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.

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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.

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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

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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,

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(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. The chemical equation imposes a strong constraint on the changes in the numbers of moles of each component; for the forward reaction, each time a mole of B reacts, 2 of A are used up and 3 of C and one of D are produced. Mathematically, dN dN dN dN A = B = C = A ≡ dξ , −2 −1 3 1 where the extent of the reaction is characterized by the quantity ξ (the Greek letter,pronounced kse)called the progress variable. For an arbitrary chemical reaction involving r chemical components, the last expression generalizes to dN dN 1 = ... = r ≡ dξ ,(22) ν ν 1 r ν where i is the stoichiometric coefﬁcient for the i’th component in the reaction (by convention, it is negative for reactants). Forconstant temperature and pressure and total mass (for each element) conditions, the reaction can proceed until the Gibbs free energy is a minimum with respect to all allowed variations in the state of the system. Forﬁxedtotal mass, temperature, and pressure, the only variations which can be considered are those which change ξ .That is the reaction moveseither to the right or left until G(ξ )isaminimum. Gcan be a minimum with respect to changes in the progress variable only if  ∂G  = 0(23)  ∂ξ  T,P,Ntotal and

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 ∂2G  >0.(24)  ∂ξ 2  T,P,Ntotal By using the differential form for the change in the free energy together with Eq. (22) we ﬁnd that r =− + + ν µ ξ dG SdT VdP Σ i i d ,(25) i=1 which when used in Eq. (23) gives r ν µ = Σ i i 0(26) i=1 µ at equilibrium. Since the i are the partial molar Gibbs free energies, Eq. (26) is equivalent to ∆G = 0. At equilibrium the free energy change in the reaction per mole vanishes. (Indeed, this is the principle we applied in the "reaction" between graphite and diamond). From the deﬁ- nition of the Gibbs free energy (G = H -TS), it follows that ∆H ∆S = T at equilibrium. What happens if we change temperature or pressure by a small amount? Which way will the equilibrium shift? To answer this note the following Maxwell relations:  ∂µ   ∂S  i =− =−S (27a)  ∂T   ∂N  i P,N j i P,T,N j≠i and  ∂µ   ∂V  i = = V (27b)  ∂P   ∂N  i T,N j i P,T,N j≠i which followfrom the Gibbs free energy.Thus the changes in the chemical potential associated with temperature or pressure are related to the partial molar entropies or volumes, respectively. Next consider  ∂µ   ∂µ   ∂(Σν µ ) d(Σν µ ) = Σν i dT + Σν i dP + i i dξ i i i ∂T  i ∂P   ∂ξ  i i P,N j i P,N j T,P,Ntotal

∂∆G =−∆SdT +∆VdP + dξ ,(28) ∂ξ ∆ ≡ ν ∆ ≡ ν where, cf. Eqs. (27), S Σ i Si and V Σ iVi are the entropyand volume changes per mole or reaction. Equation (28) shows howthe free energy change per mole of reaction changes when we change T,P,or ξ . What happens if we change, T or P in a system where chemical reaction is possible? The progress variable will change until Eq. (26) is again valid. Since both the initial and ﬁnal states satisfy Eq. (26), the change in ∆G must vanish. From Eq. (28) this implies that

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∆SdT −∆VdP dξ = .(29)  ∂∆G   ∂ξ  T,P,Ntotal Moreover, the denominator of the right hand side of the equation is positive,cf. Eq. (24). Equa- tion (29) can be rewritten by noting [cf. Eq. (26)] that ∆S =∆H/T;i.e., ∆H dT −∆VdP dξ = T .(30)  ∂∆G   ∂ξ  T,P,Ntotal Equation (30) is a mathematical statement of LeChatellier'sprinciple.For reactions which lead to an increase in the volume (∆V >0,increasing (decreasing) the pressure will shift the equilibrium to the reactant (dξ <0)[product (dξ >0)] side of the equation. The reverse is true if the volume change is negative.Similarly,increasing the temperature shifts the equilibrium to the reactant side for reactions which are exothermic (∆H <0)and to the product side for reactions which are endothermic.

3. Chemical equilibria in dilute gases Foraone-component gas, the pressure dependence of the chemical potential (free energy per mole) is easily obtained from Eq. (20); i.e., P µ = µ + ′ ′ 0(T) ∫ V(T, P )dP . 1 µ Where 0(T)isthe standard Gibbs free energy of formation at one atm and temperature T.Ifthe gasisideal, then V=RT/P and we ﬁnd that µ(T, P) = µ0(T) + RT ln(P), (31) where P is the pressure in atmospheres. This result can be generalized to gas mixtures if we recall our discussion of Dalton’slaw of partial pressures. There we considered a gas mixture where one of the components could dif- fuse in and out of the system through a selective,porous ﬁlm into a container containing a pure sample of that component. At equilibrium, the pressure in the pure sample was Pi,the partial pressure of the i’th component in the mixture. If we viewthe process as the following "chemical reaction" Component i in mixture ←→ Component i in puresample. The equilibrium condition becomes: µ ... = µ = µ0 + i,mixture(T, P, x1, , xr−1) i,pure(T, Pi) i (T) RT ln(Pi). (32) Hence, the form of the chemical potential in a gas mixture is very similar to that in a pure sample, with the exception that the pressure is not the total pressure of the gas, but is the partial pressure of the component in question.

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Knowing this, we are ready to discuss chemical equilibria in gases. From Eq. (26), the equilibrium condition becomes: ν ν ν = ν µ0 + 1 2 ... r 0 Σ i i (T) RT ln(P1 P2 Pr )(33a) i or

ν ν ν   −∆ 1 2 ... r = ≡ − ν µ0 = G/RT P1 P2 Pr KP(T) exp  Σ i i (T)/RT e ,(33b) where KP(T)iscalled the pressure equilibrium constant. Notice that it is only a function of temperature, the stoichiometric coefﬁcients, and properties of the pure (i.e., unmixed) gases.

4. Examples of Chemical Equilibrium Calculations

4.1. Free Energy and Entropy of Mixing Perhaps the simplest process were twosamples of different pure gases are isothermally mixed as depicted below

A(P,T,V) + B(P,T,V) A+B (P,T,2V) As you might expect, this process always occurs spontaneously.The total pressure and temperature remain constant during the process. What is the Gibbs free energy change? From Eq. (32), it follows that the free energy of the ﬁnal state is

= µ0 +  + µ0 +  G final N1 1(T) RT ln(P1) N2 2(T) RT ln(P2).(34a) Similarly,the Gibbs free energy of either of the pure samples is

= µ0 +  Gi Ni i (T) RT ln(P) (34b) ∆ and hence, the free energy of mixing per mole of mixture, Gmix,is ∆ = + Gmix RTx1 ln(x1) RTx2 ln(x2), (35) = where xi is the mole fraction of i and where Dalton’slaw ofpartial pressures, Pi Pxi,was used. It is easy to generalize this result to arbitrary mixtures of ideal gases ∆ = Gmix RT Σ xi ln(xi). (36) i

Since 0 < xi <1,the free energy change is negative and the mixing occurs spontaneously. Equation (36) can be used to calculate the entropyand enthalpyofmixing. By using Eq. (14) it follows that  ∂∆G  ∆S =− mixing ,(37) mix  ∂T  P,xi

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which when used in Eq. (36) gives ∆ =− Smix R Σ xi ln(xi)>0.(38) i Moreover, since ∆ =∆ + ∆ Hmix Gmix T Smix, it follows that the heat of mixing associated with the mixing of ideal gases is zero. No heat is absorbed or released for the mixing of ideal gases. The process is drivenentirely by entropy. As we shall see later,asimilar result holds for the mixing of dilute solutions. Finally,note that there is no volume change for mixing ideal gases (see Eq. (14) and takeapressure derivative). In abinary mixture, what composition has the most negative free energy of mixing?

4.2. Determination of Free Energies of Formation There are a number of ways in which to measure the standard free energies of formation of acompound. Consider the formation of ammonia, 1 3 N (g) + H (g) ←→ NH (g), 2 2 2 2 3 ∆ 0 at STP.The free energy change for the reaction is G f (NH3, g). If we measure the equilibrium constant,

P 0 NH3 −∆G (NH ,g)/RT = e f 3 , P1/2 P3/2 N2 H2 then we can easily compute the free energy of formation of ammonia.

4.3. Determination of the Extent of a Reaction Reconsider the reaction ←→ N2O4(g) 2NO2(g). The extent of the reaction is easily measured by measuring the apparent deviation from the ideal α gaslaw.Asbefore, let be the fraction of N2O4 dissociated. If there were N0 moles of N2O4 − α α initially,then there are (1 )N0 and 2 N0 moles of N2O4 and NO2 at equilibrium, respectively.The corresponding partial pressures can be computed from Dalton’slaw. When the result is used in the equilibrium constant condition we ﬁnd that P2 4α 2 N RT 4α 2 P K = NO2 = 0 = , P − α − α 2 P N2O4 1 V 1 where P is the total pressure on the system. This can be solved for the fraction dissociated, with the result that  K 1/2 α = P  +  KP 4P Thus, if we calculate the equilibrium constant from a table of free energies, the degree of dissoci- ation is easily found. Note that the result depends on both T and P.

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5. Coupled Reactions In some cases, the direct formation of a certain compound by direct reaction is thermody- namically forbidden. An example is the formation of titanium tetrachloride from common TiO2 ore; i.e., + → + TiO2(s) 2Cl2 TiCl4(l) O2(g). It turns out that ∆G =+152. 3kJ/mole.Nonetheless, we can makethe reaction go by coupling it to one which pulls it along. Forexample, suppose we use the produced oxygen to burn carbon; i.e., + → C(s, graphite) O2(g) CO2(g), where here ∆G =−394. 36kJ/mole.The free energy change for the coupled processes is -394.36 +152.3 = -242.1 kJ/mole, and thus the coupled reaction can proceed. The burning carbon sup- plies the needed free energy to makethe desired reaction work.

6. TemperatureDependence of K p

The equilibrium constant, K p,isonly a function of the temperature. From its deﬁnition, cf. Eq. (33b), ∆ d ln(KP) =−d( G/RT) dT dT 1  d(∆G)  =− T −∆G RT 2  dT  ∆ +∆ = T S G RT 2 ∆H = ,(39) RT 2 where the second to last equality follows when Eqs. (27a) and (33b) are used. Thus, T  K (T ) 2 ∆H ln  p 2  = dT (40) ∫ 2  K p(T1) RT T1

∆   ≈ H 1 − 1  .(41) R T1 T2 Equation (41) follows if we assume that ∆H is approximately constant. Note that as in our discussion of LeChatellier’sprinciple, the equilibrium will shift to the product side when the temperature is raised if ∆H >0

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