davissmch05

Chapter 5

Exercise 1:

 part a) The full BET equation is as follows:  P  V ads ( P o − P )

=

 c − 1   P     +  cV m  cV m   P o 1

(1)

After dividing both the numerator and denominator of the left hand side of equation (1)  by P   by P o,  P   c − 1   P   P o 1    = +   P  cV  cV  V ads (1 − ) m   m   P o  P o

By plotting

 P   P o V ads (1 −

 P   P o

)

(2)

 vs  P  for sample 1, the following can be concluded (only the  P o

following data points were used in these calculations 0.05 ≤  P  ≤ 0. 3 ):  P o The regressed slope of the resulting line =

c −1 cV m

The regressed y-intercept of the resulting line =

-3

 = 0.035 g cm 1 cV m

-5

-3

 = 9.15x10  g cm

2

R   value of the fitted line = 0.99 3

-1

Therefore c = 386.4 and V m = 28.3 cm  g

The one-point BET equation is as follows:

  1    P  =     V ads ( P o − P )  V m   P o  P 

(3)

where we assume that c>>1. Choose one point to calculate V m. Choose P/P  Choose P/P o = 0.3.

5.1

3

-1

At P/P  At P/P o = 0.3, V ads ads = 39.1 cm  g . 3

-1

3

-1

Therefore V m= V (1(1- P/P o) = (39.1cm  g )(0.7) = 27.4 cm  g .

 Note that assuming c >>1 was a good assumption, as seen from the calculations done using the full BET equation.  Now we need to calculate the surface area, S  g , for each solution. The surface area is related to the number of molecules adsorbed in a monolayer, N  monolayer, N , in the following way. S  g  =  N α   

(4)

where α is the cross-sectional area of the adsorbing gas (dinitrogen in this case).  N  =  = # molecules adsorbed per monolayer =

V m N o

 where N   where N o is Avagadro’s number and V  V  3 -1 is the volume of gas at STP (1 atm and 273 K) per mol, 22400 cm  mol .

For dinitrogen, α = 0.162 nm . 2

For the full BET equation 3

−1

28.3 cm g (6.02 x10

23

mol 22400 cm 3 mol

S  g  =

2

  1 m   )(0.162) 9   10 nm  = 123 m2g-1. −1

molecules

Similarly for the one-point BET equation, 2 -1

S  g  =  = 119 m g .

These values are essentially the same (3% difference).

 part b) Repeat the same analysis as in part a) except on Sample 2. Full BET equation: c = 22.4 3 -1 V m = 0.689 cm g .

5.2

 Note that this sample has a smaller BET constant than sample 1 (c is not much greater than 1 here). For the one-point equation: 3 -1 3 -1 V m = V (1- P/P o) = (0.89cm  g )(1-0.3) = 0.623 cm g . 2

-1

The surface areas for the full BET equation and the one-point equation are 3.00 m  g 2 -1 and 2.7 m  g , respectively. Due to the small BET constant, the two methods do not give the same answer (10% difference).

Exercise 2:

 Note that: # moles H chemisorbed irreversibly (strong sites) = #moles H chemisorbed in Run 1- # moles H chemisorbed in Run 2

  H     . First, plot  H/Pt  vs P  for each run; find the y-intercepts   Pt     P =0  Dispersion= H/Pt irreversibly adsorbed =

 H 

−

 H 

 Pt  P =0, Run1  Pt  P =0, Run2

= 0.93 − 0.34 = 0. 59

The average particle size is estimated using the following relationship, assuming that the  particles are spherical: Particle diameter (nm)=1/dispersion. In this case the particle diameter is 1.69 nm.

Exercise 3:

The reaction investigated by A. Peloso is CH3CH2OH  CH3CHO + H2  (Et) (Ad) (DH)

(1)

=

5.3

The proposed mechanism is: r  =

k [ P   Et  − ( P   Ad  P   DH  ) / K e ]

[1 + K  Et  P  Et  + K  Ad  P  Ad ]2

 

(2)

To formulate a mechanism the following clues can be gained from inspection of the  proposed expression: st

1)  P  DH  is raised to the 1  power, so the adsorption and desorption of DH is equilibrated (it is not dissociative adsorption). nd 2) The denominator is raised to the 2  power, so surface sites are involved in the RDS. The following mechanism is proposed: 1

Et + *

2

Et* + *

3

Ad*

 DH*

Et* Ad* + DH* Ad + * DH + *

The following analysis is used to derive the expression in equation (2).  K  Et  =

[ Et ∗] [ Et ][∗]

, K  Ad  =

[ Ad ∗] [ Ad ][∗]

, K  DH  =

[ DH ∗] [ DH ][∗]

 

(3)

From the RDS, r =

k 1[ Et ∗][∗] − k −1[ Ad ∗][ DH ∗] [∗]0

 

(4)

The site balance is [∗] o = [∗] + [ Et ∗] + [ Ad ∗] + [ DH ∗]  

(5)

After substituting equation (3) into equation (5)

5.4

[∗]o [∗]

= 1 + K  Et [ Et ] + K  Ad [ Ad ] + K  DH [ DH ]  

(6)

After substituting equation (6) into equation (4), with the assistance of equation (3), the  proposed rate expression can be derived. Note that the constants are all lumped into k  and K e. In addition, the assumption that K  DH [ DH ]>1 or

[ N 2 ∗] [∗]

>>1

− k −1[∗]o  

(10)

[ H 2 ]3 K 2

Since equation (10) and equation (9) are identical it is probably not possible to distinguish between the two rate expressions experimentally unless an extremely wide range of accurate data is used.

Exercise 10:

The proposed mechanism is:  N 2O + ∗   N 2O∗ 

N 2O∗  N 2 + O∗  (1)

CO + ∗  CO∗  + O∗ 

CO∗  CO2 + 2*

The adsorption of N2O and CO on the surface is equilibrated. Where the overall reaction is  N 2O + CO ⇒ N 2 + CO2 

(2)

 part a) The site balance for this mechanism is [∗] o = [∗] + [ N 2 O∗] + [CO∗] + [O∗]  

(3)

However, if the surface coverage of oxygen is assumed to be very small, the site balance is reduced to [∗] o = [∗] + [ N 2 O∗] + [CO∗]  

(4)

The equilibrium relationships are as follows:

5.14

 K 1 =

[ N 2 O∗] [∗][ N 2 O]

  and  K 3 =

[CO∗] [CO ][∗]

 

(5)

After substituting equation (5) into the site balance and rearranging [∗] =

[∗]o 1 + K 1[ N 2 O ] + K 3 [CO]

 

(6)

The rate of reaction is either the rate of N 2O reaction on the surface or the rate of CO2 formation. r = k 2 [ N 2 O∗] = k 4 [CO∗][O∗]  

(7)

After substituting the site balance and the equilibrium relationships into equation (7), the rate expression is then determined. The rate expression is as follows: r =

k 2 K 1[∗]o [ N 2 O] 1 + K 1[ N 2 O] + K 3 [CO ]

 

(8)

In addition, let k 2 K 1[∗]o = K 2 .  part b) First, linearize the rate expression in equation (8) from part a).  P   N 2O r 

Plot

Plot

=

1  K 1

 P   N 2O r   P   N 2O r 

+

 K 2 P   N 2O  K 1

+

 K 3 P CO  K 1

 

(1)

vs  P   N 2O at constant P CO to get slope =

vs PCO at constant  P   N 2O

 K 2  K 1

 and the intercept =

1  K 1

+

 K 3 P CO  K 1

 K 3 1  K 2 P   N 2 O to get slope =  and the intercept = +  K 1  K 1  K 1

The results of the regression analysis are shown below:

5.15

Linear Regression (Holding

Pco  Constant)

2500 2000   r    /      O      2      N

     P

1500

y = 8.7x + 1486.6

1000

2

R  = 0.9

500 0 0

20

40

60

80

P N2O 

Linear Regression (Holding

P N2O  Constant)

5000 4000   r    /      O      2      N

     P

3000 2000

y = 45.3x + 460.4

1000

2

R  = 1.0

0 0

20

40

60

80

P CO 

There are now four equations to solve with three unknowns. Therefore, there are four different ways to solve the set of overdetermined equations. The average of the 4 cases was taken here. -1

 K 1 = 0.007 s torr  -1

 K 2 = 0.069 torr 

-1

 K 3 = 0.320 torr 

 part c) The results of the non-linear regression using the rate expression in part a) and the data in  part b) yielded the following results. -1

 K 1 = 0.004 s torr  -1

 K 2 = 0.025 torr 

5.16

-1

 K 3 = 0.168 torr 

The answers in parts b) and c) will be different because two different methods were used. The answers in this case are within a factor of 3 of each other.  part d) Since K 3 is large compared to K 1 and K 2 where  K 1 =

[ N 2 O∗] [∗][ N 2 O]

 and  K 3 =

[CO∗] [CO ][∗]

and K 2 from the regression analysis was equal to the equilibrium constant K 1, The coverage of CO on the surface must be larger than the coverage of N 2O on the surface.

Exercise 11:

The proposed mechanism is the following:  H 2O + ∗  V  H 2 + O∗  (1) CO + O∗  V CO2 + ∗ 

Since we cannot assume that either step is the RDS, the steady state approximation is used. r = r 1 = r 2 = k 1 [ H 2 O][∗] − k −1 [ H 2 ][O∗] = k 2 [CO][O∗] − k − 2 [CO2 ][∗]  

(2)

Use equation (2) to solve for [O∗ ] [O∗] =

k 1 [ H 2 O][∗] + k −2 [CO2 ][∗] k 2 [CO ] + k −1 [ H 2 ]

 

(3)

The site balance for this system is the following: [∗] o = [∗] + [O∗]  

(4)

After substituting equation (3) into the site balance and rearranging

5.17

[∗] = 1+

[∗] o   k 1 [ H 2 O] + k − 2 [CO2 ]

(5)

k 2 [CO ] + k −1 [ H 2 ]

 Next, substitute equations (5) and (3) into equation (2). r  =

k 1[ H 2 O ][∗]o (k 2 [CO] + k −1[ H 2 ]) − k −1[ H 2 ][∗]o (k 1[ H 2 O] + k −2 [CO2 ]) k 2 [CO ] + k −1 [ H 2 ] + k 1[ H 2 O] + k −2 [CO2 ]

 part b)

The Langmuir-Hinshelwood type mechanism for the WGS reac tion is: CO + ∗ 

CO∗ 

 H 2O + 3∗ 

2H ∗   + O∗  (1)

CO∗  + O∗  2H ∗ 

CO2 + 2* H 2 + 2∗ 

The rate of the RDS is: r =

k 3 [CO∗][O∗] [∗]o

−

k −3 [CO2 ][∗]2 [∗]o

 

(2)

The adsorption of CO on the surface, the desorption of H 2 from the surface and the dissociation of water are all assumed to be equilibrated. Therefore the following relationships are used.  K 1 =

[CO∗] [CO][∗]

,  K 2 =

[ H ∗] 2 [O∗] [∗]3 [ H 2 O]

  and  K 4 =

[ H 2 ][∗] 2 [ H ∗] 2

 

(3)

The site balance for this mechanism is: [∗] o = [∗] + [CO∗] + [O∗] + [ H ∗]  

(4)

After substituting the equilibrium expressions into the site balance and rearranging 5.18

[∗]o

[∗] = 1 + K 1[CO ] +

 K 2 K 4 [ H 2 O] [ H 2 ]

+

[ H 2 ]1 / 2

 

(5)

 K 41 / 2

Substituting equation (5) and the equilibrium expressions into equation (2) results in the following rate expression: r =

 K 1 K 2 K 4 k 3 [CO ][ H 2 O ][∗]o − k −3 [CO2 ][∗]o [ H 2 ]

 

 K 2 K 4 [ H 2 O]

 

[ H 2 ]

[ H 2 ]1 + K 1[CO ] +

1/ 2

+

[ H 2 ]

 K 41 / 2

    

2

 

(6)

Exercise 12:

The overall reaction is the following CH 3CH 2CHO + H 2 = CH 3CH 2CH 2OH

 part a) A reasonable sequence of reaction steps for this reaction is the following: CH 3CH 2CHO + ∗   H 2 + 2∗ 

CH 3CH 2CHO∗ 

2H ∗ 

CH 3CH 2CHO∗  + H ∗ 

CH 3CH 2CH 2O∗  + ∗  

CH 3CH 2CH 2O∗  +H ∗ 

CH 3CH 2CH 2OH ∗   + ∗ 

CH 3CH 2CH 2OH ∗ 

(1)

 + CH 3CH 2CH 2OH

∗ 

 part b) The rate of consumption of CH 3CH 2CHO is (since the adsorption of CH 3CH 2CHO is the RDS):

− d [CH 3CH 2CHO] dt 

= r = k 1[CH 3CH 2CHO][∗] − k −1[CH 3CH 2CHO*]  

5.19

(2)

Since the adsorption of CH 3CH 2CHO is the RDS, all other steps are quasi-equilibrated. Therefore, the following equilibrium relationships are used.  K 2 =

[ H ∗] 2 [ H 2 ][∗]

 K 3 =

2

[CH 3 CH 2 CH 2 O∗][∗] [CH 3CH 2 CHO∗][ H ∗]

, (3)

 K 4 =

[CH 3 CH 2 CH 2 OH ∗][∗] [ H ∗][CH 3 CH 2 CH 2 O∗]

 

 K 5 =

and

[∗][CH 3CH 2 CH 2 OH ] [CH 3 CH 2 CH 2 OH ∗]

The site balance for this mechanism is the following: [∗] o = [∗] + [CH 3 CH 2 CHO∗] + [ H ∗] + [CH 3CH 2 CH 2 O∗] + [CH 3 CH 2 CH 2 OH ∗]   (4) After substituting the equilibrium expressions into the site balance and then plugging the result into the rate expression in equation (2), the rate of consumption of CH3CH2CHO is the following:

− d [CH 3CH 2 CHO] dt 

  k  [CH 3 CH 2 CH 2 OH ]   k 1[CH 3CH 2 CHO] − −1 [∗]o  K 5 K 4 K 3 K 2 [ H 2 ]    

= 1+

[CH 3 CH 2 CH 2 OH ]  K 5 K 4 K 3 K 2 [ H 2 ]

+ K 21 / 2 [ H 2 ]1 / 2 +

[CH 3 CH 2 CH 2 OH ]  K 4 K 5 K 21 / 2 [ H 2 ]1 / 2

+

[CH 3CH 2 CH 2 OH ]

 part c)

If  P   H 2 (or [ H 2]) is large and  P  CH 3CH 2CH 2OH  (or [CH 3CH 2CH 2OH ]) is small, the reaction rate reduces to k  [∗] [CH  CH  CHO] r = 1 o 1 / 2 3 12/ 2  K 2 [ H 2 ] Therefore, in order to reduce to r =

kP   prop 0.5  P   H 2

, H 2 must be large and that of CH 3CH 2CH 2OH 

must be small.

5.20

 K 5

Exercise 13:

Use the steady state approximation since we cannot assume that one step is the RDS. Therefore, r = r 1 = r 2 = k 1 [ RH 2 ][2O −2 ] = k 2 [O2 ][2 á vac]

(1)

Solving for the concentration of vacancy pairs yields, [2á vac]=

k 1 [ RH 2 ][2O −2 ] k 2 [O2 ]

 

(2)

The site balance for this system is [∗] o = [ 2O −2 ] + [2á vac]

(3)

After substituting the expression for the concentration of vacancy pairs into the site -2  balance and solving for the concentration of O , [ 2O − 2 ] = 1+

[∗]o   k 1 [ RH 2 ]

(4)

k 2 [O2 ]

Lastly, substitute equation (4) into the rate expression in equation (1). r =

k 1[ RH 2 ][∗]o   k 1[ RH 2 ] 1+ k 2 [O2 ]

(5)

After simplification, the rate expression in equation (5) becomes 1 r 

=

1 k 1[∗]o [ RH 2 ]

+

1 k 2 [∗]o [O2 ]

where K 1 = k 1[∗]o and K 2 = k 2[∗]o

5.21