Solutions Lecture 6 Clapeyron Equation Consider two phases
- Slides: 23
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 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)Al. Si 3 O 8 is a solution (at high temperature). o A mixture of orthoclase (KAl. Si 3 O 8) and albite (Na. Al. Si 3 O 8) 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: P i = h i. X i 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 O 2 dissolved in seawater. o This tell us how the volume of water changes for an addition of dissolved O 2 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 (d. G = 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.
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 • 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 X 1 = (1 X 2). Then: ∆G = µ 1[(1 -X 2) +µ 2 X 2 = µ 1 + (µ 2 - µ 1)X 2 • This is the equation of a straight line on a G-bar–X 2 plot. • We can use it to extrapolate to chemical potentials.
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