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

`DeltaG_"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:

`color(blue)(DeltaG_"mix") = nRT(chi_1lnchi_1 + chi_2lnchi_2)`

`= ("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"))`

`~~ -"12800 J"`

`~~ color(blue)(-"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`.
• `barG_i` is the molar Gibbs' free energy of component `i` mixed.
• `barG_i^"*"` is the molar Gibbs' free energy of component `i` unmixed.

Recall that for a chemical potential `mu_i = barG_i`, we can write a change away from a standard state in the liquid phase as:

`mu_i = mu_i^(@) + int_(P^@)^(P_i) ((delmu_i)/(delP))_TdP`

where `P^@ = "1 bar"` is the standard pressure, and `mu_i^(@)` is the standard chemical potential for the liquid phase defined at `"1 bar"` and some temperature `T`.

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

`dmu = dbarG = -barSdT + barVdP`

And so,

`((delmu_i)/(delP))_T = barV_i`

This gives:

`mu_i = mu_i^(@) + int_(P^@)^(P_i) barV_idP`

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^(@) + RTint_(P^@)^(P_i) 1/P_idP`

`= mu_i^(@) + RTln(P_i/P^@)`

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

Then, `mu_i^(@) -= mu_i^"*"`, and `P^@ -= P_i^"*"`, where `P_i^"*"` is the pure vapor pressure of component `i`. This gives us the relationship:

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

`= mu_i^"*" + RTlnchi_i`

where we used Raoult's law, `P_i = chi_iP_i^"*"`. `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 `DeltaG_"mix"`:

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

`= RTsum_i n_ilnchi_i`

For a two-component solution, we then have:

`DeltaG_"mix" = RT(n_1lnchi_1 + n_2lnchi_2)`

Now, if we multiply the right-hand side by `n/n`, where `n` is the total mols, the change in Gibbs' free energy of mixing is then given by:

`color(blue)(DeltaG_"mix" = nRT(chi_1lnchi_1 + chi_2lnchi_2))`