Chemistry 2402 - Thermodynamics Lecture 8 : Converting Heat to Work Lecture 9 : Free Energies Lecture 10: Phase Diagrams

The Intensive Parameters (again) The intensive parameters temperature, pressure and chemical potential were previously defined in terms of changes in the energy U by We will also want to know how entropy changes when we change energy, particle number and volume – to this end we need:

How can entropy change? Now we can write the most general expression for a change in entropy dS in terms of changes is U, V and N as or, making use of the intensive parameters, Now we can relate how systems tend towards equilibrium (i.e. to maxima in S) in terms of T, P and  and the kinds of changes (i.e. dU, dV and dN) that we chose to allow.

Does behave like a real temperature? Here’s the basic problem in thermodynamics. UI VI NI UII VII NII ? ? After Before In an isolated system, we remove an internal constraint. What happens? Let’s consider the case when we allow the internal wall to conduct heat between subsystem I and subsystem II.

Does behave like a real temperature? The total entropy is S = SI + SII and dS = dSI +dSII As the only quantity allowed to change in either region is the energy, we have dS = dUI/TI + dUII /TII The system is isolated, i.e.the total energy is conserved, so dUI = -dUII - i.e. what ever flows into I must have flown out of II. So now we have dS = dUI (1/TI - 1/TII) Equilibrium corresponds to the state at which S is a maximum, i.e. dS = 0. According to the equation, equilibrium corresponds to the state at which TI = TII. In any spontaneous process dS > 0. If TI > TII then, according to our equation relating dS to dU, dUI<0, i.e. heat must flow from the high temperature region to the low temperature one.

Thermodynamics in Open Systems U V N ? ? ? An isolated system is characterised by the energy, volume and particle number which are fixed What can one say about on open system? What properties are fixed?

Constraints and Reservoirs When we allow exchange of a system quantity with the surroundings we shall assume that the surroundings are so much larger than our system, that the state of the surroundings remains unchanged – i.e. the surroundings represents infinite reservoirs. If a system can exchange heat with the surroundings then energy is no longer fixed but, at equilibrium, the temperature of the system must equal that of the surroundings and so we have fixed the temperature. If our system can expand or contract, volume is no longer fixed but at equilibrium the system pressure is equal to that of the surroundings. Similarly, if we allow particle exchange, we no longer have a fixed number of particles, but we are effectively setting the system chemical potential to be equal to that of the surroundings.

Equilibrium at constant temperature System with fixed V and N qin Surroundings or bath T At equilibrium, dS = dSsys+dSbath =0 dSbath = -qin/T, where T is the temperature of the bath. And dUsys = qin because V is fixed so no work can be done So our equilibrium condition is dSsys - dUsys/T = 0 where T is the bath temperature. Note that we are no longer maximising Ssys. .

So what ARE we maximising? dSsys - dUsys/T = 0 → d(S-U/T) = 0 at constant T So we could invent a new entropy  = S-U/T to be maximised. Perfectly sound, but not the way it was done. Instead, the equilibrium condition was rewritten to be dUsys - TdSsys = 0 and we defined A = U-TS such that equilibrium at constant temperature was defined as the state that minimised A at constant T, i.e. dA = dU-TdS = 0 This A is a modification of the energy U and is called a free energy. This one is called the Helmholtz or canonical free energy. Instead of U=U(S,V,N) we have A = A(T,V,N)

What about equilibrium at constant T and P? Equilibrium at constant T and P, the most useful for most chemical processes, minimizes the Gibbs free energy G = G(T,P,N) with G = U -TS +PV (Note enthalpy H = U+PV) = H - TS = A + PV = N And There are 7 possible free energies, each defined by a different combination of constraints. We will work with the Gibbs free energy from now.

Solid, Liquid and Vapour Image source: Mila Zinkova & CillanXC http://en.wikipedia.org/wiki/Image:Iceberg_with_hole_edit.jpg

Equilibrium between phases The equilibrium state at a fixed temperature and pressure is that which minimises G. Let’s look at a sketch of G versus temperature for the solid, liquid and gas phase of a substance at a single pressure. Ssolid < Sliquid < Sgas

Pressure Effects on Phase Equilibrium When pressure is decreased, G decreases too, but it decreases a different amount for each phase. Phases with bigger the molar volume (gases) are decreased more. In this case the liquid is now no longer the most stable phase at any temperature. Ts is the sublimation point where gas and solid are in coexistence.

The Critical Point: where liquid and gas merge Increasing the pressure sees the liquid and gas states become increasingly similar. G T solid liquid gas Tf Tb At a particular value of P and T, the critical point, the liquid and gas phase become indistinguishable.

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.

Flash Quiz! Why can’t we use easily use hydrogen or methane as a fuel? P T solid gas liquid critical point triple point supercritical fluid Why can’t we use easily use hydrogen or methane as a fuel? substance Ttp (°C) Tc (°C) hydrogen –259.4 –240.2 methane –182.5 –82.7 propane (LPG) –187.9 96.7 hexane (petrol) –95.1 234.5 data from http://www.lakeshore.com/pdf_files/Appendices/LSTC_appendixI_l.pdf

Answer Why can’t we use easily use hydrogen or methane as a fuel? P T solid gas liquid critical point triple point supercritical fluid Why can’t we use easily use hydrogen or methane as a fuel? Unless we cool them heaps, they are always above their critical point, which means that we cannot exploit a phase transition to a small-volume liquid. If we want small volumes, we would need extreme pressures. If we want manageable pressures, we would need to store a large volume to contain enough to provide the chemical energy we need. substance Ttp (°C) Tc (°C) hydrogen –259.4 –240.2 methane –182.5 –82.7 propane (LPG) –187.9 96.7 hexane (petrol) –95.1 234.5 data from http://www.lakeshore.com/pdf_files/Appendices/LSTC_appendixI_l.pdf

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-TS=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 II

Sample exam questions from previous years Provide a brief explanation of the following terms: Gibbs free energy Triple point Water ice melts at 0 °C at 1 atm, with an enthalpy change of 6.0095 kJmol‑1. The volume decrease on melting is 1.65×10‑6 m3mol‑1. What change in the bulk melting temperature will be observed if an ice skater applies a pressure of 13 atm? Is this effect enough to explain the melting of ice while skating? Under standard laboratory conditions, the melting point of iron is 1811 K. The inner core of the earth is primarily composed of iron. It is solid, despite being at a temperature of around 5600 K. Give a brief explanation of how this is possible. Calculate an estimate of the pressure at the surface of the inner core, where solidification first occurs. You can assume that the volume increase upon melting, 0.055 cm3mol−1, and the enthalpy of fusion, 13.8 kJmol-1, remain constant. Napthalene melts at 353 K at 1 atm with an enthalpy change of 19 kJmol-1. The volume increase on melting is 19 × 10-6 m3mol-1. What change in the melting temperature will be observed if the pressure is raised by 100 atm?

Summary Next Lecture You should now Be able to calculate the entropy change of an equilibrium system as a function of changes in U, V, and N. Be able to explain the concept of free energies, and in particular what the Gibbs free energy represents Be able to draw and read a P-T phase diagram, understand its meaning, and understand why the different phases occur under the conditions they do Be able to explain the meaning of coexistence, a triple point, and a critical point Understand and apply the Clayperon equation to coexistence lines Next Lecture Phase diagrams and solubilities