What is chemical potential?

2 Answers
Jan 30, 2016

Answer:

#mu_i = frac{del G}{del n_i}#

Explanation:

Physically it is the change in the Gibbs free energy of the whole system if we added 1 mol of chemical #i# to a huge amount of the mixture that is the system.

Nov 11, 2016

Chemical potential is the differential change in the molar Gibbs' free energy of the system at constant #T#, #P#, and #n_(j ne i)# for component #i#. That came from the Maxwell relation:

#dG = -SdT + VdP + sum_i mu_idn_i#

In simple terms, it is analogous to the potential energy of a system: chemical potential runs downhill.

Now, if one divides the equation by #dn_i# and holds #T# and #P# constant, the relation is obtained for each component #i#:

#color(blue)(mu_i = ((delG_i)/(deln_i))_(T,P,n_(j ne i)))#

The chemical potential is where many of the thermodynamic relationships come from for ideal, ideally-dilute, and nonideal solutions.

The following discussion will give an overview of the first two types of solutions. By no means is this exhaustive of the type of information you can learn about these systems, but I will take a look at #DeltaV#, #DeltaG#, #DeltaS#, and #DeltaH#.


IDEAL SOLUTIONS

In an ideal solution, the interactions between solute (#i#) and solvent (#j#) molecules are identical. For these solutions, we have the following major quantities describing ideal mixtures:

Change in volume due to mixing

#bb(DeltaV_"mix"^"id" = sum_i n_i(barV_i - barV_i^"*") = 0)#

where #((delmu_i)/(delP))_(T,n_(j ne i)) = barV_i# is the molar volume of component #i# in solution, and #((delmu_i^"*")/(delP))_(T,n_(j ne i)) = barV_i^"*"# is the molar volume of component #i# by itself (i.e. not in solution).

Since the components don't interact with each other in solution, the volume of the solution is perfectly additive. Hence, #DeltaV_"mix"^"id" = 0# for an ideal solution, i.e. #barV_i = barV_i^"*"#.

Change in Gibbs' Free Energy, Entropy, and Enthalpy due to mixing

These are NOT zero. They do, however, cancel out for ideal solutions to give #DeltaH_"mix"^"id" = 0#:

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

#= sum_i n_i(mu_i - mu_i^"*")#

But if we note that #mu_i = mu_i^"*" + RTlnchi_i# for component #i# in an ideal solution, then:

#bb(DeltaG_"mix"^"id" = RTsum_i n_ilnchi_i)#

Similarly, the entropy of mixing also comes from the chemical potential.

#DeltaS_"mix"^"id" = sum_i n_i(barS_i - barS_i^"*")#

#= sum_i n_i(-((delmu_i)/(delT))_(P,n_(j ne i)) - ((delmu_i^"*")/(delT))_(P,n_(j ne i)))#

#= -sum_i n_i((del[mu_i - mu_i^"*"])/(delT))_(P,n_(j ne i))#

Plugging in #mu_i = mu_i^"*" + RTlnchi_i#, we get:

#bb(DeltaS_"mix"^"id") = -sum_i n_i((del[RTlnchi_i])/(delT))_(P,n_(j ne i))#

#= bb(-Rsum_i n_ilnchi_i)#

which if you notice, means that the enthalpy of mixing also follows:

#DeltaG_"mix"^"id" + TDeltaS_"mix"^"id" = bb(DeltaH_"mix"^"id")#

#= RTsum_i n_ilnchi_i - RTsum_i n_ilnchi_i = bb(0)#

where #chi_i = (n_i)/(n_1 + n_2 + . . . + n_N)# is the #"mol"# fraction f #i# in solution.

IDEALLY-DILUTE SOLUTIONS

In this scenario, there is very little solute and a lot of solvent. In this case, we differentiate between the solute and solvent and explicitly write solute as #i# and solvent as #A#, giving us similar quantities as before.

For the solutes:

#bb(DeltaV_"mix"^"id") = sum_(A) n_Acancel((barV_A - barV_A^"*"))^0 + bb(sum_(i ne A) n_i(barV_i - barV_i^"*"))#
#bb(DeltaG_"mix"^"id") = sum_(A) n_Acancel((mu_A - mu_A^"*"))^0 + bb(RTsum_(i ne A) n_ilnchi_i)#
#bb(DeltaS_"mix"^"id") = sum_(A) n_Acancel((-((delmu_A)/(delT))_(P, n_(A ne B)) - -((delmu_A^"*")/(delT))_(P, n_(A ne B))))^0 + bb( -Rsum_(i ne A) n_ilnchi_i)#
#bb(DeltaH_"mix"^"id") = sum_(A) n_Acancel((barH_A - barH_A^"*"))^0 + bb(sum_(i ne A) n_i(barH_i - barH_i^"*"))#

For the solvents:

They behave like the pure #A# (i.e. by itself, not in solution), since in an ideally-dilute solution, there is mostly solvent. So, they follow the same equations as in ideal solutions, just with a different subscript, #A# instead of #i# or #j#.