I am using an approximation of density matrix in a high order series expansion to predict the total internal energy of chemical systems. During my research,it was found out that the precision of such algorithm is accurate up to 0.00001 Hartree by optimising the independent parameters, which is roughly 0.026kJ/mol's error fluctuation. I wonder in reality, what can 1 kJ/mol do in our daily life, what phenomena can 1 kJ/mol of energy possibly trigger? I'm quite a mathematical person and haven't really studied enough chemistry nor physics. if someone could give me some physical intuitions that would be greatly appreciated!

And I meant to leave it as a broad question. Since we are not picking a specific chemical system to chew on here, any unbiased thoughts are welcomed and shall be considered in this post! A collective answer from the various point of view is probably a better way to get one's head around rather than a single aspect answer.

  • $\begingroup$ When I'm confronted with heat figures either from calorimetry work or molecular modelling studies I ask myself the same question. What I've tried to do, over the years is to try and calibrate myself. For example, what Wattage is your kettle? How long does it take to heat up some water from room temp to boiling? But not simply timing this process but getting a feel for it. $\endgroup$ Commented Oct 27, 2016 at 7:37
  • $\begingroup$ The average kinetic energy of molecules is $3RT/2$ which is $\approx 3.7 \pu{kJ mol^{-1}}$at $300$ K. There is also internal vibrational & rotational energy of molecules which adds $RT/2$ for each squared term in the energy. So your $1 \pu{kJ mol^{-1}}$ does not account to much as its going to be overwhelmed by thermal energy. Of course at $30$ K its different. $\endgroup$
    – porphyrin
    Commented Oct 27, 2016 at 7:37
  • $\begingroup$ @porphyrinas I agree with you. Lower the temperature, more quantum effects would be involved, the predicted result must certainly require a higher level of accuracy... that's totally reasonable to think for. $\endgroup$
    – Gvxfjørt
    Commented Oct 27, 2016 at 9:59

1 Answer 1


For this reaction

$$\ce{A -> B}$$

a $\Delta G$ value of $1\ \mathrm{kcal\,mol^{-1}}$ will result in an equilibrium value

$$K = e^{-\frac{\Delta G}{RT}}$$

of approximately 0.18 favoring reactant. Every additional increment or decrement by this value will change the relative ratio by roughly 5.


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