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Unformatted text preview: 245 7.1. ISOTHERMAL, ISOCHORIC KINETICS
Thus, we once again ﬁnd
ρO + 2ρO2 = ρO + 2ρO2 = constant. (7.117) As before, we can say
ρO
ρO2 = =ρ ρO
1
+
−1
ρO2
2
=D b
=ρ ξO . (7.118) =ξ This gives the dependent variables in terms of a smaller number of transformed dependent
variables in a way which satisﬁes the linear constraints. In vector form, the equation becomes
ρ = ρ + D · ξ. (7.119) Once again φ · D = 0.
7.1.1.2.1.2 Thermodynamics Equations (7.1037.104) are supplemented by an expression for the thermodynamicsbased equilibrium constant Kc,13 which is:
Kc,13 = Po
exp
RT −∆Go
13
RT . (7.120) Here Po = 1.01326 × 106 dyne/cm2 = 1 atm is the reference pressure. The net change of
Gibbs free energy at the reference pressure for reaction 13, ∆Go is deﬁned as
13
∆Go = 2g o − g o 2 .
O
O
13 (7.121) We further recall that the Gibbs free energy for species i at the reference pressure is deﬁned
in terms of the enthalpy and entropy as
o g o = hi − T so .
i
i (7.122) o It is common to ﬁnd hi and so in thermodynamic tables tabulated as functions of T .
i
We further note that both Eqs. (7.103) and (7.104) are in equilibrium when
ρe 2 ρe =
OM 1 eee
ρρρ .
Kc,13 O O M (7.123) We rearrange Eq. (7.123) to ﬁnd the familiar
Kc,13 = ρe ρe
OO
=
ρe 2
O [products]
.
[reactants] (7.124) If Kc,13 > 1, the products are preferred. If Kc,13 < 1, the reactants are preferred.
CC BYNCND. 18 November 2011, J. M. Powers. 246 CHAPTER 7. KINETICS IN SOME MORE DETAIL Now, Kc,13 is a function of T only, so it is known. But Eq. (7.124) once again is one
equation in two unknowns. We can use the element conservation constraint, Eq. (7.117) to
reduce to one equation and one unknown, valid at equilibrium:
Kc,13 = ρe ρe
OO
1
ρO2 + 2 (ρO − ρe )
O (7.125) Using the element constraint, Eq. (7.117), we can recast the dynamics of our system by
modifying Eq. (7.103) into one equation in one unknown:
dρO
= 2a13 T β13 exp
dt −E 13
RT 1
1
1
1
× (ρO2 + (ρO − ρO )) (ρO2 + (ρO − ρO ) + ρO ) −
ρO ρO (ρO2 + (ρO − ρO ) + ρO ) , 2
2
Kc,13
2
=ρO2 =ρM =ρM (7.126) 7.1.1.2.2 Example calculation Let us consider the same example as the previous section with T = 5000 K . We need numbers for all of the parameters of Eq. (7.126). For O ,
we ﬁnd at T = 5000 K that
o erg
,
mole
erg
= 2.20458 × 109
.
mole K hO = 3.48382 × 1012 (7.127) so
O (7.128) So
erg
erg
− (5000 K ) 2.20458 × 109
,
mole
mole K
erg
= −7.53908 × 1012
.
mole go =
O 3.48382 × 1012 (7.129) For O2 , we ﬁnd at T = 5000 K that
o erg
,
mole
erg
= 3.05406 × 109
.
mole K hO2 = 1.80749 × 1012 (7.130) so 2
O (7.131) So
erg
erg
− (5000 K ) 3.05406 × 109
mole
mole K
13 erg
= −1.34628 × 10
.
mole go 2 =
O 1.80749 × 1012 CC BYNCND. 18 November 2011, J. M. Powers. (7.132) ...
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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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