Thermal energy in calories

A short note. In chemistry and related fields we quite often describe energies in units of calories, instead of using Joules from the SI system. From physics, we know that the thermal energy at room temperature is about 25 meV. The question is: how does the thermal energy relate to an interaction energy (e.g. between two molecules) if that energy is provided in units of kcal/mol?

The thermal energy in Joule is k_B*T. That energy is “accessible” to any given molecule, i.e. for comparison with a value normalized on a mole (as an energy provided in units kcal/mol obviously is) we need to multiply it with the Avogadro constant A, yielding the thermal energy in Joule/mol. About 4190 Joule comprise one kcal (1000 calories), i.e. the thermal energy in kcal/mol is k_B * T * A / 4190. Given that, we can relate any given interaction energy to the thermal energy. A quick Python implementation:

import sys
# Thermal energy for 300 K:
# T_300 = 4.14 * 10^-21 J
# (k_B * 300 K, with k_B = 1.38 * 10^-23 J/K)
# The relation between Joule and (kilo)calories:
joule_per_kcal = 4190
# Thermal energy in kcal/mol:
# T_300_kcalpermol = T_300 * A / joule_per_kcal
# with A = 6.02 * 10^23 (Avogadro constant)
T_300_kcalpermol = 4.14 * 6.02 * 100 / joule_per_kcal
print "Thermal energy (at 300 K): %.2f kcal/mol" % (
# Read actual energy in kcal/mol.
energy_kcalpermol = float(sys.argv[1])
energy_per_thermal_energy = energy_kcalpermol / T_300_kcalpermol 
print "%.3f kcal/mol divided by the thermal energy: %.1f" % (
    energy_kcalpermol, energy_per_thermal_energy)

So how does an interaction energy of 2 kcal/mol relate to the thermal energy?

$ python 2
Thermal energy (at 300 K): 0.59 kcal/mol
2.000 kcal/mol divided by the thermal energy: 3.4

First of all, the thermal energy is 0.59 kcal/mol — something to keep in mind when dealing with kcal/mol-based energy values on a regular basis. We learn that an interaction energy of 2 kcal/mol is already more than three times larger than the thermal energy, i.e. this kind of interaction may easily dominate diffusion and stands out of the thermal noise.

Leave a Reply

Your email address will not be published. Required fields are marked *

Human? Please fill this out: * Time limit is exhausted. Please reload CAPTCHA.