Unformatted text preview: 5.5. CHEMICAL KINETICS OF A SINGLE ISOTHERMAL REACTION 187 Now use Eq. (5.350) to eliminate the density derivative in Eq. (5.338) to get
dρi
ρ ρRT r
= νi r − i
dt
ρ
P N
k =1 νk ρ RT
νi
= r
−i P
reaction eﬀect , (5.352) νk , k =1
N (5.353) expansion eﬀect or dρi
= r
dt νi reaction eﬀect − yi ∆n
expansion eﬀect . (5.354) There are two terms dictating the rate change of species molar concentration. The ﬁrst, a
reaction eﬀect, is precisely the same term that drove the isochoric reaction. The second is
due to the fact that the volume can change if the number of moles change, and this induces
an intrinsic change in concentration. Note that the term ρi RT /P = yi, the mole fraction.
Example 5.14
Study a variant of the nitrogen dissociation problem considered in an earlier example, see p. 147,
in which at t = 0 s, 1 kmole of N2 exists at P = 100 kP a and T = 6000 K . In this case, take the
reaction to be isothermal and isobaric. Consider again the elementary nitrogen dissociation reaction
N2 + N2 ⇌ 2 N + N2 , (5.355) which has kinetic rate parameters of
a = 7.0 × 1021 β = cm3 K 1.6
,
mole s −1.6, E = (5.356)
(5.357) cal
224928.4
.
mole (5.358) In SI units, this becomes
3 a = E = m3 K 1.6
1m
1000 mole
cm3 K 1.6
= 7.0 × 1018
,
mole s
100 cm
kmole
kmole s
cal
J
kJ
kJ
1000 mole
4.186
= 941550
224928.4
.
mole
cal
1000 J
kmole
kmole
7.0 × 1021 (5.359)
(5.360) At the initial state, the material is all N2 , so PN2 = P = 100 kP a. The ideal gas law then gives at
CC BYNCND. 18 November 2011, J. M. Powers. 188 CHAPTER 5. THERMOCHEMISTRY OF A SINGLE REACTION
t=0
P
ρN 2 t=0 = PN2 = ρN2 RT,
P
,
=
RT
100 kP a
,
=
kJ
8.314 kmole K (6000 K )
kmole
= 2.00465 × 10−3
.
m3 (5.361)
(5.362)
(5.363)
(5.364) Thus, the initial volume is
V t=0 = nN2 t=0
1 kmole
=
ρN2 t=0
2.00465 × 10−3 kmole
m3 = 4.9884 × 102 m3 . (5.365) In this isobaric process, one always has P = 100 kP a. Now, in general
P = RT (ρN2 + ρN ); (5.366) therefore, one can write ρN in terms of ρN2 :
ρN =
=
= P
− ρN 2 ,
RT
100 kP a
− ρN2 ,
kJ
8.314 kmole K (6000 K )
2.00465 × 10−3 kmole
m3 (5.367)
(5.368) − ρN 2 . (5.369) Then the equations for kinetics of a single isobaric isothermal reaction, Eq. (5.353) in conjunction
with Eq. (5.280), reduce for N2 molar concentration to
dρN2
dt = aT β exp −E
RT ′ ′ (ρN2 )νN2 (ρN )νN 1− 1
(ρ )νN2 (ρN )νN
Kc N2 νN2 − ρN2 RT
(νN2 + νN ) .
P =r (5.370)
′
′
Realizing that νN2 = 2, νN = 0, νN2 = −1, and νN = 2, one gets dρN2
dt = aT β exp −E
RT ρ2 2 1 −
N 1 ρ2
N
K c ρN 2 −1 − ρN2 RT
P . (5.371) =k ( T ) The temperature dependency of the reaction is unchanged from the previous reaction:
k (T ) =
=
= aT β exp −E
RT 7.0 × 1018 , m3 K 1.6
kmole s 7.0 × 1018
exp
T 1.6 (5.372) −1.1325 × 105
T CC BYNCND. 18 November 2011, J. M. Powers. kJ
kmole
kJ
kmole K T −941550 T −1.6 exp 8.314
. , (5.373)
(5.374) ...
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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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