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))#