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Thermodynamics filled in class notes_Part_98

Thermodynamics filled in class notes_Part_98 - 203 5.7...

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Unformatted text preview: 203 5.7. SIMPLE ONE-STEP KINETICS In terms of the mole fractions yi , one then has y A + y B = 1. (5.504) The reaction rate r is then 1 ρB , Kc ρA 1 ρB /ρo 1− Kc ρA /ρo 1 yB , 1− Kc y A r = k ρA 1 − = k ρo ρA ρo = k ρ o yA ˆ = k ρo (1 − ζ ) 1 − (5.505) , (5.506) (5.507) ˆ 1ζ . ˆ Kc 1 − ζ (5.508) ˆ ˆ ˆ Now r = (1/V )dζ/dt = (1/V )d(no ζ )/dt = (no /V )d(ζ )/dt = ρo dζ/dt. So the reaction dynamics can be described by a single ordinary diﬀerential equation in a single unknown: ρo ˆ ˆ 1ζ dζ ˆ = k ρo (1 − ζ ) 1 − , ˆ dt Kc 1 − ζ ˆ ˆ dζ 1ζ ˆ = k (1 − ζ ) 1 − . ˆ dt Kc 1 − ζ (5.509) (5.510) Equation (5.510) is in equilibrium when ˆ ζ= 1 1 + ... 1 ∼1− Kc 1 + Kc (5.511) ˆ As Kc → ∞, the equilibrium value of ζ → 1. In this limit, the reaction is irreversible. That is, the species B is preferred over A. Equation (5.510) has exact solution ˆ ζ= 1 − exp −k 1 + 1+ 1 Kc 1 Kc t . (5.512) For k > 0, Kc > 0, the equilibrium is stable. The time constant of relaxation τ is τ= 1 k 1+ 1 Kc . (5.513) For the isothermal, isochoric system, one should consider the second law in terms of the Helmholtz free energy. Combine then Eq. (4.394), dA|T,V ≤ 0, with Eq. (4.311), dA = CC BY-NC-ND. 18 November 2011, J. M. Powers. 204 CHAPTER 5. THERMOCHEMISTRY OF A SINGLE REACTION N i=1 −SdT − P dV + µi dni and taking time derivatives, one ﬁnds N dA|T,V = −SdT − P dV + dA dt − ≤ 0, (5.514) µi dni ≤ 0, dt (5.515) µi dρi ≥ 0. dt (5.516) µi dni i=1 T ,V N = T ,V 1 dA V =− T dt T i=1 N i=1 This is exactly the same form as Eq. (5.482), which can be directly substituted into Eq. (5.516) to give 1 dA − T dt N T ,V = −RV k (T ) ρi ′ νi i=1 1 1− Kc N ρi νi ln i=1 1 Kc N ρi νi i=1 ≥ 0, (5.517) dA dt N = RV T k (T ) T ,V ρi ′ νi i=1 1− 1 Kc N ρi νi ln i=1 1 Kc N ρ i νi i=1 ≤ 0. (5.518) For the assumptions of this section, Eq. (5.518) reduces to dA dt T ,V ˆ 1ζ ˆ = RT k ρo V (1 − ζ ) 1 − ln ˆ Kc 1 − ζ ˆ 1ζ ˆ = kno RT (1 − ζ ) 1 − ln ˆ Kc 1 − ζ ˆ 1ζ ˆ Kc 1 − ζ ˆ 1ζ ˆ Kc 1 − ζ ≤ 0, ≤ 0. (5.519) (5.520) Since the present analysis is nothing more than a special case of the previous section, Eq. (5.520) certainly holds. One questions however the behavior in the irreversible limit, 1/Kc → 0. Evaluating this limit, one ﬁnds lim 1/Kc →0 dA dt T ,V ˆ = kno RT (1 − ζ ) ln >0 1 Kc →−∞ ˆ ˆ ˆ ˆ +(1 − ζ ) ln ζ − (1 − ζ ) ln(1 − ζ ) + . . . ≤ 0. (5.521) ˆ → 1; that is the reaction goes to completion, Now, performing the distinguished limit as ζ ˆ one notes that all terms are driven to zero for small 1/Kc . Recall that 1 − ζ goes to zero ˆ faster than ln(1 − ζ ) goes to −∞. Note that the entropy inequality is ill-deﬁned for a formally irreversible reaction with 1/Kc = 0. CC BY-NC-ND. 18 November 2011, J. M. Powers. ...
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