How do you measure Gibbs free energy?

Jul 30, 2017

You can't measure it directly, and $\Delta G$ is non-arbitrarily defined only for isothermal processes. Here are two ways to investigate the possibilities.

SCENARIO 1: DISSOLVING SOLUTE INTO SOLVENT

One way to measure it indirectly is through the chemical potential at constant temperature.

Let's say we had a solvent $j$ and we add solute to it. Then:

${\mu}_{j} = {\mu}_{j}^{\text{*}} + R T \ln {\chi}_{j}$

where:

• ${\chi}_{j} = \frac{{n}_{j}}{{n}_{t o t}}$ is the mol fraction of solvent $j$ in solution.
• ${\mu}_{j} = {G}_{j} / {n}_{j}$ is the chemical potential, or the molar Gibbs' free energy of substance $j$.
• ${\mu}_{j}^{\text{*}}$ is the chemical potential for some reference point. In this case, it is for an unmixed solvent $j$.

We define the change in molar Gibbs' free energy as:

$\Delta {\mu}_{j} = {\mu}_{j} - {\mu}_{j}^{\text{*}} = R T \ln {\chi}_{j}$

$= \Delta {\overline{G}}_{j} = {\overline{G}}_{j} - {\overline{G}}_{j}^{\text{*}} = \frac{\Delta {G}_{j}}{n} _ j$,

where the bar signifies molar quantities.

So, the change in the Gibbs' free energy of the solvent for adding solute into solution (starting with no solute) would be given by:

$\overline{\underline{| \stackrel{\text{ ")(" "DeltaG_j = n_jRTlnchi_j" }}{|}}}$

SCENARIO 2: MEASUREMENT OF NATURAL VARIABLES

Let's attempt to derive a functional form using only natural variables ($V , T , P$, etc). Consider the Maxwell Relation of $G$ in a thermodynamically-closed system, i.e. a conservative system:

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$ $\text{ } \boldsymbol{\left(1\right)}$

Now, if we write the total differential of $G$, there is another form of the Maxwell Relation:

$\mathrm{dG} = {\left(\frac{\partial G}{\partial T}\right)}_{P} \mathrm{dT} + {\left(\frac{\partial G}{\partial P}\right)}_{T} \mathrm{dP}$ $\text{ } \boldsymbol{\left(2\right)}$

We know the second term to be $V$ by inspection of $\left(1\right)$. For ideal gases we can use the ideal gas law, $V = \frac{n R T}{P}$, at constant temperature.

But unfortunately the first term has no physical meaning in thermodynamics, as the entropy at absolute zero is arbitrarily defined (Physical Chemistry by Levine). This is referring to...

$\Delta G = \Delta H - \Delta \left(T S\right)$

$= \Delta H - {T}_{1} \Delta S - {S}_{1} \Delta T - \Delta S \Delta T$

However, ${S}_{1}$ is arbitrary in the field of thermodynamics. As such, $\Delta G$ is only defined in a non-arbitrary way for isothermal processes, i.e. in

$\overline{\underline{| \stackrel{\text{ ")(" "DeltaG = DeltaH - TDeltaS" }}{|}}}$

Otherwise, it cannot be measured.