Thermodynamic Stability: Free Energy and ... - McGill University

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Free Energy -12- Chemistry 223principle we expect that the reaction should shift to the left (i.e., less dissociation) as pressure isincreased. Our final equation for α shows this. For this reaction, as we shall see in the next section,K P (T )decreases as temperature increases, which when used with our expression for α isagain consistent with LeChatellier’s principle.5.3. Temperature Dependence of K PThe equilibrium constant, K P ,isonly a function of the temperature. From its definition,cf. Eq. (34),d ln(K P )dT=− d(∆G(0) /RT )dT=− 1 ⎛RT 2 ⎜T d(∆G(0) )−∆G (0)⎞ ⎟ = T ∆S(0) +∆G (0)=⎝dT ⎠RT 2∆H(0)RT 2 ,(39)where the second to last equality follows when Eqs. (27a) and (34) are used. This is known asthe Gibbs-Helmholtz equation. Thus, by integrating we find thatln ⎡ K P (T 2 ) ⎤⎢ ⎥ =⎣K P (T 1 ) ∫ T 2⎦(0)∆H ⎛ 1dT ≈ − 1 ⎞RT2R ⎝ T 1 T 2 ⎠ . (40)T 1∆H (0)Equation (40) follows if we assume that ∆H (0) is approximately constant with respect to temperature,or equivalently, that ∆C P ≈ 0(which also implies that ∆S (0) is constant. Why?). Indeed,with this approximation, Eq. (40) simply states thatK P (T 2 ) ⎛K P (T 1 ) = exp ⎜− ∆G(0) (T 2 )+ ∆G(0) (T 1 ) ⎞⎟, (41)⎝RT 2 RT 1 ⎠where ∆G (0) (T ) =∆H (0) − T ∆S o ,asusual. Finally, note, that as in our discussion of LeChatellier’sprinciple, the equilibrium will shift to the product side, i.e., K P increases, when the temperatureis raised if ∆H (0) >0.There is a simple graphical way in which to apply the Gibbs-Helmholtz equation. Byexpressing Eq. (39) as a differential, it follows thatd ln(K P ) =∆H(0)(0)dT =−∆HRT2Rd⎛ 1 ⎞⎝ T ⎠ ;hence, plotting ln(K P )versus 1/T will give a curve whose slope at any point is −∆H (0) /R, and tothe extent that ∆H (0) is independent of temperature, will give a straight line. This is a powerfulway todetermine enthalpy (and entropy) changes without having to do calorimetry.5.4. Free Energy and Entropy of MixingPerhaps the simplest process is one where two samples of different pure gases are isothermallymixed as depicted in the figure belowFall Term, 2014
Free Energy -13- Chemistry 223A(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 temperatureremain constant during the process (at least for an ideal gas). What is the Gibbs free energychange?From Eq. (32), it follows that the free energy of the final state isG final = N 1⎡⎣µ (0)1 (T ) + RT ln(P 1) ⎤ ⎦ + N 2 ⎡ ⎣ µ(0) 2 (T ) + RT ln(P 2) ⎤ ⎦ . (42)Similarly, the Gibbs free energy of either of the pure samples isG i = N ⎡ i µ (0)⎣i (T ) + RT ln(P) ⎤ ⎦(43)and hence, the free energy of mixing per mole of mixture, ∆G mix ,is∆G mix = RTx 1 ln(x 1 ) + RTx 2 ln(x 2 ), (44)where x i is the mole fraction of i and where Dalton’s law of partial pressures, P i = Px i ,wasused. It is easy to generalize this result to arbitrary mixtures of ideal gases∆G mix = RTΣ x i ln(x i ). (45)iSince 0 < x i 0. (47)i∆H mix =∆G mix + T ∆S mix, (48)it follows that the heat of mixing associated with the mixing of ideal gases is zero. No heat isabsorbed or released for the mixing of ideal gases. The process is driven entirely by entropy.As we shall see next term, a similar result holds for the mixing of dilute solutions. Similarly,‡ For this to happen, it is necessary that the pressures in the unmixed state be identical; hence,PRT ≡ N A≡ N B= N A + N BV A + V B,V A V Bwhere the first two equalities are conditions on the initial state, while the last one follows from the precedingtwo.Fall Term, 2014
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