Several energy units come up frequently in chemistry; it is important to know how they are related. These are

Erg: The erg is the basic unit in the centimeter-gram-second (cgs) system of units. We work up to the erg from more fundamental units of length, mass, and time using two relationships:

The cgs unit of force is the dyne. From Newtons second law,

The cgs unit of work is the erg, since work is energy. From the work equation

This gives the erg in terms of fundamental units.

Joule: The joule (J) is the analog of the erg in the meter-kilogram-s (mks) system, so

We easily connect joules and ergs.

Thus 1 J = 10⁷ ergs. Of course 1 kJ = 10³J!

Calorie: The calorie is an odd unit associated with heat. When James Joule showed that mechanical work and heat were 2 different forms of energy, he also showed that

This conversion is experimentally det'd, in contrast to the J/erg conversion above, which follows from definitions. Of course, 1 kcal = 10³ cal!

L-atm: The liter-atmosphere (L-atm) is also an odd unit associated with pressure volume work. We relate it to joules.

1 atm = 760 mm Hg. The pressure exerted by a column of Hg 76.0 cm tall is the force due to the mass of Hg in the column per cm² of cross-sectional area of the column. This is a volume of 76.0 cm³, giving a mass of

The force exerted by this is given by Newtons 2nd law, F = ma

The pressure is thus 1.013 * 10⁶ dynes/cm².

It follows that 1 L-atm = (10³cm³) * (1.013 * 10⁶ dynes/cm²)

1 L-atm = 1.013 * 10² J.

Using this, we calculate R in J units

(Conversion among L-atm, J (kJ), cal (kcal) are often carried out using values of the ideal gas constant, R.

For example to convert from cal to J, one would multiply by 8.314J/1.987cal (which is 4.184!))

eV: The electron-volt (eV) is a unit used in connection with ionization potential measurements, where electrical potentials are used to strip e from atoms. 1 eV is defined as the energy acquired by an electron when it is accelerated thru a potential difference of 1V (1V = 1 joule/coulomb (J/C) by definition). This energy is the product of the e^- charge and the voltage.

1 eV = 1.602 * 10^-19 J

This is easily remembered, because it is numerically the same as the charge of the e^- in coulombs.

cm^-1: The reciprocal centimeter (cm^-1) is used by spectroscopists, who commonly measure wavelengths. To report an energy, they invert the wavelength (1/l). We relate ergs to cm^-1 using the equation for the energy of a photon.

We can ask what value of 1/l corresponds to an E of 1 erg:

Thus 1 erg = 5.0345 * l0¹⁵ cm^-1

To sum up:

You should be able to get these numbers by remembering the few basic equations from which they are derived.

Finally, there is the matter of whether an energy is on a per molecule or per-mole basis. Two of the units, eV and cm^-1, are naturally on a per molecule basis. In the case of the eV, this is because we are dealing with accelerating a single e- thru a potential difference, not a mole of e^-. The eV is in fact a very small amount of energy (1.6 x 10^-12 erg!) The cm^-1 is the reciprocal of the wavelength of a photon, and is the same no matter how many photons we have (1 or a mole). Unless specifically told otherwise, assume energies in eV or cm^-1 to be on a per molecule basis.

Ergs, joules, and calories are commonly used on both a per molecule and per mole basis. In the series of equivalencies above, everything is on a per molecule basis, but we easily convert to a per mole basis using Avogadro's #:

Thus 1 eV (per molecule) = 23061 cal/mole = 23.061 kcal/mole.

A similar calculation can be performed for ergs and joules.

Two forms of Coulomb's Law are commonly seen:

1. F = q₁q₂/4pe_or² (and V = -q₁q₂/4pe_or) This is the SI accepted form.

To use this, charges must be expressed in coulombs, and e_o, the electrical permittivity of a vacuum, has the value 8.854185 * 10^-12 J^-1 C² m^-1 = 8.854185 * 10^-21erg^-1 C² cm^-1

2. F = q₁q₂/r² (and V = -q₁q₂/r)

To use this, charges must be in esu (electrostatic units).

Applied to the H atom, the two equations for potential energy are

This gives V = -(1.602 * 10^-19C)²/4p(8.854185 * 10^-21 erg^-1 C² cm^-1)r = -2.3066 * 10^-19/r, where r is in cm.

the two expressions for V as follows.

Solving for e, we obtain e = 4.803 * 10^-10 erg^1/2 cm^1/2

By definition, 1 esu = 1 erg^1/2 cm^1/2 = g^1/2 cm^3/2 s^-1.

In conclusion, there are two values for the electronic charge, depending on which form of Coulombs Law is used:

Energy Relationships

An understanding of the meaning of energy, relationships among the energies of a series of related states or species, the connection between energy and stability, and the use of energy diagrams is essential for a scientist. A glossary of energy-related terms and phrases follows:

• kinetic energy
• potential energy
• enthalpy
• free energy
• bond energy
• stable and unstable. These are adjectives referring to a single species or state, relative to some reference state.
• high energy
• low energy
• decrease in energy (phrase referring to process involving at least 2 species or states)
• increase in energy (phrase referring to process involving at least 2 species or states)
• endothermic
• exothermic
• positive energy change
• negative energy change

Several types of energy are defined in thermodynamics:

Chemists use the word energy loosely. Depending on context, it can mean any of the first four types of energy above. However, chemists most often mean by energy the potential energy of chemical bonds; that is, energy due to the relative positions of atoms, which attract each other with electrical forces. Chemical reactions result in changes of relative positions of atoms, and corresponding changes in bond potential energy.

For chemical reactions, changes in enthalpy and internal energy are related by equation C-1:

For reaction at constant temperature of 298 K, the last term has the value 2.51 Dn_gas kJ. For typical values of Dn_gas, this is on the order of 5 kJ. Since internal energy changes for chemical reactions usually exceed 40 kJ and are more often on the order of 400 kJ, it follows that

The second equality holds because at constant T, DKE = 0. Since the changes in enthalpy and PE in a chemical process are nearly the same, we are justified in using the term energy when we actually mean enthalpy.

Now we consider the terminology for energies of states and energy changes of processes. Consider a process which takes a system from State 1, energy E₁, to State 2, Energy E₂:

(State 1 might be the reactants in a chemical reaction, State 2 the products.) We use a question/answer approach to illustrate some of the terms in the glossary.

1) Which state is more stable, 1 or 2? This depends on whether E₁ < or > E₂. If E₁ < E₂, state 1 has lower energy and is more stable; if E₁ > E₂, state 2 has lower energy and is more stable. From here on, we suppose E₁ < E₂, and state 1 is more stable. It is a lower energy state than state 2. Thus higher stability is associated with lower energy.

2) Does energy increase or decrease in the process? Calculate the change in energy of the process, DE = E₂ - E₁. (NOTE: the change in a quantity is always the final value minus the initial value.)

Since we have supposed that E₁ < E₂, DE = E₂ - E₁ > 0, and energy increases in the process. An energy increase is associated with a decrease in stability.

3) Is the process endothermic or exothermic? An endothermic process is one in which heat is absorbed by the system. If heat energy is absorbed by a system during a process, the energy of the system must increase; i.e., E_final > E_initial and DE > 0. An endothermic process has DE > 0. Similarly, heat is given out by the system during an exothermic process. If heat is produced by the system, the energy of the system must decrease; E_final < E_initial, DE < 0. An exothermic process has DE < 0.

For an endothermic process, the final state is less stable (higher energy) than the initial state; for an exothermic process, the final state is more stable (lower energy) than the initial state. For our choice of E₁ < E₂, the process is endothermic since DE > 0.

4) How can the energies of the two states be represented on an energy level diagram? An energy level diagram is a plot of the relative energies of states on a vertical axis. For E₁ < E₂, the diagram is in Figure C-1. The left vertical arrow is the axis, labelled with energy (E). Energy increases as we move up the axis. States 1 and 2 are represented by horizontal lines placed at heights on the axis representing their energies. State 1 is lower than state 2, consistent with E₁ < E₂. (NOTE: E₁ < E₂ in the algebraic sense. If E₁ and E₂ are both negative, a common situation, the negative number with larger magnitude is lower. Thus -200 < -100.) Consider energies to be on a number line, on which a number is less than any number to the right of it and greater than any number to its left (Figure C-1).

5) How is DE represented on the diagram? This is done using an arrow leading from state 1 to state 2, just as in the equation representing the process. The arrow has been put in Figure C-1. The length of the arrow is the magnitude of DE, and the direction gives its sign. If the arrow points up, DE > 0; if down, DE < 0. In Figure C-1, DE > 0.

We now consider bonds and energy, and ultimately bond energy, BE (or D). Two atoms, A and B, exert attractive forces on each other. These forces lead to chemical bond formation. Formation of chemical bonds gives a more stable (lower energy) state than when the atoms are separated. Generally, the greater the number of bonds formed, the greater the energy lowering and the greater the amount of stabilization.

AB is more stable than A + B. Bond formation causes a lowering of energy. The greater the number of bonds formed between A and B, the greater the energy lowering.

As a particular example, consider the following data:

A carbon-oxygen triple bond is more stable (stronger) than a carbon-oxygen double bond, which is more stable (stronger) than a single bond, with respect to the separated atoms. But consider the statement, "a CºO bond has higher bond energy than a C-O bond." This is a statement that a chemist might routinely make. Based on what we have said above, this statement is confusing, because it seems to imply that a CºO bond has a higher energy, and must therefore be less stable, than a C-O bond. This directly contradicts our conclusions above. The problem is in the definition of the term bond energy. The term does not refer to the actual energy of the state in which the atoms are bonded. It refers instead to the energy difference between the non-bonded and bonded states. A bond energy is an energy difference. The confusion arises because the D symbol meaning change is not used with bond energy. The bond energies of the C-O, C=O, and CºO bonds are defined as the potential energy changes of the following reactions in the gas phase:

The BE of the C-O bond is the energy which must be put in to break the C-O bond. All bond energies are positive. A higher bond energy means that more energy must be put in to separate the atoms; the bonded atoms have lower energy, and are more stable, than the separated atoms. High bond energy thus means high stability. There is no contradiction between the concept of bond energy and the concepts discussed above, as long as the definition of bond energy is kept in mind. We represent BE(C-O), BE(C=O) and BE(CºO) on an energy level diagram in Figure C-3. The diagram shows that CºO is more stable than C=O, that BE(CºO) > BE(C=O), and that all bond energies are positive (arrows up).

We close with a challenging problem in energy relationships. Consider equation C-2 and its experimental DH_R^o:

Equation C-3 in Figure C-4 shows that S₆ is a non-planar ing of S atoms, singly bonded. One of the three resonance forms of SO₃ is shown on the product side. The O₂ molecule is best regarded as containing a double bond.

a) Calculate DH_R^o for reaction C-3 using average bond energies: BE(S-S) = 213 kJ/mole; BE(O=O) = 498 kJ/mole; BE(S-O) = 268 kJ/mole; BE(S=O) = 523 kJ/mole.

b) The calculated and experimental enthalpy changes disagree because the calculation neglects both the strain energy of real S₆(g) and the resonance stabilization energy of real SO₃(g).

Taking into account strain and resonance energies, place the following states on a relative energy level diagram:

From reaction C-2, products (state B) have lower energy (are more stable) than reactants (state A) by 2477 kJ. State B lies 2477 kJ below state A on the energy level diagram. This is shown in Figure C-5. DH_R^o(exp) for reaction C-2 is represented by an arrow leading from state A to state B. This is shown in Figure C-6. The arrow points down, consistent with DH < 0.

From reaction C-3, products (state D) have lower energy than reactants (state C) by 594 kJ. State D lies 594 kJ lower than state C on the diagram shown in Figure C-7. DH_R^o(calc) is represented by an arrow from state C to state D.

The problem now is to put all 4 states on the same diagram. Consider A and C first. Which of these is more stable? State A includes real S₆(g), which is strained. State C includes hypothetical S₆(g), which has no strain. Which is lower in energy (more stable), strained or unstrained S₆? Strain is associated with instability, so state C is lower than state A. Now consider states B and D. Which has lower energy (is more stable), real SO₃, a resonance hybrid, or the single resonance form in equation C-3? The true structure of a molecule is always more stable than any contributing resonance form. Thus real SO₃ has lower energy (is more stable) than the single resonance form in C-3. State D has higher energy than B, and lies above it on the energy level diagram.In combining the two diagrams, we must end up with state C below state A, and state D above state B. The result is in Figure C-8.

c) S₆(g) is considered to have strain energy of 8.4 kJ per mole of S-S bonds. Calculate the resonance stabilization energy of one mole of SO₃(g).

State A is higher than state C by 8.4 kJ for each mole of S-S bonds. 1 mole S₆ contains 6 moles S-S bonds, for a total strain energy of 50.2 kJ. State A is 50.2 kJ higher than state C.

Figure C-8 shows that the energy difference between states D and B is the resonance stabilization energy of 6 moles SO₃. We calculate this difference as follows:

d) Setting the energy of state A to zero, assign relative energies to the other 3 states, B, C, and D.

We've already done the work in part c). Since states B, C, and D are all below state A on the diagram, their energies must be lower; to be lower than 0, they must be negative.

A final note on the terms stability and instability. Strictly speaking, these words should be used in conjunction with free energy changes, which incorporate both energy and entropy changes. Both energy and entropy are important in determining the direction of spontaneous change. Free energy and stability are related as follows:

In practice, chemists use the words stability and instability to refer to both free energy and energy (enthalpy), assuming that context will make the meaning clear. Since it is entirely possible for a process to have DE > 0 and DG < 0 (because DS > 0), this careless use of the term stability is dangerous. To be perfectly clear, the following terminology should be used: