. But if the final products absorb energy, the amount of energy remaining in the system will fall and fewer molecules will have enough energy to leap the activation energy barrier, so the reaction rate may fall (perhaps low enough to stop the reaction). In which case you can add incrementally more external energy to compensate for the internal loss and keep the reaction rate up.

But, at no point can you add the exact "right amount" because energy is always distributed among all the molecules and the issue is what proportion have enough energy for the reaction to happen.

So, in the bulk, it doesn't matter how the energy is added (partwise or all at once) as long as there is enough for some proportion of the molecules to react.

  • $\begingroup$ +1 certainly. But I think it's also important to emphasize that adding a little energy at a time is not necessarily going to do what you think because of distribution into other vibrational / rotational / translational modes. $\endgroup$ Dec 27, 2022 at 21:35

In principle, you could add 50% energy to the motion that goes over the reaction barrier, which moves you up some of the vibrational energy levels .. and then another 50%. But in reality, that's going to be hard because in most molecules, there are dissipation mechanisms -- so it would be very hard to "keep" the energy in the motion (vibration) needed to overcome the reaction barrier.

Sometimes it's useful to use metaphors when talking about kinetics.

Let's consider that you need to jump over a barrier to get from a starting point to another place:

cartoon of stick figure trying to jump a wall

If the figure attempts to jump 50% of the height, it's not that they can jump another 50% to get over the wall.

In detailed chemical kinetics theory, molecules will have a distribution of different energies (the Boltzmann distribution) but either they have the energy to go "up and over" or they do not and stay on one side of the reaction barrier.

Now this metaphor isn't perfect because some chemical processes (e.g., heating water) can use repeated additions of energy. Also, few chemical reactions are one dimensional and some reactions occur via catalysts or quantum tunneling, so they're not really "jumping over" a barrier.

You might wonder, what happens to the first 50 kJ/mol. In the metaphor, the figure "falls back" because it does not get over the wall. There are a few mechanisms in which the molecule may dissipate energy, for example going into translational or rotational kinetics, or other low-energy vibrational modes.


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