# Why does potential energy relate to exothermic reactions in this way?

I am not understanding how potential energy (which is the energy that results from position or configuration) relate to exothermic reactions. We say that exothermic reactions give off heat and so lose energy but I don't see how I can use this definition of potential energy to say that potential energy is lost from the system of an exothermic reaction. It might be my definition for potential energy that is wrong, I don't know.

• If you got any system in a state $q_1$ with potential energy $U(q_1)$ and it changed into state $q_2$ with potential energy $U(q_2)$, then if $\Delta U = U(q_2) - U(q_1) < 0$, energy has been lost to the surroundings. Does that go in the direction you want? – Felipe S. S. Schneider Jan 28 '17 at 16:30
• @FelipeSchneider i understand that the system lost heat energy but i dont understand saying the system lost potential energy so i think i have the wrong definition for potential energy – Sam19KY Jan 28 '17 at 17:09

Energy is a useful concept for rationalising the capability of doing work or heating; other than that, it is purely a theoretical concept. As Feynman brilliantly pointed out,

It is important to realise that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and when we add it all together it gives “28”—always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.

But let's get back to something answerable.

Energy is (a) conserved and (b) convertible into different forms. If you got any system in a state $q_1$ with potential energy $U(q_1)$ and it changed into state $q_2$ with potential energy $U(q_2)$, then if $\Delta U = U(q_2) - U(q_1) < 0$, energy has been lost to the surroundings. This loss could be lots of things, most of the time it is just heat (kinetic motion). This loss, in the big picture, is just a transfer or conversion.

Let's get to a very simple example, hydrogen combustion:

$$\ce{2 H2 + O2 -> 2 H2O}$$

Since energy can't be produced, we say this reaction releases large amounts of heat. But, where does it come from?

All that the equation above tells us is the atomic rearrangement that has been taken place: in the left hand side we have two $\ce{H-H}$ bonds and a single $\ce{O=O}$ bond, in the right hand side we got four $\ce{O-H}$ bonds. So quantifying the bond energies in each side gives us:

     |  Average bond energy (kJ/mol)
-----|-------------------------------
|        |   Left    |   Right
Bond | Single | hand side | hand side
-----|--------|-----------|----------
H-H  |  -432  | 2 x -432  |    0
O=O  |  -495  |  -495     |    0
O-H  |  -467  |    0      | 4 x -467
--------------|-----------|----------
Total      |   -1359   |  -1868


This means we get $-1868 - (-1359) = -509$ kJ for two moles of $\ce{H2}$, i.e., $-254.5$ kJ/mol of heat is released during the combustion of hydrogen.

Wikipedia says the correct value is $-242$ kJ/mol, so our calculation has an error of around 5%.

So heat energy has been released and it came from bond breaking and forming. The energy stored in a bond could be called latent or potential, if you will.

It all boils down to conservation and convertibility.

The potential energy of a system is converted in to "heat energy" during an exothermic reaction. Heat is a measurement of molecular vibrations.

When you have a molecule with a potential energy, this reflects the strength of its molecular bonds. To break this bond, energy is first applied to the molecule, to reach an "activation energy". At this activation energy, the molecule will enter an unstable transition-state (which has a potential energy even higher than that of the starting molecule).

Since the transition state is unstable, the molecule will naturally move towards a more stable configuration. To do so, the molecule will form a bond with another molecule. This bond will be even stronger than the bond of the first molecule, and therefore has a lower potential energy. The stronger bond, will not vibrate as strongly as the bond from the starting material.

As the first molecule is converted into the second molecule, its motions (literally, movement) throughout the solution will provide energy, or heat, to nearby molecules (and the container in which the reaction takes place).