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:

• And

Solutions• Solutions are defined as homogenous phases

produced by dissolving one or more substances in another substance.

• Mixtures are not solutionso 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

• 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:

• Where Pi is the partial pressure exerted by gas i:

• 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

• 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:

• 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:

• Now consider exchange of component i between two phases, α and βholding all else constant; then:

• For a closed system:

• At equilibrium (dG = 0), then: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:

• Equating with the earlier equation:

• We can rearrange this as the Gibbs-Duhem Equation:

Interpreting Gibbs-Duhem

• 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:

• This equation further tells us that the chemical potentials do not vary independently, but change in a related way.

In spontaneous processes, components or species are distributed between phases

so as to minimize the chemical potential of all

components.

Final point about chemical potential:

Chemical Potential in Ideal Solutions

• In terms of partial molar quantities

• For an ideal gas:

• 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:

Volume and enthalpy changes of solutions

• Water–alcohol is an example of a non-ideal solution. The volume is expressed as:

• ∆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:

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:

o Note similarity to configurational entropy.o Note negative sign. How will entropy change?

• The total entropy of an ideal solution is then:

Free Energy Change of Solution

∆Gmixing = ∆Hmixing - T∆Smixing

• For an ideal solution, ∆Hmixing = 0

• And

• 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:

Total Free Energy of an Ideal Solution

• If we substitute

• Into

• And rearrange, we have:

• And

• 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.