How the Gibbs free energy of a chemical compound is theoretically calculated from scratch?

When saying that the Gibbs free energy of NaCl is -384.1 kJ/mol, this value is related to the chemical bonds of Na-Cl.

We should be able to calculate this value from the sum of energies in chemical bonds of one mole of NaCl? Right? How?

  • $\begingroup$ In theory, one can use quantum mechanics based codes to calculate bonding energies from first principles. However, constructing new potentials usually involves making sure that the potential satisfies basic materials properties, such as getting the melting point right. So it can be a little circular. Usually in something like DFT you are interested in defect formation of diffusivity, and tailor a potential to get the bulk thermodynamic properties right. $\endgroup$ – Jon Custer Sep 22 '14 at 21:44

The value you have reported is the standard Gibbs free energy change of formation, which has a specialized definition. A standard free energy change of formation is the free energy change that occurs when 1 mole of substance is formed by reaction of its constituent elements in their standard states at 1 atmosphere of pressure and 25 degrees Celsius.

Thus, for sodium chloride:

$$\ce{Na(s) +\frac{1}{2} Cl2(g) -> NaCl(s)}\ \ \ \Delta_f G^\circ =-384.1 \text{ kJ/mol}$$

Now, all we need to be able to do is measure $\Delta G$ for this reaction at standard temperature and pressure.

In practice, we will separately detemrine $\Delta_f H^\circ$ and $\Delta_f S^\circ$ and plug them into $\Delta_f G^\circ = \Delta_f H^\circ -T\Delta_f S^\circ$.

At constant pressure (which we are by definition operating at), $\Delta H$ is equal to the heat transfer $\Delta H = q = -C_p\Delta T$, which is equal to the heat capacity of the surrounding (in practice the calorimeter, which is reasonably isolated from the universe) multiplied by the change in temperature.

$\Delta S$ is still tricky to measure. At equilibrium, for a reversible process at constant temperature $\Delta S = \frac{q}{T} = \frac{\Delta H}{T}$. If the process is not reversible or you are not at equilibrium, then... then it is challenging.

  • $\begingroup$ Keep in mind that the $\Delta G$ for the elements at standard state, is by definition 0. So for this particular reaction, you only need to measure the right side - how much energy goes into the $\ce{NaCl(s)}$. $\endgroup$ – Geoff Hutchison Sep 23 '14 at 13:34

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