How do you calculate the Gibbs' free energy of mixing?

Sep 14, 2017

$\Delta {G}_{\text{mix}}$ for an ideal solution is defined as the change in Gibbs' free energy due to mixing two components of the ideal solution together.

For a two-component ideal solution, we have:

$\textcolor{b l u e}{\Delta {G}_{\text{mix}}} = n R T \left({\chi}_{1} \ln {\chi}_{1} + {\chi}_{2} \ln {\chi}_{2}\right)$

= ("6.00 mols" + "2.50 mols")("8.314472 J/mol"cdot"K")("298 K")("6.00 mols"/"8.50 mols" ln ("6.00 mols"/"8.50 mols") + ("2.50 mols")/("8.50 mols")ln("2.50 mols"/"8.50 mols"))

$\approx - \text{12800 J}$

$\approx \textcolor{b l u e}{- \text{12.8 kJ}}$

This confirms that the mixing is spontaneous (as it should be for two nonpolar organic substances).

I derive this result below.

DISCLAIMER: DERIVATION BELOW!

The initial state is the unmixed state, and the final state is the mixed state:

DeltaG_"mix" = sum_i n_i barG_i - sum_i n_i barG_i^"*" = sum_i n_i(barG_i - barG_i^"*"),

where:

• ${n}_{i}$ is the mols of component $i$.
• ${\overline{G}}_{i}$ is the molar Gibbs' free energy of component $i$ mixed.
• ${\overline{G}}_{i}^{\text{*}}$ is the molar Gibbs' free energy of component $i$ unmixed.

Recall that for a chemical potential ${\mu}_{i} = {\overline{G}}_{i}$, we can write a change away from a standard state in the liquid phase as:

${\mu}_{i} = {\mu}_{i}^{\circ} + {\int}_{{P}^{\circ}}^{{P}_{i}} {\left(\frac{\partial {\mu}_{i}}{\partial P}\right)}_{T} \mathrm{dP}$

where ${P}^{\circ} = \text{1 bar}$ is the standard pressure, and ${\mu}_{i}^{\circ}$ is the standard chemical potential for the liquid phase defined at $\text{1 bar}$ and some temperature $T$.

The Maxwell Relation for the molar Gibbs' free energy shows:

$\mathrm{dm} u = \mathrm{db} a r G = - \overline{S} \mathrm{dT} + \overline{V} \mathrm{dP}$

And so,

${\left(\frac{\partial {\mu}_{i}}{\partial P}\right)}_{T} = {\overline{V}}_{i}$

This gives:

${\mu}_{i} = {\mu}_{i}^{\circ} + {\int}_{{P}^{\circ}}^{{P}_{i}} {\overline{V}}_{i} \mathrm{dP}$

The change in chemical potential due to a change in pressure occurs above a liquid, and can reasonably be based on ideal gases. So:

${\mu}_{i} = {\mu}_{i}^{\circ} + R T {\int}_{{P}^{\circ}}^{{P}_{i}} \frac{1}{P} _ i \mathrm{dP}$

$= {\mu}_{i}^{\circ} + R T \ln \left({P}_{i} / {P}^{\circ}\right)$

Now, suppose we redefine the standard state to be the pure (unmixed) state of one of two liquid phases.

Then, ${\mu}_{i}^{\circ} \equiv {\mu}_{i}^{\text{*}}$, and ${P}^{\circ} \equiv {P}_{i}^{\text{*}}$, where ${P}_{i}^{\text{*}}$ is the pure vapor pressure of component $i$. This gives us the relationship:

mu_i = mu_i^"*" + RTln(P_i/P_i^"*")

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

where we used Raoult's law, ${P}_{i} = {\chi}_{i} {P}_{i}^{\text{*}}$. ${\chi}_{i}$ is the mol fraction of $i$, and is understood to be in the liquid phase.

Now, we can plug in this result to obtain $\Delta {G}_{\text{mix}}$:

DeltaG_"mix" = sum_i n_i (mu_i - mu_i^"*")

$= R T {\sum}_{i} {n}_{i} \ln {\chi}_{i}$

For a two-component solution, we then have:

$\Delta {G}_{\text{mix}} = R T \left({n}_{1} \ln {\chi}_{1} + {n}_{2} \ln {\chi}_{2}\right)$

Now, if we multiply the right-hand side by $\frac{n}{n}$, where $n$ is the total mols, the change in Gibbs' free energy of mixing is then given by:

$\textcolor{b l u e}{\Delta {G}_{\text{mix}} = n R T \left({\chi}_{1} \ln {\chi}_{1} + {\chi}_{2} \ln {\chi}_{2}\right)}$