Bài tập Home work II

Student’s name: DO THI KIM HUE
Class: K52 Advanced Chemistry
**********
Home work II

Problem 7.1
a.
SSA states that d[O]/dt = 0 => r
1
= r
2


r =
3
[]dO
dt


3
[]dO
dt

= r = 2r
1
=2k
1
[O
3
][M] or
2

[]dO
dt

= 3k
1
[O
3
][M]
b.
Cannot assume Quasi-Equilibrated Adsorption (No Langmuir isotherm)
r =
 
2
d CO
dt
; at steady state, so r
1
= r
2
, Site Balance: L =
 
SO
+ [S]

         
1
1 2 2 2
2
[]
k N O S k CO S O S O N O

[]
k
S
k CO
    


  
  
   
 
1 2 2
2
1 2 2
Lk k CO N O
r k CO S O
k N O k CO
  


c.

Step 1 is a quasi equilibrated, but step 2 and 3 isn’t, so no Langmuir isotherm, then
we use SSA to determine surface concentration. We also assume single-site adsorption
and C
2
H
2(ad)
is MARI.
We assume: d[CH

2(ad)
]/dt = 0 → d[CH
2
]/dt = 2k
1
[C
2
H
2
][H
2
] - k
2
[CH
2(ad)
][H
2
] = 0
→ [CH
2(ad)
] = 2k
1
[C
2
H
2
] /k
2
→ 2k
1

= k
2
(1)
r = d[CH
4
]/dt = k
2
[CH
2(ad)
][H
2
] (2)
Site balance for C
2
H
2(ad)
is MARI: L = [C
2
H
2(ad)
] + [S] or 1 = θ
C2H2
+ θ
v

→ [S] = L - [C
2
H
2(ad)
] (3)

From eq. QE 1: K =
[CH 2(ad )][H2]2
[C2H6][S]
→ [C
2
H
2(ad)
] =
 C2H6][S]
[H2]2

(4)

Replace (3) into (4), we obtain: [C
2
H
2(ad)
] =
 [C2H6]
[H2]2
( L - [C
2
H
2(ad)
])
→ [C
2
H
2(ad)
] =

 [C2H6]
(1+[26/[2]2])

H2

2
(5)
Replace (5) into (2), we have the rate:
r = d[CH
4
]/dt = k
2
[CH
2(ad)
][H
2
] =
k2 [C2H6]


26

2


+

H2


Problem 7.2

 
3
[ ] [ ]
r k B S
d A d B
dt dt
    

12
[ ] [ ]
K , K
[ ][ ] [ ]
A S B S
A S A S




Site balance:
L=[S] +[A-S] + [B-S]
a) If [B-S] is the MARI, then L= [s] + [B-S] and

          
 
 
2 1 2
12

L S K A S S K K A S S
1 K K A
L
      


r= k
3
K
2
[A-S]= [K
1
K
2
k
3
[A][S] =
 
 
 
3 1 2
12
Lk K K A
1 K K A

b) If [A-S] is the MARI, then L= [s] + [A-S] and
L= [S] + K
1
[A][S] ] =>
 

 
 
1
S
1 K A
L



r= k
3
K
2
[A-S]= [K
1
K
2
k
3
[A][S] =
 
 
 
3 1 2
1
Lk K K A
1 K A

the mathematical forms are identical.

Problem 7.4

r = -
     
2 2 2
d H O d H d CO

dt dt dt


r = k
3
[H
2
O*] = k
4
[O*][CO*] and K
1
=
 
 
co
CO*
P*

L = [*] + [CO*] + [H
2
O] + [O*] (from site balance)

Steady – state approximations:

On H
2
O* : k
2
P
H2O
[*]= k
3
[H
2
O*]
On O* : k
3
[H
2
O] = k
4
[O*][CO+]
       
2
3
2
22
34
H O* ( )P & O* H O* / CO*
k
HO
k
k
k




         
3
2
1 CO H2O 2
34
L K P * P * * H O / CO*
k
k
kk
   

the last term can be negligible, so

 
2
2
1 CO H O
3
L K P P 1 *
k
k

  



   
 
22
2
3 2 2 H O
32
3
1 CO H O
. Lk P
r k H O* *
1 K P k'P
HO
k k P
k
  



Problem 7.5
a. A + S
1
1
k
k



A S reversible RDS (For an ideal gas)
B + S
2

K


B S
A S + B S
3
K


C S + D S
C S
4
K


C + S

5
K
D S D S




A + B


C + D

The rate of reaction r = k

1
P
A
[S] – k
-1
[A S]
From site balance: L = [S] + [A S] + [B S] + [C S] + [D S]
Moreover
 
 
2
4
5
[ ] [ ]
[]
[]
B
C
D
B S K P S
PS
CS
K
PS
DS
K








and
 
  
 
3
C S D S
AS
K B S





 K =
CD
AB
PP
PP
= K
1
K
2
K
3
K
4
K

5

[A S] =
 
CD
5 4 3 2 B
P P S
K K K- K P

So we can obtain:
L = [S]
CD
2 3 4 5 B 2 B C D
45
1 P P
11
K K K K P K P P P
KK

  

Thus r =
 
1 C D
1A
2 3 4 5 B
Lk P P
Lk P
K K K K P
S







=
1 C D
1A
2 3 4 5 B
CD
2 3 4 5 B 2 B C D
45
Lk P P
Lk P
K K K K P
1 P P
11
K K K K P K P P P
KK






  


b.

A + S
1
K


A S
B + S
2
K


B S
A S + B S
3
K


C S + D S
C S
4
K


C + S
5
5
k
k
D S D S






Reversible RDS

A + B


C + D
The rate of the reaction:
r = k
5
[D S] – k
-5
P
D
[S]

Using site balance: L = [A S] + [B S] + [C S] + [D S] + [S], we also have
some relations:

 
 
2
1
4
[ ] [ ]
[]
[]

B
A
D
B S K P S
A S K P S
PS
DS
K







and
  
  
3
C S D S
K
A S B S




So [D S] =
  
 
 

3
1 2 3 4 A B
C
K K K K P P
S
P
K A S B S
CS




L =
 
C 1 2 3 4 A B
1 A 2 B
4C
P K K K K P P
1 K P K P
KP
S

   



Thus r =
5 1 2 3 4 A B
5D
C

C 1 2 3 4 A B
1 A 2 B
4C
Lk- K K K K P P
– Lk P
P
P K K K K P P
1 K P K P
KP

   

In this case, K = K
1
K
2
K
3
K
4
K
5
and K
1
= K
A
, K
2
= K
B

,
4
1
K
= K
C
,
5
1
K
= K
D
=
5
5
k
k

, K
1
K
2
K
3
K
4
=
5
55
kK

K
Kk



Substituting on the rate equation, we get:
r =
5 A B
5D
C
AB
1 A 2 B
C
Lk- KP P
– Lk P
P
KK P P
1 K P K P P K
P
D
CC

   


c.
A
2
+ 2S
1

1
k
k



2A S reversible RDS
2[B + S
2
K


B S]
2[A S + B S
3
K


C S + D S]
2[ C S
4
K


C + S]
A
2
+ 2B



2C

The rate equation is: r = k
1
P
A2
[S]
2
– k
-1
[A S]
2
,
Applying site balance: L = [A S] + [B S] + [C S] + [S]

 
2
4
[ ] [ ]
[]
B
C
B S K P S
PS
CS
K





and
  
  
3
C S S
K
A S B S




 [A S] =
 
  
 
3
*
*
*
C S S
AS
K B S

, K = K
1
K
2
2
K
3

2
K
4
2
or K
1/2
= K
1
1/2
K
2
K
3
K
4

Then, [A S] =
 
2 3 4 B
K K K P
C
PS
. And L =
 
2
2 3 4 4
1
CC
B
B

PP
S K P
K K K P K

  



So we can calculate that:
r =
 
2
2
22
11
2 3 4 B
2
2
2 3 4 4
22
K K K P
1
C
A
CC
B
B
PS
ZZ
L k P L k

LL
PP
KP
K K K P K





  



with
2
Z
L
= site-pair probability and Z = coordination number
Moreover, K
2
K
3
K
4
= K
1/2
/K
1
1/2
then:

r =
2
2
'
2
1
2
2
2 3 4 4
1
C
A
B
CC
B
B
P
Lk P
KP
PP
KP
K K K P K





  



d.
A
2
+ 2S
1
K


2A S
2[B + S
2
K


B S]
2[A S + B S
3
K


C S + D S]

4
4
2
k
k
C S C S








reversible RDS
A
2
+ 2B 2C
The rate equation is:
r = k
4
[C S] – k
-4
P
C
[S]
Also we have: [A S]
2
=
 
2
2
1A
K P S
; [B S] = K
2
P
B

[S], and

  
  
3
C S S
K
A S B S



[C S] =
 
 
 
2
11
22
3 1 2 3
[A ]
K
AB
S B S
K K K P P S
S



From site balance: L = [S] + [A S] + [B S] + [C S]
=

 
 
1/2 1/2 1/2 1/2
1 A2 2 B 1 2 3 A2 B
1 K P K P K K K P PS   

Moreover, we also have: K = K
1
K
2
2
K
3
2
K
4
2
K
1
1/2
2
1 2 3
4
K
K K K
K


So
2 1/2 1/2 2

4 2 3 A B 4 C
1/2 1/2 1/2 1/2
1 A2 2 B 1 2 3 A2 B
L k K K K P P L k P
22
1 K P K P K K K P P
ZZ
LL
r



  

With K
1
= K
A2
, K
2
= K
B
,
4
1
K
= K
C
 K = K
A

K
B
2
K
3
2
K
4
2

So
 
1/2 1/2
4 A2 B C
1/2 1/2 1/2
A2 A2 B B C A2 B
L k K P P – P
2
1 K P K P KK P P
Z
r


  

Problem 7.6
a. The rate equation is: r =
2 4 2
26
3

[]
C H H
d C H
Lk P
dt




2 4 2
25
C H H
2
C H H
P
P
k




Applying steady-state approximation on total surface carbon atoms, we obtain
another equation for
25
CH

, dθ
C(total)
/dt

= 0. Additionally, because steps 2 and 4 are very
rapid:

2 6 2 5 2 4 2
C(total)
1 C H 1 C H 3 C H H

d
k P – Lk – Lk P 0
dt






2 6 2 4 2 5
1 C H 3 C H H2 1 C H H
k P – Lk P Lk
  


 
2 5 2 4
25
1 C H 3 C H H2
C H
1
k P – Lk P

Lk
H







 
2 6 2 4 2
24
2
1 C H 3 C H
2
CH
1

H
H
H
H
k P Lk P
K
P
Lk











=
2 6 2 2 4
2 1 C H 1 H 2 3 C H 1
K k P / Lk P – LK k / Lk




26
24
2
1 2 C H
23
CH
1 1 H
k K P
Kk
1
k Lk P



  





 
 
2 6 2
24
12
C H H
1
CH
2 3 1
kK
P / P
Lk

1 K k k









and

2 2 6
2

26
3 H 1 2 C H
1H
1 2 3 C H
C2H6
2 3 2 3
11
Lk P k K P

Lk P
k K k P
r k’P
K k K k
1 1
kk





  
   

   
   

but θ
C2H4
can be rewritten to give:

 
 
2 6 2
25
1
C H H
3
C H
1 2 3
k
P / P
k

L 1 k / K k







so that
26
1 C H
1
23
k P
r
k
1

Kk







b.
C
2
H
6
+ 2S
1
1
k
k



C
2
H
5
S + H S
C
2
H
5

S + H S
2
K


C
2
H
4
S + S + H
2

C
2
H
4
S + H
2
+ S
3
k

2CH
3
S
4
3 2 4
2 2 2
K
CH S H CH S






C
2
H
6
+ H
2


2CH
4
The rate of reaction can be defined as:
r =
 
26
d C H
dt

= k
3
[C
2
H
4
S][S]
2

H
P

K
2
= [C
2
H
4
S][S]
2
H
P
/ [C
2
H
5
S][H S]
Steady-state approximation on all surface C atoms gives:

k
1
26
CH
P
[S]
2
– k
-1
[C

2
H
5
S][H S] – k
3
[C
2
H
4
S][S]
2
H
P
= 0
k
1
P
C2H6
[S]
2
– k
3
[C
2
H
4
S][S]
2
H
P

= k
-1
[C
2
H
5
S][H S] and
[C
2
H
5
S] =
 
  
 
26
2
2
1 C H
3 2 4
11
k P S
k [H ]
H
k C H S S P
S k H S






[C
2
H
4
S] =
 
 
   
 
26
2
2
1 3 2 4
2
1
CH
H
k P S k C H S
K H S
S P k H S
 



26
2
1 2 C H
23

24
1
1H
k K P
Kk
[C H ] 1
k
kP
S



  





2 6 2
12
C H H
1
24
23
1
kK
( )P / P
k
[C H ]
Kk

1
k
S






Now, site balance to get [S] is:
L = [S] + [H S] +[C
2
H
5
S] + [C
2
H
4
S] + [CH
3
S]
(a) is obtained only if [S]~ L, it means the surface is essentially free of all adsorbed
species or θ
H
, θ
C2H5
, θ
C2H4
, θ
CH3

<< 1. This is a questionable assumption.

Problem 7.7

Plotting ln rate vs. lnP
i
using a power rate law gives the following reaction orders:

T(K)
Reaction Order

N
2
O
O
2

N
2
623
0.08
-0.31
0
653
0.24
-0.12
0
673
0.31
-0.07

0

The simplest L-H model would be for unimolecular decomposition:
(1) 2[N
2
O + *
2
NO
K


N
2
O *]
(2) 2[N
2
O*
k

N
2
+ O*]
(3) 2O*
2
1/
O
K


O

2
+2*
2N
2
O ==> 2N
2
+ O
2

r
m
=
1
m

dN
N
2
dt
=
1
m



2


= k[N
2

O*]
From(1) : K
N2O
=
[N
2
O

]


2

[]
, so [N
2
O*] = K
N2O
P
N2O
[*]
From(3) : K
O2
=
[]
2


2
[]

2
, so [O*] = K
O2
1/2
P
O2
1/2
[*]
Site balance gives: L=[*] +[N
2
O*] +[O*] thus
L =[*] +

2



2

[*] + 

2
1/2


2
1/2
[*] , and [*] =

(1 + 


2



2

+ 

2
1/2


2
1/2
)
,
consequently,
r = k K
N2O
P
N2O
[*] =


2



2


1+ 

2



2

+ 

2
1/2


2
1/2

Arrhenius plots of the fitting parameters listed in Table 2 provide the following
values :
For K
N2O
:∆H
ad
o
= -17 kcal mole
-1
and ∆S
ad
o

- 21 cal mole
-1
K
-1
(e.u.)
For K
O2
:∆H
ad
o
= -25 kcal
1
and ∆S
ad
o
- 35 e.u.
For k : E
RDS
= 57 kcal mole
-1

The enthalpy and entropy values for adsorption fulfill all the guidelines in Table 6.9,
thus they are consistent.
From either a linear extrapolation of the high-P portions of the two isotherms in
Figure 1 or using the difference between the two at 100 Torr CO pressure, the irreversible
uptake is 580 µmole CO g
cat
-1
. The dispersion of Cu is : D
Cu

= Cu
s
/ Cu
tot
, and with CO
ad

/Cu
s
= 1,
D
Cu
=
580 µmole Cu
s
g
cat
1

0.0456 g Cu g
cat
1


mole Cu
63.55 g Cu

(10
6

µmole
mole
= 0.81
Under differential reaction conditions, P
O2
O and can be ignored; therefore, an
easy way is to choose a known differential rate and correct for temperature , for example
:


823 

673 
=

823

673
=

36200 cal /mole /(1.987cal /mole .K )(823K )

36200 cal /mole /(1.987cal /mole .K)(673K )
= 139
thus
TOF
823K
=

823 



=
(12.6 µmole /s.g) (13 atm 1)(0.0666 atm )(139)
[1+ (13 atm 1)(0.0666 atm )](580 µmole Cu s/g)
= 1.4 s
-



Problem 7.8
Step 2 defines the rate:
r
m
=
1
m
d
N
2
dt
=
1
m
(-
d
N
2
dt
) = k[N

2
O*]
Step 1 gives : K
N2O
=
[N2O]
PN2O []
, so [N
2
O*] = K
N2O
P
N2O
[*]
Assuming all surface species are included, a site balance gives
L =[*] +[N
2
O*] + [O*]
To remove the unknown [O*], the SSA must be used:

d
[O ]
dt
= k[N
2
O*] + k
-1
P
O2
[*]

2
– k
1
[O*]
2
= 0
And [O*] =(
k[N2O] + k
1
PO2 []
2

k1
)
1
2


Then L = [*] + K
N2O
P
N2O
[*] +
k[N2O] + k
1
PO2 []
2

k1
)

1
2
= z[*] + (x[*] + y[*])
1/2

Where z = 1 + K
N2O
P
N2O

x = k K
N2O
P
N2O
/ k
1

y = k
-1
P
O2
/ k
1

Rearranging and squaring each side gives :
x[*] + y[*]
2
= (L-z[*])
2
= L

2
– 2Lz[*] + z
2
[*]
2

and (y-z
2
) [*]
2
+ (x +2Lz) [*] – L
2
= 0
The solution for this quadratic expression is
[*] =
(x + 2Lz ) ± (x2 + 4Lxz + 4yL2)
1
2
2(yz
2
)

Substituting back and using the positive root gives :
[*] =
kK
N 2O
P
N 2O
/ k
1

+ 2L(1 + K
N 2O
P
N 2O
)
2[k
1
P
O
2
/k
1
(1 + K
N 2O
P
N 2O
)
2
+
+
(

2

2

1
)
2
+

4
2

2
(1+
2

2
)

1
+
4
2

1

2

1
)
1
2

2[k
1
P
O
2
/k

1
(1 + K
N 2O
P
N 2O
)
2

Because the rate is: r
m
= k[N
2
O*] = K
N2O
P
N2O
[*] , the final rate expression is :
r
m
=

2
(1+(+)
2
+[
2
+ (
2
+)
2

+
2
]
1
2

)
(1+
2
)
2

2


Where a = kK
N2O
/ k
1

b = k
1
/k
-1

c = K
N2O

d = LkK
N2O

These four parameters can be combined in various ways to give the values in Table
1.
An Arrhenius plot of the adsorption equilibrium constant, K
N2O
, gives :∆H
ad
o
= -
25.2 kcal mole
-1
and ∆S
ad
o
- 33 e.u. , which satisfy the guidelines in Table 6.9.
The rate constant Lk represents a unimolecular decomposition reaction on the
surface, so guideline 2 in Table 6.10 can be applied :
r
a
= r
m
/A
cu,g

1.1
. .

1  
2


354 



10
15



2
= 2x10
12

molecule/s.cm
2

or with correction for activation energy:
LA
d
e
-34400/1.987-843
= 2x10
12
= 1.2 x 10
-9
LA
d
and LA
d
= 2x 10

21
molecule/s.cm
s

Both values are below 10
28


.
2
.

Problem 7.10

a, The rate of the reaction as:


2
2
2
[]
1 [ ]
[ *]
2
dN
d NO
r k N
dt dt
   

Moreover, there are 3 quasi-equilibrated steps, so we have:

22
1
11
22
[ *] [*]
[ *][ *] [ *][*]
[ *] [*]
NO NO
OO
NO K P
N O K NO
O K P




So:

22
22
11
1
11
22
[*] [*]
[ *][*]
[ *]
[ *] [ *]

[*]
NO NO NO NO
OO
K K P K K P
K NO
N
OO
KP
  

Using the balance gives: L = [N*] + [O*] + [NO*] + [*] then:

22
22
22
22
11
1
22
11
22
11
1
22
11
22
[*] [*] [*] [*]
[*]
1
NO NO

NO NO O O
OO
NO NO
NO NO O O
OO
K K P
L K P K P
KP
L
K K P
K P K P
KP
   

  

But, for site pairs:
2
Z
L
is probability factor, thus:



22
2 2 2 2
22
2 2 2 2 2
2 2 2 2
' 2 2 2

21
21
11
22
1
11
22
2
2
1 1 1 1
2 2 2 2
21
[*]
2
(1 )
NO NO
NO NO
NO NO
OO
O O NO NO O O
OO
NO
O O O NO O NO O O NO NO
ZL
k K K P
Lk K K P
r
K K P
KP
K P K P K P

KP
kP
r
K P K K P P K P K K P

  

  

b, If [NO*] is assumed to be very lowed compared to the other surface
species, it means [NO*] ~ 0.




2 2 2
2
2
11
22
21
NO
O O O O NO NO
kP
r
K P K P K K P



c, If [NO*] is assumed to be the MARI



2 2 2 2
2
2
1 1 1 1
2 2 2 2
NO
O O O NO O NO
kP
r
K P K K P P



d, If [O*] is assumed to be very low compared to the other surface species or
[O*] ~ 0.



2 2 2 2
2
2
1 1 1 1
2 2 2 2
1
NO
O O O NO O NO NO NO
kP

r
K P K K P P K K P



e, If [O*] is assumed to be the MARI



2 2 2
2
2
11
22
2
NO
O O O O
kP
r
K P K P



f, If [N*] is assumed to be the MARI:



22
2
2

11
22
1
NO
O O NO NO
kP
r
K P K K P



g, If [N*] is not only the MARI, but the surface is also nearly saturated with
N atoms:
 [N*] >> [*] and [N*] ~ L
So

2
22
2
Z
r k L ZLk
L


In this case, over a La
2
O
3
catalyst at 923K, this reaction exhibited a reaction order
on NO of about 1.2 with no O

2
in the feed and approximately 1.5 with O
2
, and the rate
decrease significantly with the present of O
2
. the calculation of e, and g, will be rejected
because:
- (e) Order of P
NO


2

- (g) The observed reaction order is not zero.

Problem 7.11

The rate of the reaction can be defined as:

24
11
N CH
m
dN dN
r
m dt m dt

  




The general reaction can be written as:
CH
4
+ 2O
2
 CO
2
+ 2 H
2
O
With dissociative oxygen adsorption, we can obtain the steps of the reaction as
below:
(1)
2
2
2* 2 *
O
K
OO



(2)
4
44
**
CH
K

CH CH



(3)
1
43
* * *
k
CH O CH OH  

(4)
2
32
* * * *
K
CH O CH O H



(5)
3
2
* * * *
K
CH O O HCO H



(6)

4
* * * *
K
HCO O CO OH



(7)
5
2
* * * *
K
H OH H O



(8)
6
2
* * * *
K
CO O CO



(9)
2
1
22
**

CO
K
CO CO 



(10)
7
2
2 * * *
K
OH H O O



(11)
2
1
22
2 * *
HO
K
H O H O







(12)
1
**
CO
K
CO CO





4 2 2 2
22CH O CO H O  

The last step (12) shows the desorption of CO* to get the product of CO. It is easy
to see that the RDS of this sequence is step (3), thus:
  
14
**
m
r k CH O

However, using the steps (1) and (2), we have
From step (1):
   
22
11
22
**
OO

O K P

From step (2):
   
44
4
**
CH CH
CH K P

 
4 2 4 2
11
2
22
1
*
m CH O CH O
r k K K P P

Moreover, the site balance gives that:
           
4 2 2
* * * * * *L CH O CO H O CO     

Continuously, using the information of steps (9), (11) and (12) we obtain:

 
 
 

22
22
2
2
* [*]
* [*]
* [*]
CO CO
H O H O
CO CO
CO K P
H O K P
CO K P




Then
4 4 2 2 2 2 2 2
11
22
[*]
1
CH CH CO CO O O CO CO H O H O
L
K P K P K P K P K P

    



4 2 4 2
4 4 2 2 2 2 2 2
11
22
1
11
22
'
1
CH O CH O
m
CH CH O O CO CO H O H O CO CO
Lk K K P P
r
K P K P K P K P K P

    

Using the data of table 1 and table 2 adding to Arrhenius relation, we can calculate
that:
For CH
4
:
1
20 . ; 13 .
oo
ad ad
H kcal mol S eu


     

For O
2
:
1
30 . ; 25 .
oo
ad ad
H kcal mol S eu

     


Problem 7.12

a, NO disappearance:
The rate of the reaction can be defined as:

11
2
NO
i NO
m
i
dN dN
r
m v dt m dt


  



And
* 0 * 1
NO
m s NO s NO HNO
r L L k L L k
   

in which L
*
and L
s
is the density of * and S
sites, respectively.
Applying SSA on HNO* gives:
 
* 0 * 1
*
0
s NO s NO HNO
d HNO
L L k L L k
dt
   
  
and r
3

= r
4
, then

 
*0
*
2
NO
m s NO H
d HNO
r L L k
dt

  

Site balance for [*] sites is: L
*
= [*] + [NO*] and
 
 
*
*
NO
NO
NO
K
P


 We have:
 
 
 
 
 
*
*
*
[*]
1
*
1
*
1
NO NO
NO NO
NO NO
NO NO
NO
NO NO
L
KP
L K P
NO
KP
NO
KP
L K P









Site balance on S sites: L
s
= [S] + [H – S] and
 
 
2
2
2
2
H
H
HS
K
PS




 


 



22
22
22
22
11
22
11
22
11
22
11
22
1
1
S
HH
HH
H H H
S
HH
L
S
KP
KP
HS
KP
L
KP





  


when θ
H
<< 1
Thus:
22
11
22
*0
2
1
NO
S NO H H NO
m
NO NO
L L k K K P P
r
KP



b, N
2
formation:
The rate:

2
2
*2
1
NO
N
m N O
dN
r L k
m dt



Using SSA on N
2
O* site:

 
2 2 2
2
* 1 * 2 * 2 * 3 *
0
S HNO NO N O N O N O
d N O
L L k L k L k L k P
dt
    

    

Then
 
     
  


22
2 2 2 2
2
2 2 2
2
* 2 * 3 * 1 * 3 *
11
22
0 3 * 0 3
2 3 2 3 2 3
11
*2
22
03
23
11
1
N
N O S HNO NO N O
S NO H N O S NO H NO H N O
NO
NO NO NO NO
m S NO H NO H N O
NO NO

L k L k L L k L k P
L k k P L k K K P P k P
k k k k K P k k K P
Lk
r L k K K P P k P
k k K P
   
  




  

   
    
  


c, N
2
O formation:
Similarly a, b we have:

 
2
22
2
2 2 2 2
2 2 2 2

2
* 3 * 3 *
11
22
0 3 * 3
*3
23
11
22
0 3 3 3 3 3 2 3
*
23
1
03
*
23
1
11
1
NO
NO
m N O N O
S NO H NO H N O N O
NO NO NO NO
S NO H NO H N O N O
NO NO
S NO H
dN
r L k L k P
m dt

L k K K P P k P L k P
Lk
k k K P K P
L k k K K P P k k P k k k k P
L
k k K P
L k k K K
L
kk




  





  


  








22
1
22
3 2 3
1
NO H N O
NO NO
P P k k k P
KP








Problem 7.14

The specific rate is:


=
1


2

=
1



2

= 
1
Ѳ
2

The steps 1and 2:

2
=
Ѳ
2

2
Ѳ


And 

=
Ѳ



Ѳ


SSA on O*:
Ѳ


= 
1
Ѳ
2
 
2
Ѳ

Ѳ


 Ѳ

=

1
Ѳ
2

2
Ѳ

=

1


2

2
Ѳ


2




Ѳ


Site balance:
1 = Ѳ

+ 
2

2
Ѳ

+ 



Ѳ

+


1

2

2

2







1 

1

2

2

2





= Ѳ



1 + 
2

2
+ 







= 
1

2

2
Ѳ

=

1

2

2


1 

1

2

2

2






1 + 
2

2
+ 






Or 

=


µ
1

1

=

2


2
2



1+ 
2

2
+ 






where = 
1

2

and =

1
2

2
2

2




An Arrhenius plot of LK
1
gives E
1
= 26kcalmole
-1
while similar plots for K
N2O
and
K
CO
give:
- For N
2
O: 



= 13 kcalmole
-1
and 


= 29 calmole
-1
K
-1
(e.u.)
- For CO: 


= 10 kcalmole
-1
and 


= 23e.u.
All these thermodynamic values satisfy the criteria in Table 6.9. The TOF at 573K =
2100µ2
1

1

769µ
1

0.91


= 3.0
1
. Thus the pre-exponential factor per site is 3.0
1
=

0


26000
1.987
(573)
 
0
= 2.4 x 10
10
s
-1
which is less than 10
13
s
-1
and satisfies criterion 2
in table 6.10 for a unimolecular reaction. Also, the heat of adsorption of 10kcalmole
-1
for
CO on Cu is consistent with value of 7-10 kcalmole
-1
reported in the literature.

Problem 7.15

= 



4


= 
1

4




1

2 


2
= 
7
[2]

From step (6): [
2
O*] = 

6


2




/[]
From step (4): [OH*] = 
4


2




/[]
From step (2): [
2
*] = 
2

2
[]
From step (8): [CO*] = 
8



[]
 Therefore:
= 
7

6

2 



/[] = 
7

6

4
[
2
][
2
]/[]

= 
7

6

4


2

2
[
2
][
2
]/
8
[]

SSA from surface CH
2
* (and CH
2
O*) species:
[
2
···]

= 
1

4




1


2 


2

7

2

= 0




2


=


1

4




1

2



2


7


From step (6):


2


=


2


[]

6
[]
=


2



[]

6

4
[
2
]
=

8


2




[]

2

4

6

2
[]
=



8



2

4

6

2

[
2
]

From SSA:

1

4
[] = (
7
+

1

8




2

2

4

6

2
)


2



Hence,


2


=


1

7



4
[]
1 +

1

8



2

7

6

4

2





Site balance with [CH
2
O*] as MARI: =




+[CH
2
O*]=




1 +

1

4

7
1 +

1

8



2

7

6


4

2

2
=




1 +

1

8



2

7

6

4

2

2
+


1

4

7
1 +

1

8



2

7

6

4

2

2







=
1+

1

8



2

7

6

4

2

2

1+

1

8




2

7

6

4

2

2
+

1

4

7


= 
7


2


=

7



1

7
 
4



1 +

1

8



2

7

6

4

2



=

1

4
1 +

1

8



2

7

6

4

2

2
+

1

4


7



 =

1

4

2


2
+

1

8



2

7

6

4


2
+

1

4

2

7



The TOF for step 7 is
5.35 µ
1

1

2.4µ 
2

1

2µ

µ
2
= 1.1
1

Rate per site: TOF=1.1
s-1
=Ae
-E/RT
=Ae
-38000/1.987 (723)
=A( 3.25x10
-12
). Therefore,
A= 1.1/3.25x10
-12
= 3.4x10
11
s
-1
, which is less than 10
13
(Criterion 2 in Table 6.10)
Problem 7.16
2 2
1 1 1 1 1
22
N
i NO NO
m
i
dN
dN dN dN

dN
d
r
m dt m dt m dt m dt dt dt



     

2
1
2 [ *]
NO
dN
k NO
dt

,
[ *]
[ *] [*]
[*]
NO NO NO
NO
NO
K NO K P
P
  

L= [*] + 2[*(NO
2

)*] + [N
2
O*] + [O*] + [NO*] = [NO*]+ [O*]
2 2 2
2
2
1/2 1/2
[ *]
[ *] [*]
[*]
O O O
O
O
K O K P
P
  

22
1/2 1/2
[*](1 )
NO NO O O
L K P K P  
and
22
1/2 1/2
[*] / (1
NO NO O O
L K P K P  

So

22
22
2 2 2 2
1
1
21
1/2 1/2 2
2
[ *] [*]
2 (1 )
NO NO
N NO NO
NO NO O O
Lk K P
k
r NO k K P
K P K P

  


Or
2
32
[ *]
N
r k N O
and SSA on N
2
O* gives k

2
[*(NO
2
)
2
*]=k
3
[N
2
O*]
So
2
22
[* *]
N
r k N O
and SSA on[* N
2
O*] gives k
1
(NO
*
)
2
=k
2
[*(NO)
2
*]
So

2 2 2
2 2 2 1/2 1/2 2
11
[ *] / (1 )
N NO NO NO NO O O
r k NO Lk K P K P K P   

b) From the site balance
k
3
[N
2
O*]=k
1
[NO
*]
2
and [N
2
O*]=(k
1
/k
3
)K
NO
P
NO
[*]
2

and k
2
[*(NO)
2
*]=k
1
[NO*]
2
and [*(NO)
2
*] = (k
1
/k
2
)K
NO
P
NO
[*]
2
then
4
3 2 4
2 2 2
K
CH S H CH S




Consequently, a quadractic expression for [*] is obtained and the solution for such
an equation is so complicated.

Problem 7.17

The series of elementary steps is:
(1) MCH + * MCH*
(2) MCH* MCX* +H
2
(3) MCX* MCD* +H
2

(4) MCD* TOL* + H
2

(5) TOL* TOL + * (RDS)

MCH TOL
Or
MCH + * TOL* + 3H2 where K’=K
1
K
2
K
3
K
4
TOL* TOL + *
Step 5 is RDS, so r = k
5

[TOL*]


L = [*] + [MCH*] + [MCX*] + [MCD*] + [TOL*]
If [TOL*] is MARI, then L = [*] + [TOL*] = [*](1+




2
3
) and [*] =

1+




2
3

Then, r = k
5
[TOL*]=k
5




2

3
=

5


(1+




2
3
)

2
3

or, r =

5


(

2
3
+

)

This is not consistent with the behavior because it contains
a 

2
3
term.

Problem 7.18

      
22
22
2( )
1 3 2
2
g
OO
dO
r k P S k HCO S O S kP S
dt


      


     
2
L S HCO S O S    

(1)
 
 
 
2
2
2
2
H CO
K P O S
HCO S
OH S





(2)
     
22
1/2 1/2
1/2 1/2
H O H O
OH S K P O S S  


(3)
 
 
   

   
22
22
22
2 1/2 1/2
22
2
1/2 1/2
1/2 1/2
1/2 1/2
H CO H CO
H O H O
H O H O
K P O S K P O S S
HCO S
KP
K P O S S


  



(4) At steady-state:
 
 
 
2
2
1

32
O
k P S
OS
k HCO S


(Balance on O-S species)

(5) Substitute (4) in (3):

 
 
22
22
3/2 3/2 5/2
21
2
3/2
3/2 1/2 1/2
32
O
H CO O
H O H
K P k P S
HCO S
k HCO S K P





, then

(6)
 
 
22
22
2/5 2/5 3/5 3/5
21
2
3/5 1/5 1/5
3
H CO O
H O H O
K P k P S
HCO S
k K P



(7)
 
 
2 2 2
2
2/5 1/5 1/5
1
2/5 2/5 3/5 2/5
3 2 1

O H O H O
H CO
k P S K P
OS
k K k P



(8)
 
   
2 2 2 2 2
2 2 2
2/5 3/5 2/5 3/5 2/5 2/5 1/5 1/5
2 1 1
3/5 1/5 1/5 2/5 2/5 2/5
3 3 2
H CO O O H O H O
H O H O H CO
K k P P S k P K P S
LS
k K P k K P
  


(9)
 
 
2 2 2 2 2 2
2/5 3/5 1/5 2/5 2/5 1/5

1 ' ''
H CO O H O O H CO H O
L
S
K P P P K P P P




Assume P
H2O
is approximately constant and power of 0.2 is low. Then,

 
2
2 2 2 2
1
2
0.4 0.6 0.4 0.4
1 "' ''''
O
H CO O O H CO
Lk P
r
K P P K P P




If θ
HCO2
<<1 and θ
O
<<1, then r = kP
O2
If θ
O
is MASI, i.e., θ
O
>> θ
HCO2
then

   
2 2 2
2 2 2 2
0.8
1
22
0.4 0.4 0.4 0.4
'
1 '''' / ''''
O O H CO
O H CO H CO O
Lk P k P P
r
K P P P K P

