Solutions
Lecture 6
Clapeyron Equation
• Consider two phases graphite & diamond–of
one component, C.
• Under what conditions
does one change into
the other?
• It occurs when ∆G for
the reaction between
the two is 0.
• Therefore:
d∆ Gr = ∆ Vr dP -∆ Sr dT = 0
• And
dP ∆ Sr
=
dT ∆ Vr
Solutions
• Solutions are defined as homogenous phases
produced by dissolving one or more substances in
another substance.
• Mixtures are not solutions
o Salad dressing (oil and vinegar) is not a solution, no matter how much you
shake it.
o The mineral alkali feldspar (K,Na)AlSi3O8 is a solution (at high temperature).
o A mixture of orthoclase (KAlSi3O8) and albite (NaAlSi3O8) will never be a
solution no matter how much you grind and shake it.
Molar Quantities
• Formally, a molar quantity is simply the quantity per
mole.
• For example, the molar volume is
V
=V
N
• Generally, we will implicitly use molar quantities and
not necessarily use the overbar to indicate such.
• Another important parameter is the mole fraction:
Xi = Ni/ΣN
Raoult’s Law
• Raoult noticed that the vapor pressures of a
ethylene bromide and propylene bromide solution
were proportional to the mole fractions of those
components:
Pi = Xi Pi o
• Where Pi is the partial pressure exerted by gas i:
Pi = Xi Ptotal
• Raoult’s Law states that the partial pressure of an
ideal component in a solution is equal to the mole
fraction times the partial pressure exerted by the
pure substance.
Ideal Solutions
• Turns out this does not hold in the exact and is only
approximately true for a limited number of solutions.
• Such solutions are termed ideal solutions. Raoult’s
Law expresses ideal behavior in solutions.
• In an ideal solution, interactions between different
species are the same as the interactions between
molecules or atoms of the same species.
Henry’s Law
• As we’ll see, most substances approach ideal
behavior as their mole fraction approaches 1.
• On the other end of the spectrum, most substances
exhibit Henry’s Law behavior as their mole fractions
approach 0 (Xi -> 0).
• Henry’s Law is:
Pi = hiXi
o where hi is Henry’s Law ‘constant’. It can be (usually is) a function of T and
P and the nature of the solution, but is independent of the concentration
of i.
Vapor Pressures in a
Water-Dioxane Solution
Partial Molar
Quantities and the
Chemical Potential
Partial Molar Quantities
• Now that we have introduced the mole fraction, X,
and variable composition, we want to know how
the variables of our system, e.g., V, S, change as we
change composition.
o These are partial molar quantities, usually indicated by the lower case
letter.
• For example:
o Such that
æ ¶V ö
vi = ç
è ¶ni ÷ø T,P, j¹i
åv = V
i
i
• This is the partial molar volume of component i. For
example, the partial molar volume of O2 dissolved in
seawater.
o This tell us how the volume of water changes for an addition of dissolved
O2 holding T, P, and the amounts of everything else constant.
Partial Molar Volumes of
Ethanol and Water
• If you add a shot (3 oz)
of rum to 12 oz of Coca
Cola, what will be the
volume of your ‘rum ‘n
coke’?
• Less than 15 oz! Blame
chemistry, not the
bartender.
Chemical Potential
• The chemical potential is defined as partial molar Gibbs
Free Energy:
æ ¶G ö
µi = ç
è ¶ni ÷ø T ,P,ni ,i¹ j
• The chemical potential tells us how the Gibbs Free
Energy will vary with the number of moles, ni, of
component i holding temperature, pressure, and the
number of moles of all other components constant.
• For a pure substance, the chemical potential is equal to
the molar Gibbs Free Energy (also the molar Helmholtz
Free Energy).
• The total Gibbs free energy of a system will depend upon
composition as well as on temperature and pressure. The
Gibbs free energy change of a phase of variable
composition is fully expressed as:
dG = VdP - SdT + å µi dni
i
• Now consider exchange of component i between two
phases, α and βholding all else constant; then:
dG = µia dnia + µib dnib
• For a closed system:
dnia = -dnib
• At equilibrium (dG = 0), then:
µia = µib
In a system at equilibrium, the chemical potential of every component
in a phase is equal to the chemical potential of that component in
every other phase in which that component is present.
Gibbs-Duhem Equation
• The Free Energy is the sum of chemical potentials:
• Differentiating:
G = å ni µi
i
dG = å ni dµi + å µi dni
• Equating with the earlier equation:
å n dµ + å µ dn = VdP + SdT + å µ dn
i
i
i
i
i
i
• We can rearrange this as the Gibbs-Duhem
Equation:
VdP + SdT - å ni dµi = 0
Interpreting GibbsDuhem
VdP + SdT - å ni dµi = 0
• In a closed system at equilibrium, net changes in
chemical potential will occur only as a result of
changes in temperature or pressure. At constant
temperature and pressure, there can be no net
change in chemical potential at equilibrium:
å n dµ = 0
i
i
• This equation further tells us that the chemical
potentials do not vary independently, but change
in a related way.
Final point about chemical potential:
In spontaneous processes,
components or species are
distributed between phases so as to
minimize the chemical potential of
all components.
Chemical Potential in
Ideal Solutions
æ ¶G ö
çè
÷ø = V
¶P T
• In terms of partial molar quantities
• For an ideal gas:
æ ¶ µi ö
çè
÷ø = vi
¶P T
RT
æ ¶µi ö
=
çè
÷
¶P ø T ,ideal
P
µiP - µio = RT ln
P
P˚
• Integrating from P˚ to P:
• Where P˚ is the pressure of pure substance in
‘standard state’ and µ˚ is the chemical potential of i
in that state. In that case, P/P˚ = Xi and:
µi,ideal = µio + RT ln Xi
Volume and enthalpy changes
of solutions
• Water–alcohol is an example of a non-ideal
solution. The volume is expressed as:
V = å Xi Vi + ∆ Vmixing
i
• ∆V term may be negative, as in rum ‘n coke.
• Similarly, mix nitric acid and water and the solution
gets hot. Enthalpy of solutions is expressed as:
H = å Xi H i + ∆ H mixing
o The ∆H term is positive in the nitric acid case.
• For ideal solutions, however, ∆Vmixing = ∆Hmixing = 0
Entropy changes of
solution
• What about entropy and free energy changes of
ideal solutions?
• Even in ideal solutions, there is an entropy change
(increase) because we have increased the
randomness of the system.
• The entropy change of ideal solution is:
∆ Sideal mixing = -Rå Xi ln Xi
i
o Note similarity to configurational entropy.
o Note negative sign. How will entropy change?
• The total entropy of an ideal solution is then:
S = å Xi Si + ∆ Sideal mixing = å Xi Si - Rå Xi ln Xi
i
i
i
Free Energy Change of Solution
∆Gmixing = ∆Hmixing - T∆Smixing
• For an ideal solution, ∆Hmixing = 0
• And
∆ G ideal mixing = RT å Xi ln Xi
i
• Because the log term is always negative, ideal
solutions have lower free energy than a mixture of
their pure constituents and ∆G decreases with
increasing T. This is why things are usually more
soluble at higher T.
• Total Free Energy of an ideal solution is:
∆ Gideal solution = å Xi Gi + RT å Xi ln Xi
i
i
Total Free Energy of an
Ideal Solution
• If we substitute
• Into
µi,ideal = µio + RT ln Xi
∆ Gideal solution = å Xi Gi + RT å Xi ln Xi
i
i
• And rearrange, we have:
• And
G ideal = å Xi µio + RT å Xi ln Xi
G ideal = å Xi µi
• The Free Energy of a solution is simply the sum of the
chemical potentials of the components times their mole
fractions.
Free Energy of Mixing in
an Ideal Solution
Method of Intercepts
• Consider a two
component ideal solution
so that X1 = (1-X2). Then:
∆G = µ1[(1-X2) +µ2X2 =
µ1 + (µ2 - µ1)X2
• This is the equation of a
straight line on a G-bar–X2
plot.
• We can use it to
extrapolate to chemical
potentials.