Why does the Born equation give the Gibbs free energy of solvation rather than enthalpy of solvation?

Question

The Born equation gives the difference in energy required to charge a particle in a vacuum and in solution which results in the work required to transfer an ion from a vacuum into solution. It is derived thus:

$$w=\int_0^Q\frac{Q}{4\pi\epsilon r}dQ-\int_0^Q\frac{Q}{4\pi\epsilon_0 r}dQ$$ Where $\epsilon=\epsilon_r\epsilon_0$ ($\epsilon_r$ is the relative permittivity of the medium/solvent - a dimensionless quantity). $w$ is work done.

$$w=\frac{1}{4\pi\epsilon r}\int_0^QQdQ-\frac{1}{4\pi\epsilon_0 r}\int_0^QQdQ$$ $$w=\frac{Q^2}{8\pi\epsilon r}-\frac{Q^2}{8\pi\epsilon_0 r}$$ $$w=\frac{z^2e^2}{8\pi\epsilon r}-\frac{z^2e^2}{8\pi\epsilon_0 r}$$ So for a mole of ions the work done is equal to: $$\frac{N_Az^2e^2}{8\pi\epsilon r}-\frac{N_Az^2e^2}{8\pi\epsilon_0 r}=\triangle_{solv}G^{\theta}$$ My textbook seems to suggest that the last step makes it Gibbs free energy of solvation rather than just work done. I have two issues with this. Firstly, why is it Gibbs free energy and not, say, enthalpy of solvation or internal energy of solvation? Secondly, why would scaling the number of ions to one mole make it into Gibbs free energy? Isn't Gibbs free energy an extensive property with units of Joules? I know that the Gibbs free energy (and enthalpy) are often given with units of $Jmol^{-1}$ but isn't this strictly molar gibbs free energy/chemical potential?

Answer 1

1) Because $-\Delta G$ is the maximum possible nonexpansion work done by a system in a constant temperature and pressure process. You can write $$\Delta G \le w_{non PV}\quad (\text{constant T and P, closed system})$$ where the equality sign holds for reversible changes.

Here's proof: $$\begin{align*} H &= U + P V&\text{(definition of enthalpy)}\\ dH &= dq + dw + d(PV) &\text{(because dU = dq + dw)}\\ dG &= dH - T dS - S dT &\text{(because G = H - TS)}\\ &=dq + dw + d(PV) - T dS -S dT &\\ &= dq + dw + d(PV) - T dS & \text{(when the change is isothermal)}\\ &= dw_{rev} + d(PV) &\text{(if reversible,}\ dq_{rev} = T dS\text{)}\\ &= dw_{electrical,rev} + dw_{expansion,rev} + P dV + V dP & \text{(partitioning the work)}\\ &= dw_{electrical,rev} + V dP & (dw_{expansion,rev} = - P dV)\\ &= dw_{electrical,rev} & \text{(if pressure is constant)}\\ \end{align*}$$

2) It isn't the scaling to one mole that makes the result a Gibbs free energy (see 1); that just makes it a molar Gibbs free energy. You can use either extensive or intensive Gibbs free energies in calculations; whatever is most convenient. Just make it clear (either by context or notation) which one you're using.