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Thermodynamic Stability: Free Energy and Chemical Equilib- Rium Thermodynamic Stability: Free Energy and Chemical Equilib- rium Chemistry CHEM 213W David Ronis McGill University All the criteria for thermodynamic stability stem from the Clausius inequality; i.e., for any possible 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 that using the first lawas − = + − µ d Q dE Pop dV dN > TdS,(3) which implies that + − µ − dE Pop dV 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. Winter Term 2001-2002 Free Energy -2- Chemistry CHEM 213W Of course, it is not always easy to see howtohold the entropyconstant in real experi- ments. (When is the entropyconstant?) A more common situation is when the temperature of the system 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) 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 + µdN.(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 ∂ ∂ .(9) V T,N T V,N 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. Winter Term 2001-2002 Free Energy -3- Chemistry CHEM 213W 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 (12a) = (d− Q − TdS) − (d− W − PdV) ≤−(d− W − PdV), (12b) where the last inequality follows from the Clausius inequality.For finite changes in state, we find 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 that can be obtained from a constant T,P,N process. If you are manufacturing elec- tric batteries you probably don’tcare 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 + µdN.(14) 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 the last handout. One final point, note that the partial molar Gibbs free energy is ∂G = = µ G ∂ ,(16) N T,P 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 µ is i,and hence, from Euler’stheorem = µ G Σ Ni i.(17) i As we discussed earlier,inorder that Eqs. (17) and (14) be consistent, a Gibbs-Duhem relation must hold; i.e., = − + µ 0 SdT VdP Σ Ni d i,(18) i which shows that the changes in temperature, pressure and chemical potentials are not all Winter Term 2001-2002 Free Energy -4- Chemistry CHEM 213W independent. The various stability results are summarized in the following table. Criteria for Stable Equilibrium State Stable Equilibrium Held Fixed Definition Differential Function Criterion dQ E,V,N S - dS = rev maximum T S,V,N E - dE = TdS − PdV + µdN minimum S,P,N H H ≡ E + PV dH = TdS + VdP + µdN minimum T,V,NAA ≡ E − TS dA =−SdT − PdV + µdN minimum T,P,NGG ≡ H − TS dG =−SdT + VdP + µdN minimum 1. 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/mole (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. Onthe other hand, a small amount of light at the right frequencywill catalyze the reaction which then proceeds explosively! Next consider the reaction between graphite and diamond, C(graphite, s) → C(diamond, s). Now ∆G =2.90 kJ/mole. The reaction does not proceed as written (too bad). What is perhaps more troubling is that the reverse reaction should proceed spontaneously at STP.(So whyinv est in diamonds?) What happens at other temperatures or pressures. To answer this note that from Eq. (14), for anycompound, Winter Term 2001-2002 Free Energy -5- Chemistry CHEM 213W (T,P) ∆ =∆ 0 + − + G f (T, P) G f ∫ S(T, P)dT V(T, P)dP, (298K, 1 atm) where anyconvenient path can be chosen. Thus if we raise the pressure, (298K,P) ∆ =∆ 0 + ∆ Grxn(T, P) Grxn ∫ Vrxn dP.(20) (298K, 1 atm) ∆ 3 At STP, Vrxn =-1.9 cm /mole.Hence, increasing the pressure decreases the Gibbs free energy change. If we assume that the molar densities of carbon are roughly independent of pressure, we can calculate the pressure at which the reaction will proceed as written. In this approximation, ∆ ≈ − × −9∆ Grxn(T, P) 2. 90 1. 9 10 P (kJ/mole) Hence, the reaction begins to be possible when P ≈ 1. 530 × 109 Pa or about 15,000 atm. 2. Chemical Equilibrium Avery important example of thermodynamic equilibrium is that of chemical equilibrium at constant pressure and temperature. Consider the following general chemical reaction: 2A + B ←→ 3C + D.
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