3 44. Use collision theory to estimate the preexponential factor in the ...

Assign 3 13: 44, 46, 56, 67; 14: 3
44. Use collision theory to estimate the preexponential factor in the rate constant for the
elementary reaction
NO 2 + NO 2 N 2 O 4
at 500K. Take the average dia to be 2.6x10 -10 m and use the steric factor p from table
13.1.
the kinetic theory rate constant is given by eq 13.7,
! RT
E
k = 2d 2 a
N 0
M e" RT
! RT
The preexponential factor is just k = 2d 2 N 0
M
=2x(2.6x10 -10 ) 2 x6.02x10 23 xsqrt(πRT/M)=
2x(2.6x10 -10 ) 2 (m 2 /molec) x6.02x10 23 (molec/mol) xsqrt(πRT/M)(m/s)=4.33x10 7 m 3 /mol/s
if p=5x10 -2 , the prexeponential factor is A=2.17x10 6 m 3 /mol/s=2.17x10 9 L/mol/s
46. The conversion of dissolved CO 2 in blood to HCO - 3 and H 3 O + is catalyzed by the
enzyme carbonic anhydrase. The Michaelis-Menten constants for this enzyme and
substrate are K m =8x10 -5 mol/L and k 2 =6x10 5 s -1 .
a) What is the maximum rate of reaction of CO 2 if [E 0 ]=5x10 -6 M
b) At what CO 2 concentration will the rate of decomposition be 30% of a)
a)The mechanism of the reaction is
E+SES k 1
ES E + S k -1
ESE + P k 2
d[ES]
= 0 = k 1
[E][S] ! k !1
[ES] ! k 2
[ES]
dt
total amount of E = E 0
= [E] + [ES]
k
[ES] = 1
[E 0
][S]
k 1
[S] + k !1
+ k 2
d[P]
dt
( ) = [E 0
][S]
( )
= k 2
[ES] = k 2 [E 0 ][S]
[S] + K m
[S] + k !1 + k 2
k 1
= [E 0 ][S]
[S] + K m
maximum rate is for very large value of S. Then max rate =k 2 [E 0 ]=3 mol/L/s
b) a rate of 30% of above is obtained when
d[P]
dt
= k 2
[ES] = k 2 [E 0 ][S]
[S] + K m
= 0.3k 2
[E 0
]
[S]
[S] + K m
= 0.3
( )
[S] = 0.3 [S] + K m
0.7[S] = 0.3K m
[S] = 3.4x10 !5 M

Assign 3 13: 44, 46, 56, 67; 14: 3
56. For the decomposition of N 2 O 5 the rate constant is a function of pressure where k eff
as shown below for various total pressure where N 2 has been added:
P(atm) k eff (s -1 )
9.21 0.265
5.13 0.247
3.16 0.248
3.03 0.223
0.625 0.116
0.579 0.108
0.526 0.104
0.439 0.092
0.395 0.086
Use this data to estimate the value of k eff at very high total pressure and the value of k 1 .
According to the steady state approximation
Rate = k 1 k 2 [M ][N 2 O 5 ]
k 2
+ k !1
[M ]
k eff
= k 1k 2
[M ]
k 2
+ k !1
[M ]
= k eff
[N 2
O 5
]
1
= k + k [M ]
2 !1
1
=
k eff
k 1
k 2
[M ] k 1
[M ] + k !1
k 1
k 2
So a plot of 1/k eff vs 1/[M] should give a st line with slope of 1/k 1 as shown below
1/k
12
Data 1
y = 3.3161 + 3.3207x R= 0.99889
10
8
1/k
6
4
2
0 0.5 1 1.5 2 2.5 3
1/P
The intercept is 1/k eff in the limit of high P ≈ 1/3.32 so k eff = 0.30 -1 s -1 and the slope is
3.32=1/k 1 so k 1 = 0.3 L/mol/s
67. The gas phase reaction between H 2 and I 2
H 2 + I 2 = 2HI

Assign 3 13: 44, 46, 56, 67; 14: 3
Has kf=240 L/mol/sec and an activation E = 165kJ/mol at 1000K. Using this and data
from appendix D, calculate Ea and kb for the reverse reaction at 1000K, assuming ∆H
and ∆S are temperature independent.
The activation energies for the forward and reverse reactions are related to the overall
∆E for the reaction as shown in fig 13.11:
This picture is for a different reaction, but it shows that E a,b =E a,f -∆E. From Appendix D
we have ∆H=+26.48*2-62.44=-9.5 kJ/mol. Since ∆n=0, ∆E=∆H. Thus E a,b =E a,f -
∆E=165+9.5=174.5.
To calculate k b , we need the equilibrium constant K=[HI] 2 /[H 2 ][I 2 ], where ∆G 0 =-RTlnK.
∆G 0 =∆H 0 -T∆S 0 ; ∆H 0 =-9.5 kJ. ∆S 0 =2(206.48)-260.58-130.57 =21.81J/deg
∆G 0 =-9500-1000*21.81=-31.3kJ
K=43=kf/kb so k b =k f /43 = 5.5 L/mol/s. (HI is not very stable stuff!)
14.3 Calculate the total binding energy in kJ/mole and MeV/atom and the binding energy
per nucleon of the following nuclides, using data from Table 14.1
a) 40 20
Ca b) 87 37
Rb c) 238 92
U
The binding energy of Ca is the energy released in the reaction
20 1 1
H + 20 1 0
n = 40 20
Ca
39.9625912 ! 20 *1.0086649158 ! 20 *1.007825032 = !.367u
! m = !931*.367 = !342MeV = !8.55keV / nucleon
The binding energy of Rb is the energy released in the reaction
37 1 1
H + 50 1 0
n = 87 37
Rb
86.909183 ! 50 *1.0086649158 ! 37 *1.007825032 = !.813u
! m = !931*.813 = !758MeV = !8.71keV / nucleon
The binding energy of U is the energy released in the reaction
92 1 1
H + 146 1 0
n = 238 92
U
238.050783 ! 146 *1.0086649158 ! 92 *1.007825032 = !1.934u
! m = !931*1.934 = !1802MeV = !7.57MeV / nucleon