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

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?

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*}
• Would you please elaborate on why $-\triangle G$ is the maximum electrical work? – RobChem Jan 12 '15 at 15:55
• OK, I added the proof to my answer... the inequality applies to either irreversible or reversible processes. For an irreversible process you'll have $\Delta G < w_{nonPV}$. Watch the signs, $w_{nonPV}$ is minus the non-PV work done by the system, so you could write $w_{by, nonPV} < -\Delta G$ for an irreversible, closed system, constant T,P process. – Fred Senese Jan 12 '15 at 16:22
• "When the change is not reversible the change is gibbs free energy is different to the electrical work done. How is this possible given that energy cannot be created nor destroyed?" Look at the line in the proof where I say "if reversible, $dq_{rev} = TdS$". To see where the inequality for an irreversible change comes in, you can change that step to "if irreversible, $dq_{irrev} \lt T dS$" and take it from there. – Fred Senese Jan 12 '15 at 16:28