Example: Chemical bonding
Prerequisites
Understanding the situation
Our simple model of atom-atom interactions provides a schematic model of a chemical bond: the two atoms interact by a potential that arises from the electric forces between the atoms' constituent parts -- electrons and nuclei. This potential energy corresponds to an attractive force that holds the atoms together. As a result of this attractive force, work has to be done in order to separate the two atoms. This work is the bond energy. Let's work through a simple example to see how this plays out.
Consider a hydrogen molecule, H_{2}. Each hydrogen atom has one electron. When the two atoms are brought together, they attract and form a covalent bond. This happens because the electrons rearrange themselves to reduce the electric repulsions among the electrons and nuclei, resulting in an arrangement that has a lower potential energy than the two atoms separately.
Since it is convenient to set the 0 of PE so that the far separated atoms have zero potential energy, the bound system has negative total PE (since you have to add a positive amount to get it up to zero). This negative value is called the binding energy.
Presenting a sample problem
For H_{2}, the binding energy of a single molecule is -4.52 eV.* For O_{2}, the binding energy of a single molecule is -5.16 eV. To break both of the bonds in a water molecule, H_{2}O, you have to put in an energy of 9.54 eV. Write the stoichiometric equation for the burning of hydrogen and associate with each molecule its binding energy. Show how you can use this to determine the amount of energy released in the burning of hydrogen.
Solving this problem
The burning of hydrogen means that chemical reaction takes place between hydrogen and oxygen resulting in water. If we write this as a chemical equation we get:
H_{2} + O_{2} ↔ H_{2}O. ??
This equation doesn't have the same number of atoms on both sides, so it doesn't balance. To fix it, we have to double the number of water molecules which means that we need to double the number of hydrogen molecules. This gives the stoichiometrically correct equation
2H_{2} + O_{2} ↔ 2H_{2}O. ??
We use a double-headed arrow since the reaction can go either way, depending on the conditions.
Now let's put in the energies. We will take our zero of potential energy when all the atoms are separated and far apart. This means that each molecule has an energy deficit (binding energy). Let's show underneath each molecule how far down it is in its potential energy well.
2H_{2} | O_{2} | ↔ | 2H_{2}O | |
Energy deficit due to binding |
2x( -4.52 eV) | -5.16 eV | 2x(- 9.54 eV) |
Notice that the information in the problem was presented in an inconsistent way. The binding energies were given for the hydrogen and oxygen molecules, but the amount of energy you had to put in to break up the water was given. We needed to recognize that means that the water molecule has a binding energy (energy deficit) of -9.54 eV.
Now let's add up the total PE deficits on each side of the equation:
2H_{2} | O_{2} | ↔ | 2H_{2}O | |
Energy deficit due to binding |
2x( -4.52 eV) | -5.16 eV | 2x(- 9.54 eV) |
Total deficit | -14.20 eV | -19.08 eV |
Just as we had to correct our original stoichiometric equation because the molecules didn't balance, we have to correct this equation because the energies don't balance. If we go from the left to the right, we drop down in potential energy by almost 5 eV, so in order to conserve energy, the difference gets "released" (typically by increasing the KE of the molecules on the right). The fully balanced equation looks like this:
2H_{2} + O_{2} ↔ 2H_{2}O + 4.88 eV
Since the -14.20 eV and the -19.08 eV are actually a part of the state of the molecules, we don't have to explicitly include them.
Schematically, we might show what is happening with the following diagram:
(This is only schematic because there really isn't only one position coordinate in a complex molecule.) If we move from the left to the right (burn hydrogen) our system goes from a binding energy of -14.20 eV to one of -19.08 eV. It's like rolling down a hill. In going down to lower PE, you gain kinetic energy. Going from left to right the reaction is exothermic. Burning hydrogen into water leads to a large increase in KE (release of heat) of 4.88 eV for every pair of hydrogen molecules burned into water.
In chemical reactions that yield energy in biology, the reactions may go a step at a time to deliver bits of the energy to various different systems. But the overall conservation of energy picture still holds.
[Note that in this problem we have discussed individual interacting atoms and molecules and treated them as if they were isolated. We have ignored that fact that in realistic situations, the molecules may be imbedded in a gas or liquid and the interactions with the environment can change some of the energetics involved.]
* The eV or "electron-Volt" is an energy scale that is commonly used for describing the states of individual atoms and molecules. 1 eV = 1.6 x 10^{-19} Joules. In chemistry, the focus is typically not on individual atoms or molecules, but on moles of them (1 mole = 6.022 x 10^{23} molecules). Therefore, the conversion between eV/molecule and Joules/mole is also useful. The conversion is: 1 eV/molecule = 96,000 Joules/mole = 96 kJ/mole.
Joe Redish 11/20/14
Last Modified: February 27, 2019