Physical Laws Examples
If you have made a stab at reading several of the linked sections, you should be asking what purpose is served? What are the uses of these disparate sections? In fact, scaling of phenomena is helpful in deciding which is the baby and which is the bath water. Let us try a few simple examples.
Example 1
A biological reaction is dominated by hydrogen bonding with a measured bond energy of 3 kcal/mol. How does this compare to the kinetic energy experienced by these molecules at a temperature of 300K and 1,000 K?
First what did you read the section on H-bonding? If you did, then you know that H-bonding is relatively weak as chemical bonds go but it is long range and is often dominant in, for example, biological reactions as the origin of the forces that self-assembles biomolecules.
As chemists measure bond energies, hydrogen bonds are typically of the order of 2 – 8 kcal/mol or about one order of magnitude less strong than normal full electron-sharing covalent bonds[1]. H-bonds arise because a hydrogen atom from one molecule is attracted towards a strongly electronegative atom of an adjacent molecule. These electronegative atoms are typically oxygen, nitrogen, fluorine etc. The H-atom loses some of its electronic charge and becomes slightly positive and the receptor electronegative atom becomes slightly negative by acquiring the partial charge borrowed from the H-atom.
A H-bond of 3 kcal/mol calculates to ~ 2 × 10-20 J/H-bond (the conscientious student needs to do this arithmetic). What is the kinetic energy of the local molecules?
According to section 2.2.1, the K.E. of a gaseous particle is ˝ kT in each of three arbitrary but mutually orthogonal directions. As an estimate we often assume that local elements of larger molecules even in a condensed state have K.E. of about kT. Thus using k = 1.3807 × 10-23 J/particle K for Boltzmann’s constant, the K.E. is ~ 4 × 10-21 J/particle at 300 K and ~ 1.4 × 10-20 J/particle at 1,000K.
The ratio of H-bond energy/K.E. per bond is thus ~ 5 at 300K and 1.4 at 1,000K. In other words, you can often break some fraction of H-bonds by heating them (if nothing has irreversibly decomposed - particularly at 1,000K in this example, they will reform on cooling).
Example 2
What if our same reaction were carried out at room temperature in an electric field of E = 109 V/m? Which forces, electrical or H-bonds, are then dominant? Assume the biological molecules have a molecular mass of 2,000 (kg/kmol).
According to section 4.2, the energy density, in SI units of J/m3, of an electric field is ˝ eE2 = where e = 8.85 × 10-12 F/m assuming a vacuum (and between 5 to 80 times higher if a liquid or solid – the factor of 80 is for water). Thus the maximum electrical energy density is ˝ × 80 × 8.85 × 10-12 × 1018 [F/m][V/m]2 = 3.5 × 108 J/m3. Assuming our medium also has the same density as water, this becomes 3.5 × 105 J/kg or 175 J/kmol = ~3 × 10-25 J/molecule.
Since H-bonding is worth 2 × 10-20 J/H bond (Example 1) the ratio of the bond energy to the electrostatic energy density is 2 × 10-20/3 × 10-25 = ~ 7 × 104, suggesting that the bond energy is dominant.
Even had the electric field been increased to a barely conceivable 1010 V/m, this ratio would still be ~ 700 in favor of the H-bonds and the electric field energy would remain unimportant.
Example 3
Let’s conduct our biological experiment in a MRI machine (magnetic resonance imager); it operates with large magnetic fields (see section 4.2).
Its energy density usually represented as 0.5B2/m0 per unit volume where m0 = 4p × 10-7 henrys/m (whatever that means!). In the MRI machine, if B = 1 tesla (again, whatever that means!), the energy density is ~ 4 × 105 J/m3. Again, if the medium had the density of water, the energy density would be ~ 4 × 102 J/kg and would be entirely negligible (since it is << than the electric energy density of the previous example and even that was negligible compared to H-bonding).
Example 4
What about the relative strength of intermolecular gravitational attraction compared to H-bonding on these biological reactions? (After all, even though the gravitational mass attraction of an atom to another atom is small, so is the distance between atoms.)
Thus if we assume that the attraction is of a molecule with the center of the Earth is negligible, we must modify our expression for gravitational potential accordingly. Instead of
where R is the radius of the Earth, we will need
where R is now the distance of approach of the hydrogen atom to its electronegative partner. Let’s assume R = 0.2 nm.
We will want to express j in joules per H-bond. Using G = 6.67 × 10-11 N-m2/kg2, and R = 0.2 × 10-9 meters, we still need the mass “m” that we must apply in this equation. Presumably it’s the mass of the molecule that is being orientated by the each hydrogen bond.
Since we know the molecular mass M of the molecules that can hydrogen bond to each other and we also know the how many hydrogen bonds “h” are likely to link per molecule, m = M/(2 × h × NAv) (the factor of 2 is so we don’t count each bond twice). If h = 1 and if M = 2,000 kg/kmol, then m = 2,000/(2 × 6.023 × 1026) = 1.7 × 10-24 kg.
Finally calculate j by direct substitution, j = ~ 10-48 J for one H-bond, i.e., a truly insignificant number compared to ~ 2 × 10-20 J per H-bond (as in example 1). Hence the ratio of H-bond energy/gravitational energy per bond is about 2 × 1028 and gravitational energy between molecules is utterly irrelevant.
Such as result should have been expected since gravitation manifests itself as the pull of large bodies on each other; however, had we compared the strength of the gravitational potential of the mass of the molecule in the Earth’s field,
to the hydrogen bond, we would have found that the gravitational potential was ~ 10-16 J for one H-bond and the ratio H-bond energy/ gravitational energy per bond would have calculated to ~ 2 × 10-4; you would have incorrectly concluded that gravitation was dominant in these reactions instead of truly trivial!
The take-home is that you must make sure you are comparing apples and apples and not apples to oranges; in this case, gravitation merely holds the reactants in place, it does not directly affect motion between molecules and your modeling must reflect this essential physics.