Chemistry 2402 - Thermodynamics Lecture 9 : Free Energies Lecture 10 : Phase Diagrams and Solubility Lecture 11: Solubility (cont.)
Phase Diagrams Liquids eke out an existence in the niche between solids and gases. P T solid gas liquid critical point triple point supercritical fluid Coexistence lines represent values of P and T where two phases can coexist.
The slope of coexistence lines. The variation in G is dG = -SdT +VdP If we want to move along the coexistence line between phases I and II, then dGI = dGII since GI = GII defines coexistence. so that -SIdT + VIdP = -SIIdT + VIIdP The temperature and the pressure must, of course, be the same in any phases at coexistence. Rearranging this equation, we have
The Clapeyron Equation But G = H-TS=0 and, therefore, S = H/T Substituting this, we arrive at the Clapeyron equation If H and V vary little with P and T then we can integrate the Clapeyron equation to give
Pressure melting or pressure freezing? From the Clapeyron equation Now ΔH < 0 for freezing. Usually, ΔV < 0 also since solids are denser than liquids. That means the dP/dT is positive as in Diagram I. When ΔV > 0 as in ice or silica, then we get Diagram II I P T solid gas liquid critical point triple point II P T solid gas liquid critical point
Example Problem or How To Use This Sausage Machine At 273.16K the enthalpy change on melting of ice is 6.0 kJ/mol and the corresponding volume change is -1.6 × 10-6 m3/mol. Estimate the temperature at which ice will melt at 1000 atm pressure (take 1 atm = 105 Nm-2). You can assume that neither the enthalpy difference nor the volume difference changes significantly with pressure. As the enthalpy and volume differences remain constant, we can use the integrated form of the Clapeyron equation P1 = 1.01 × 105 Nm-2, P2= 1.01 × 108 Nm-2, ΔH = 6.0 × 103J/mol , ΔV = -1.6 × 10-6m3/mol, T1 = 273.16K Which gives T2 = 266.0K This is a public domain image: http://en.wikipedia.org/wiki/Image:Sausage_making-H-1.jpg
Flash Quiz! Which of the following can occur? (assume T stays constant in every case) For those that can, give an example from the natural world. Gaseous hydrogen condenses as pressure is increased Liquid iron solidifies as pressure is increased Liquid water solidifies as pressure is increased Normal water-ice (ice Ih) melts as pressure is increased Gaseous water condenses as pressure is decreased.
Hints low H high H solid liquid gas Which of the following can occur? (assume T stays constant in every case) For those that can, give an example from the natural world. Gaseous hydrogen condenses as pressure is increased Liquid iron solidifies as pressure is increased Liquid water solidifies as pressure is increased Normal water-ice (ice Ih) melts as pressure is increased Gaseous water condenses as pressure is decreased. which has higher V? high V
Answers low H high H solid liquid gas Which of the following can occur? (assume T stays constant in every case) For those that can, give an example from the natural world. Gaseous hydrogen condenses as pressure is increased Liquid iron solidifies as pressure is increased Liquid water solidifies as pressure is increased Normal water-ice (ice Ih) melts as pressure is increased Gaseous water condenses as pressure is decreased. which has higher V? high V Jupiter Earth’s inner core Trick question: not to ice Ih, but there are high-density ice forms Base of (some) glaciers If you wrote “cloud formation” you assumed the gas expansion was accompanied by a temperature drop.
Reading P-V Phase Diagrams liquid solid gas V
Reading P-V Phase Diagrams critical point s + l liquid l + g solid gas s + g triple line molar V The shaded region depicts values of the pressure and total volume for which we have coexistence of two (or three at the triple point) phases. (The volume of the gas has been significantly reduced to fit it all in.)
Chemical Potential and Liquid-Vapor Equilibrium At constant T and P, equilibrium between liquid and vapour means dG = μliqdNliq + μvapourdNvapour = 0 or μliq = μvapour Generally, we can’t measure chemical potentials directly but we can get close for an ideal gas using The change in G due to a change in pressure is given by the shaded area. G(P ) - G(P ) = 2 1 shaded area V P P 1 2 P
Chemical Potential for an Ideal Gas – Introducing the Pain of Reference States Using V = nRT/P and integrating from P1 to P2 gives G2 = G1 + nRTln(P2 /P1) We have to choose a reference state. In the case of gases we shall take the reference free energy Go to be the free energy of 1 mole of gas at 1 atm. We can then write G = Go + RTln(P/atm) and, since μ = G/N, we have = o + kBTln(P/atm)
Introducing Solutions In the case of a mixture of ideal gases, each species acts as if it was alone (remember ideal gases don’t interact). This means we have the following chemical potential for species i i = oi + kBTln(Pi/atm) where Pi = xi(gas)P [ xi(gas) = Ni/N ] is the partial pressure of species i. xi is called the mole fraction. The Cunning Plan i(gas) = i(liq). liquid vapour Use the known chemical potential of a species in the gas phase to establish its chemical potential in a solution at equilibrium with the gas.
Relating Partial Pressures to Concentrations The problem is that the gas expression talks of partial pressures while concentrations are the natural way of talking about the solution. An approximate relation was proposed by Raoult in 1880. He noted that, as long as the components in solution differed little from one another - e.g. benzene and toluene – then Pi(gas)= xi(liq)Pi* Raoult’s Law P T solid gas liquid critical point triple point xi(liq) = Ni/N in solution and Pi* is the pressure at the coexistence of the liquid and vapour phases of the pure i. This pressure is called the vapour pressure of i (at the temperature of interest).
The Chemical Potential in an Ideal Solution (The Maths Lover’s Slide) A solution that follows Raoult’s Law is called an ideal solution At equilibrium i(liq) = i(gas) For an ideal gas i(gas) = oi + kBTln(Pi/atm) For an ideal solution i(liq) = oi + kBTln(xi(liq) Pi* /atm) Avoiding the gas reference state, let oi + kBTln(Pi* /atm) = i*(liq) where i*(liq) is the chemical potential of a pure liquid of i at coexistence with the vapour at the vapour pressure Pi*. So finally i(liq) = i*(liq) + kBTln(xi(liq)) we have the ideal solution chemical potential in terms of the concentration (mole fraction is directly related to concentration)
Changing the Reference State to a Standard State The chemical potential of each species will have a reference state with a different pressure Pi*. To avoid this we often assume i*(liq) oi(liq) i.e. use 1 atm instead of Pi* With this change we have, for the ideal solution, i(liq) = oi(liq) + kBTln(xi(liq))
Sample exam questions from previous years On the axes provided, draw a representative pressure-volume phase diagram (not to scale) of a typical pure substance which has a solid, liquid, and vapour phase. Label all regions, as well as the critical point. At 40.0 °C, heptane has a vapour pressure of 92.45 mmHg, and octane has a vapour pressure of 31.06 mmHg. In a 40 °C laboratory, a mixture is prepared in a sealed container, with 100.0g of liquid heptane and 60.0g of liquid octane. Assume this mixture obeys Raoult’s Law. If a liquid-vapour coexistence is established at that pressure, what will the mole fraction of octane in the vapour be? Consider a gas mixture at equilibrium with a solution of the same species. What are the conditions for coexistence in terms of the intensive properties - temperature, pressure and chemical potential.
Summary Next Lecture You should now Be able to explain pressure melting and pressure freezing, and relate it’s occurrence to ΔVmelt and ΔHmelt Be able to draw and read a typical P-V phase diagram Understand the role of reference states when handling chemical potentials or free energies Be able to explain the role of Raoult’s Law in the derivation of the properties of an ideal solution Explain and use the concept of an ideal solution Next Lecture Solubility (cont)