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Thermodynamic Stability: Free Energy and Chemical Equilibrium ©David Ronis Mcgill University

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Thermodynamic Stability: Free Energy and Chemical Equilibrium ©David Ronis Mcgill University Chemistry 223: Thermodynamic Stability: Free Energy and Chemical Equilibrium ©David Ronis McGill University 1. Spontaneity and Stability Under Various Conditions All the criteria for thermodynamic stability stem from the Clausius inequality,cf. Eq. (8.7.3). In particular,weshowed that for anypossible infinitesimal spontaneous change in nature, d− Q dS ≥ .(1) T Conversely,if d− Q dS < (2) T for every allowed change in state, then the system cannot spontaneously leave the current state NO MATTER WHAT;hence the system is in what is called stable equilibrium. The stability criterion becomes particularly simple if the system is adiabatically insulated from the surroundings. In this case, if all allowed variations lead to a decrease in entropy, then nothing will happen. The system will remain where it is. Said another way,the entropyofan adiabatically insulated stable equilibrium system is a maximum. Notice that the term allowed plays an important role. Forexample, if the system is in a constant volume container,changes in state or variations which lead to a change in the volume need not be considered eveniftheylead to an increase in the entropy. What if the system is not adiabatically insulated from the surroundings? Is there a more convenient test than Eq. (2)? The answer is yes. To see howitcomes about, note we can rewrite the criterion for stable equilibrium by using the first lawas − d Q = dE + Pop dV − µop dN > TdS,(3) which implies that dE + Pop dV − µop dN − TdS >0 (4) for all allowed variations if the system is in equilibrium. Equation (4) is the key stability result. As discussed above,ifE,V,and N are held fixed d− Q = 0and the stability condition becomes dS <0as before. What if S,V,N is held constant? From Eq. (4), the system will be stable if dE >0;i.e., the energy is a minimum. This has a nice mechanical analogy.Consider a ball rolling on a friction- less parabolic surface in a gravitational field. Clearly,ifweplace the ball at rest at the lowest point then it will stay there forever. This is the point which minimizes the energy. Of course, it is not always easy to see howtohold the entropyconstant in real experiments. (When is the entropyconstant?) A more common situation is when the temperature of the sys- tem is held fixed. What is the stability criterion? The problem and its solution are similar to those which led to the introduction of the enthalpy. If(N,T,V) are held fixed, Eq. (4) becomes − (dE)N,T,V T(dS)N,T,V >0,(5a) Fall Term, 2015 Chemistry 223 -2- Free Energy & Equilibrium or since T is constant, − d(E TS)N,T,V >0.(5b) Thus, we see that a newstate function, A ≡ E − TS,isaminimum for a stable equilibrium where (N,T,V) are not allowed to vary.This newstate function, is defined via a Legendre transforma- tion on the energy and is called the Helmholtz free energy. From the definition of A, for a general change in state (i.e., not necessarily with dT = 0, etc.) − − dA = dE − SdT − TdS = (d Q − TdS) + d W − SdT + µop dN.(6) The Clausius inequality implies that the quantity in the parenthesis is negative (or zero for a reversible process) for anyspontaneous change in the state of the system. Moreover, ifwecon- sider systems where T and N are held fixed dA ≤ d− W or − W ≤−∆A.(7) This means the −∆A is the maximum work you can get out of a process run under constant T and Nconditions (hence the name "free energy"). In addition, since A is a state function, you can get the bound without knowing anything about the path (or device)--just by knowing the initial and final states and howtocarry out a calculation similar to those we did in thermochemistry. Since A is a state function, we can always compute changes along reversible paths. In this case, dA =−SdT − PdV + µdN.(8) In addition, we pick up some newMaxwell relations, e.g., ∂S ∂P α = = ,(9) ∂ ∂ κ V T,N T V,N α = −1 ∂ ∂ κ =− −1 ∂ ∂ where V ( V/ T)P,N is the thermal expansion coefficient and V ( V/ P)T,N is the isothermal compressibility; the last equality follows by using the cyclic rule, Clearly,there are manydifferent choices of which state variables can be held constant. We will only consider twomore. First suppose (S,P,N) is held fixed. This is analogous to what we encountered with the enthalpy. Inthis case, Eq. (4) becomes + = d(E PV)S,P,N (dH)S,P,N >0 (10) for stable equilibrium; i.e., the enthalpyisaminimum. Finally,and perhaps most importantly,suppose T,P,and N are held fixed. This is the most commonly encountered case. NowEq. (4) becomes + − d(E PV TS)T,P,N >0.(11) Thus a newstate function, G ≡ E + PV − TS = H − TS,isaminimum for a stable equilibrium with fixed temperature, pressure, and mass. This state function is called the Gibbs free energy. As was the case with the Helmholtz free energy, ∆G has a direct physical interpretation. From its definition, for a constant (T,P,N) processes, dG = dE − TdS + PdV = (d− Q − TdS) + (d− W + PdV) ≤ (d− W + PdV), (12) Fall Term, 2015 Chemistry 223 -3- Free Energy & Equilibrium where the last inequality follows from the Clausius inequality.For finite changes in state, we thus find that −W − ∫ PdV ≤−∆G.(13) What does this mean? Up to now, wehav e mainly considered PV work. Of course, there are other kinds (magnetic, electrical but to name two). Hence, −∆G provides an upper bound to the non-PV work done by the system on the surroundings (i.e., −W − ∫ PdV)that can be obtained from a constant T,P,N process. If you are manufacturing electric batteries you probably don’t care about the amount of PV work which is wasted if the battery expands or contracts--all you want is the electrical work. As in the case of the Helmholtz free energy,wecan consider arbitrary changes in the Gibbs free energy along reversible paths. From its definition dG =−SdT + VdP + Σ µi dNi.(14) i As before, this givesadditional Maxwell relations, for example ∂S ∂V =− =− α ∂ ∂ V ,(15) P T,N T P,N which we obtained in a very complicated way in an earlier section. As an illustration of the usefulness of Maxwell relations, reconsider our discussion of the Joule-Thompson coefficient: 1 ∂H 1 ∂S V =− =− + =− − α µJT ∂ T ∂ V (1 T ), CP P T,N CP P T,N CP where the last equality follows by using Eq. (15) and the one before that by noting that dH = TdS + VdP + µdN.Note that the notation often givesaclue where to look for a Maxwell relation. In our example, the entropyderivative iswith respect to P,keeping T and N constant. The state function whose natural or canonical variables are T,P,and N is G, and this is where we got the Maxwell relation just used. One final point, note that the partial molar Gibbs free energy is ∂G G = = µ ,(16) i ∂N i i T,P,N j≠i where Eq. (14) was used. Hence, in a one component system, the Gibbs free energy per mole is just the chemical potential. More generally, Gi,the partial molar quantity for the i’th component is µi,and hence, from Euler’stheorem G = Σ Niµi.(17) i As we discussed earlier,inorder that Eqs. (17) and (14) be consistent, a Gibbs-Duhem relation must hold; i.e., 0 = SdT − VdP + Σ Ni dµi,(18) i Fall Term, 2015 Chemistry 223 -4- Free Energy & Equilibrium which shows that the changes in temperature, pressure and chemical potentials are not all inde- pendent. The various stability results are summarized in the following table. Criteria for Stable Equilibrium State Stable Equilibrium Simplest Held Fixed Definition Differential Function Criterion Physical Content Adiabatic* S dQrev dQrev maximum - (e.g., E,V,N) ∆S = dS = ∫ T T ∆ = + = − + ∆ = S,V,N E E Q WdE TdS PdV Σ µi dNi minimum EV,N QV,N i ≡ + = + + ∆ = S,P,N H H E PV dH TdS VdP Σ µi dNi minimum HP,N QP,N i ≡ − =− − + − ≤−∆ T,V,NAA E TS dA SdT PdV Σ µi dNi minimum WT,N AT,N i ≡ − =− + + − ≤−∆ T,P,NGG H TS dG SdT VdP Σ µi dNi minimum Wnon−PV GP,N i *E,N,V implies adiabatic in systems where only PV or chemical work is allowed, the reverse is not true if other kinds or work (e.g., electrical) are possible. 2. Examples of free energy calculations Free energy calculations are carried out in much the same as enthalpycalculations. There are tables of standard free energies of formation of compounds. Elements in their standard states areassigned zeroastheir Gibbs free energy of formation. Consider the following chemical reaction: + → H2(g) Cl2(g) 2HCl(g). Will the reaction proceed as written under constant T and P conditions? The free energy change is simply ∆ (0) −∆ (0) Σ G f (products) G f (reactants)(19) ∆ (0) which for the case at hand is just 2 G f (HCl, g)or-184.62 kJ/mol (from Barrow). Hence, a mixture of hydrogen and and chlorine can lower its free energy (by a substantial amount) by reacting to form HCl. This is an interesting example for another reason; if you mix stoichiometric amounts of H2 and Cl2,you will not see anyperceptible reaction--the rate of reaction (no matter what the ther- modynamics says) is in this case extremely slow.
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