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

##### 1 Answer

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 themolsof component#i# .#barG_i# is themolar Gibbs' free energyof component#i# .mixed#barG_i^"*"# is themolar Gibbs' free energyof component#i# .unmixed

Recall that for a chemical potential

#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^"*" + 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" = 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 **change in Gibbs' free energy of mixing** is then given by:

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