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

Question

Henry Antrim · Accepted Answer

#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.

The ideal two-component solution is as follows:

#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 demonstrates that the two nonpolar organic substances are mixing spontaneously, as they should.

I get at this conclusion below.

NOTICE: DERIVATION DOWN BELOW!

The unmixed state is the starting point, and the mixed state is the end result:

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

where

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#.

Regarding the molar Gibbs free energy, the Maxwell Relation exhibits:

#dmu = dbarG = -barSdT + barVdP#

Thus,

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

This results in:

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

This means that the change in chemical potential caused by a pressure change happens above a liquid and is logically based on ideal gases.

#mu_i = mu_i^(@) + RTint_(P^@)^(P_i) 1/P_idP#

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

Let's say that the pure (unmixed) state of one of the two liquid phases is now what we refer to as the standard state.

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#

Next, we have for a two-component solution:

#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))#

Naomi Aronson · Answer

undefined The Gibbs' free energy of mixing (ΔG_mix) can be calculated using the equation:

$$ \Delta G_{\text{mix}} = RT(x_1 \ln x_1 + x_2 \ln x_2) $$

where:
- $ R $ is the gas constant (8.314 J/(mol·K)),
- $ T $ is the temperature in Kelvin,
- $ x_1 $ and $ x_2 $ are the mole fractions of the components in the mixture,
- $ \ln $ represents the natural logarithm.

The mole fractions ($ x_1 $ and $ x_2 $) are calculated by dividing the moles of each component by the total moles in the mixture.