# What is the meaning of chemical energy, in terms of first principles?

by Ulysees
Tags: chemical, energy, meaning, principles, terms
 P: 516 We're all familiar with exothermous reactions, where chemical energy is converted to heat. However, I have not seen a satisfactory explanation of this from first principles, ie electromagnetic forces between charged electrons and nuclei. How does the force between charges and the associated dynamic and kinetic energy turn into heat? More specifically, I want to know the forces and fundamental energies (ie dynamic and kinetic, not chemical) involved in this reaction: 2 H2 + O2 = 2 H20 + heat
 Admin P: 23,718 Atoms in molecules are kept together thanks to combination of electric attraction and repulsion. There is a lot of potential energy in that system, that can be freed when nuclei and electrons rearrange in such a way that charges get closer. That's basically what is happening. Note: "charges get closer" is not synymous to "atoms get closer".
 P: 516 No role for kinetic energy? (speed of electrons) What are the maths of it, what is the quantity of energy released for every molecule of water formed? How is this heat related mathematically to the charge distribution in the H2 and O2 molecules versus the charge distribution in the H2O molecule?
 Sci Advisor HW Helper PF Gold P: 3,724 What is the meaning of chemical energy, in terms of first principles? I think you are asking a very difficult question. It is probably better to limit the question to a H - H model. Imagine two isolated atoms of hydrogen approaching each other from a distance much greater than the bond length of the hydrogen molecule. As the electrons 'see' the other hydrogen, they begin to pair up until they reach a point at which there is maximum attractive interaction between the electrons and the protons. Now the electrons 'flow' around both nuclei and thus occupy a bigger box and so are lower in energy. Where does the energy go? How is it transferred? It goes into one or more of the nearby (energetically speaking) molecular vibrational modes which can be transferred to the medium... hence heat energy flows from the nascent molecule into the system and a temperature rise is noted.
P: 516
 Quote by chemisttree Now the electrons 'flow' around both nuclei and thus occupy a bigger box and so are lower in energy.
Hang on a second, if charges occupy a bigger box, they have a higher level of dynamic energy.

Can we break a hydrogen molecule into two hydrogen atoms?
HW Helper
PF Gold
P: 3,724
 Quote by Ulysees Hang on a second, if charges occupy a bigger box, they have a higher level of dynamic energy.
No, I don't think so. Investigate "particle in a box" for an explanation.

 Can we break a hydrogen molecule into two hydrogen atoms?
Yes. It would be atomic hydrogen... a good google target!
 P: 516 Do you know where to find the maths of the hydrogen molecule, and the hydrogen atom? It seems to me that neither of them has an electron that is a particle in a box - there's no potential walls.
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PF Gold
P: 3,724
 Quote by Ulysees Do you know where to find the maths of the hydrogen molecule, and the hydrogen atom? It seems to me that neither of them has an electron that is a particle in a box - there's no potential walls.
You are right. The hydrogen molecule is not exactly a particle in a box. That treatment is only used to help visualize what is going on in reality. In reality, there is a very steep potential as the electron approaches the nucleus until it reaches a point at which it is effectively an infinite potential. As you move the electron away from the hydrogen nucleus, the potential rises and then falls off more gradually until it approximates the ionization energy of the hydrogen atom. Kind of like a particle in a one sided box with a smallish lip.
P: 516
 Quote by chemisttree As you move the electron away from the hydrogen nucleus, the potential rises and then falls off more gradually
Doesn't the potential around the proton look like this?

$$V = (1/4*pi*epsilon)* e/r$$

Positive potential means negative potential energy for the electron. So the electron potential energy increases from very negative to zero at infinity, as the electron gets further away.

Ie the bigger the spread of electrons (ie the box), the higher the potential energy. So the H2 molecule would have less energy if it's closer packed together.

But what about kinetic energy, we've ignored it so far.
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PF Gold
P: 3,724
 Quote by Ulysees Doesn't the potential around the proton look like this? $$V = (1/4*pi*epsilon)* e/r$$ Positive potential means negative potential energy for the electron. So the electron potential energy increases from very negative to zero at infinity, as the electron gets further away.
You are confusing potential with energy. The energy at r=infinity is simply the ionization energy. You are using classical physics to explain a quantum mechanical system. When you use a quantum mechanical explanation, the energy of the particle in a box becomes:

$$E_n=\frac{\hbar^2\pi^2}{2mL^2}n^2$$

In this case, 'L' is the distance between the walls of the box. Note that as L is increased, the energy of the particle is lowered. The same happens with electrons in molecules. As the wavefunction for the electron occupies more space, the energy decreases.
 P: 218 I interpret that you want to know what the degrees of freedom of the molecules is? If that is the case you can go about it like this; $$PV = RT$$ which is boyle's law. From here you get $$\frac{1}{2}mv^{2}_{x}=\frac{1}{2}kT$$ $$k = \frac{R}{N_{A}}$$ which is planks constant. Look it up in a physical table near you ;) Then the average kinetic energy of a molecule (in 3D) is $$\epsilon =\frac{1}{2}m(v^{2}_{x}+ v^{2}_{y} + v^{2}_{z}) = \frac{1}{2}kT*3$$ In other words the energy $$\frac{1}{2}kT$$ is distributed equally in all degrees of freedom. $$\frac{3}{2}N_{A}kT = \frac{3}{2}RT$$ for an atom For an atom you only get three degrees of freedom, the translations in x,y,z. For an diatomic molecule you also get rotation in two more degrees namely $$\theta$$ and $$\phi$$. If you've studied multivariable calculus you will know what this means. $$\frac{5}{2}N_{A}kT = \frac{5}{2}RT$$ for diatomic molecules.
P: 516
Thanks Fearless, but this is not what I was looking for. I'm happy with the notion that molecules "vibrate/move faster" after an exothermous chemical reaction. It is the energy of the electrons that I want to express in terms of fundamental quantities.

 Quote by chemisttree You are confusing potential with energy.
Thanks, but I am not confusing potential with energy, I am an engineer with masters in physics-based simulation.

 The energy at r=infinity is simply the ionization energy.
Ok, and I was hoping that we express everything in terms of fundamental quantities. I am not convinced that macroscopic quantities like kinetic energy, electrostatic energy, electromagnetic energy, are not defined for subatomic particles. I believe they are defined, and they are conserved. Except in nuclear reactions.

People certainly work in terms of quantum energy levels at the atomic scale (which I guess are implied to be a mixture of electromagnetic and kinetic energy, but you tell me) only because quantum laws make working this way practical.

But do we have any lower level equations, like wavefunctions from which we can derive kinetic and electromagnetic energy, and compare the before and after state, and then match the difference with the heat of the exothermous reaction?

 When you use a quantum mechanical explanation, the energy of the particle in a box becomes: $$E_n=\frac{\hbar^2\pi^2}{2mL^2}n^2$$ In this case, 'L' is the distance between the walls of the box.
Sure, but why is the electron of a hydrogen atom, living in a box between potential walls? Are you approximating this way a trough in the curve of the (total) energy of the electron as a function of the distance from the nucleus? If so, wouldn't the width of the trough L be something different from distances of the electron from the nucleus?
 P: 218 So you are trying to find out the boundry of when electrons are modelled in classical equations?
P: 516
 Quote by Fearless So you are trying to find out the boundry of when electrons are modelled in classical equations?
Not really. In my understanding the classical electromagnetic waves are the sum of lots of wavefunctions with the imaginary parts removed and the wavefunction of the electron also gives a probability density function for where the electron is, ie charge density. I think we can also derive the speed of the electron at each point from the wavefunction (ie its kinetic energy when passing through that point).
 P: 282 The original equation derives it's meaning from Thermodynamics. It is a statistical phenomenon that arises from a great many individual interactions. I do not believe it can be explained satisfactorily as arising from the behavior of a single electron or atom because it is sort of a "net" property of the system at large, but I shall do my best to try to come close. Imagine a great many H2 and O2 molecules in a closed box. Nothing happens initially because all the molecules have all their electrons in their ground state and the system does not have enough energy to excite the H2 and O2 electrons to the higher oribitals needed for a chemical reaction. Now imagine the conditions in the system change. The temperature of the system is increased if you inject a hot inert gas into the box. This gas will release it's kinetic energy by slamming into the atoms that were previously inside the box and cause both H2 and 02 atoms and electrons to move. The electrons in these atoms will be excited by the collisions into higher energy oribitals. From here one of two things can happen. The electrons can relax by settling back down to the ground level in the original H2 or 02 molecular orbitals, or if there is another atom nearby they might relax by settling down into the ground level of the molecule next door (instead of settling down into their orginal orbitals they might settle down into each others orbitals and relax as an H20 molecule). There is an equal probability that the electron will choose either path while it is in an excited state, but the key is that if one of the paths causes the electron to release more energy than the other path then there may not be enough energy locally for this electron to "climb" back up to a high enough oribital to jump back to its original molecule. A net chemical reaction occurs and the energy that is no longer capable of promoting the electrons in the H20 molecule has nowhere to go except to excite other atoms next door into a possible reaction. It is when this happens to a great many molecules in a system (i.e. H20 molecules form that can no longer use the available energy to react back to H2 and 02) that the overall effect becomes noticeable and we get a net release of heat from the system. Keep in mind that the situation of whether or not an H2 or H20 molecule forms depends on the temperature and whatever random energy might be available near a molecule. If the average energy of the system increases so that the product molecules might have enough energy most of the time to react back to the original molecules of H2 and O2, then they will do so in statistically large numbers and the heat attached to your equation above will change. The result at the atomic level would be a mixture of back and forth conversions taking place. At any given point in the system the question of what is happening to the electrons in a single H2 molecule is meaningless since the molecule is rapidly being excited and relaxing into both H2 and H20 molecules depending on how many excited atoms are relaxing next door to it, etc. etc. at any given moment (and this is changing constantly throughout the system as time goes on...the net effect of what we see coming out of the system is simply the more statistically favorable result).

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