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Unformatted text preview: 203 5.7. SIMPLE ONESTEP 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), dAT,V ≤ 0, with Eq. (4.311), dA =
CC BYNCND. 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 dAT,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 illdeﬁned for a formally
irreversible reaction with 1/Kc = 0.
CC BYNCND. 18 November 2011, J. M. Powers. ...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Parksou during the Fall '11 term at FSU.
 Fall '11
 ParkSou
 Dynamics

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