ars.els-cdn.com · Web viewThe diffraction pattern of samples calcined at 600oC showed no residual hydrotalcite or hydroxide phase is present (Figure 1b), which is due to loss of

$Page 1: ars.els-cdn.com · Web viewThe diffraction pattern of samples calcined at 600oC showed no residual hydrotalcite or hydroxide phase is present (Figure 1b), which is due to loss of$
Supporting Materials

Hydrogen dependence of the reaction mechanism and kinetics of water gas shift

reaction on Ni catalyst: Experimental and DFT study

Tayyaba Noor1,2, Yanying Qi1, De Chen*

Department of Chemical Engineering, Norwegian University of Science and Technology, Sem Sælandsvei 4,Trondheim, NO-7491, Norway

1. T. N and Y. Q are considered equal first authors and have contributed equally to the paper.

2. Present address: School of Chemical and Materials Engineering (SCME), National University of Science and Technology, H-12, Islamabad-44000, Pakistan

Catalyst Characterization:

X-Rays Diffraction:

Power X-ray diffraction (XRD) patterns study was performed at ambient temperature on a

Bruker AXS D8 Focus diffractometer using Cu Kα radiations. XRD data was collected in a

range of 2θ from 10º–90º at a scan speed of 2º/min. In the XRD analysis the prominent 2θ

peaks obtained were at ca. 11, 22, 35, 38, 46 and 62o (Figure 1a). The peaks at ca. 11, 22 and

35o corresponds to (003), (006), (009), and characteristic pair near ca. 62o of (110) and (013)

planes, as well as broad symmetric peaks at ca. 35, 38 and 46o for (012), (015) and (018)

planes are characteristic for hydrotalcites. The diffraction pattern of samples calcined at

600oC showed no residual hydrotalcite or hydroxide phase is present (Figure 1b), which is

due to loss of water and carbon dioxide from the anionic interlayer of hydrotalcite precursors.

In the case of nickel based catalyst the prominent peaks in XRD pattern after calcination

correspond to NiO, MgO and MgNiO2.The nominal ratio of Ni:Mg:Al is 1.34:1.66:1, which is

close to the ratio measured by ICP-MS.

a)

b)

Figure S1. XRD patterns of 40% Ni catalyst a) as prepared sample b) calcined sample

Nitrogen Adsorption Study:

Nitrogen physisorption measurements were performed on a MicromeriticsTristar 3000 at

-196ºC. The calcined sample was degassed overnight under vacuum at 100ºC before

adsorption. The surface area was calculated using the Brunauer–Emmett–Teller (BET)

equation in the relative pressure range from 0.01 to 0.30. The adsorption average pore width

(4 V/A) was estimated by the Barret–Joyner–Halenda (BJH) method provided by the

MicromeriticsTristar 3000 built-in software. The results of BET analysis are summarized in

table 1.

Chemisorption:

The metallic dispersion and metal surface area were determined by hydrogen chemisorption

isotherms using a Micromeritics ASAP 2010C unit at 35ºC. For the freshly calcined catalyst,

about 200 mg of the sample was weighed, placed into a quartz micro-reactor, evacuated at

35ºC for 1 h and reduced in a 5% H2/He flow at 670ºC for 10 h. After reduction, the sample

was evacuated for 0.5 h at 670ºC and for 1 h at 35ºC. At this temperature, an adsorption

isotherm was recorded. The monolayer adsorption capacity was determined by extrapolating

the straight-line portion of the total adsorption isotherm to zero pressure. The dispersion of Ni

metal was calculated, based on the assumption that two metal sites are covered by one

hydrogen molecule. By assuming spherical particles, the particle size (nm) of Ni particles

was calculated with the following formula d(Ni) (nm)= 101/D(%), where d(Ni) is the size

(nm) of Ni particles and D is the metallic dispersion of nickel. The results are summarized in

table S1.

TEM study

TEM specimens were prepared by ultrasonic dispersion of the slightly ground catalyst

samples in ethanol. Drops of the dispersions were applied on a cooper grid supported carbon

film. TEM images were taken by a JEOL 2010F Field Emission Gun (FEG) instrument

operating at 200 kV acceleration voltage.

Figure S2. TEM image of reduced Ni catalyst

Table S1. Physicochemical properties of 40% Ni catalyst

Catalystcomposition

Chemisorption N2 sorption

Metal dispersion, D

(%)

Metal surface area

(m2metal/gcatalyst)

Crystallite size,d(nm)

BETsurface area

(m2g-1)

Average Pore

diameter(nm)

40%Ni 12.4 34.1 8.1a 8.5b 141 10Ni crystal size estimated by chemisorptiona and measured by TEMb

Derivation of equations in table 3

1.0 Redox Mechanism (a)

1.1 AssumingH2Odissociation (3) as rds:

If step 3 is rds, the rate of H2O* dissociation via sequence of elementary steps shown in mechanism a, and other elementary steps are at equilibrium, is given by:

r3=k3θH2O*θ* (A1)

The equilibrium assumption for other elementary steps gives:

θCO*=K 1PCOθ* (A2)

θH2O*=K2PH 2Oθ* (A3)

θH*=√ PH 2

K7θ*

(A4)

θCO2*=PCO2

K6θ*

(A5)

θOH*=θO* θH*K 4θ*

(A6)

θO*=θCO2* θ*K 5θCO*

(A7)

By putting the values of θ CO2*andθ CO*from equation A5 and A2 in equation A7:

θO*=PCO2

K 1K5K6PCOθ* (A8)

By putting the values of θ O*andθ H*from equation A8 and A4 in equation A6:

θOH*=PCO2√PH2

K 1K 4K 5K 6√K7 PCOθ* (A9)

A site balance gives:

θ* + θCO* +θH2O* + θH*+θCO2* + θOH* +θO*= 1

(A10)

Putting values of θCO*,θH2O* , θH* ,θCO2* , θOH*, and θO* from A2- A5, A9 and A8 in equation A10 and taking θ*common:

θ*(1+K1 PCO+K2 PH 2O+PCO2

K 6+√ PH 2

K7+

PCO2

K 1K5 K6 PCO+

PCO2√PH 2

K 1K4 K5 K6 √K 7 PCO )=1

(A11)

Therefore,

θ* 2= 1

(1+K1PCO+K 2 PH 2O+PCO2

K 6+√ PH 2

K 7+

PCO2

K1K 5K 6PCO+

PCO2√PH 2

K1K 4K 5K 6 √K7 PCO )2

(A12)

By putting the values of θH2O *fromA3inequation A1:

r3=k3K 2PH 2Oθ*2 (A13)

Substituting the value of θ*2from equation A12 in equation A13:

r3=k3K 2 PH 2O

(1+K 1PCO+K2PH 2O+PCO2

K 6+√ PH 2

K7+

PCO2

K1K5K 6PCO+

PCO2√PH 2

K1K 4 K5K 6√K7PCO)2

(A14)

We assume the surface coverage of θCO2*,θCO*, and θO* species is negligible as they are involved after rds:

r3=K Ia PH2O

(1+√ PH 2

K7+K 2PH 2O+

PCO2 √PH 2

K 1K4K 5K 6 √K 7PCO )2

(A15)

Where,

K Ia=k 3K 2 (A16)

1.2 Assuming OH dissociation (4) as rds:

If step 4 is rds, the rate of OH* dissociation via sequence of elementary steps shown in mechanism a, and other elementary steps are at equilibrium, is given by:

r 4=k 4θOH*θ* (A17)

The equilibrium assumption for other elementary steps givesθCO*,θH2O* , θH* ,θCO2* and θO*by equation A2-A5 and A8. Whereas θOH* is given by:

θOH*=K 3θH2O* θ*

θH*

(A18)

By putting the values of θ H2O*andθ H*from equationA3 and A4 in equation A18:

θOH*=K 3K 2 √K7 PH 2O

√PH 2

θ*

(A19)

Putting values of θCO*,θH2O* , θH* ,θCO2* , θOH*, and θO* from A2- A5, A19 and A8 in site balance equation A10 and taking θ*common:


K 6+√ PH 2

K 7+

PCO2

K1K5 K6 PCO+K 3K2 √K 7PH 2O

√PH 2)=1

(A20)

Therefore,

θ* 2= 1


K 6+√ PH 2

K 7+

PCO2

K1K 5K 6PCO+K3K 2 √K7 PH 2O

√PH 2)

2(A21)

By putting the values of θOH*fromA19in equation A17:

r 4=k 4K 2K3 √K 7PH 2OPH 2

−0.5θ* 2(A22)


r 4=k 4K2K3 √K7 PH2O

PH 2

−0.5

(1+K 1 PCO+K2 PH 2O+PCO2

K6+√ PH 2

K 7+

PCO2

K1K5 K6 PCO+K 3K2 √K 7PH 2O

√PH 2)

2

(A23)

We assume the surface coverage of θCO2*,θCO*, and θO* species is negligible as they are involved after rds:

r 4=K IIaPH2O PH2

−0.5

(1+K 2 PH 2O+√ PH 2

K7+K3K2 √K7 PH 2O

√PH 2)

2

(A24)

Where,

K IIa=k4K 2K 3√K7 (A25)

1.3 AssumingCO2formation (5) as rds:

If step 5 is rds, the rate of CO2* formation via sequence of elementary steps shown in mechanism a, and other elementary steps are at equilibrium, is given by:

r5=k5θCO*θO*(A26)

The equilibrium assumption for other elementary steps givesθCO*,θH2O* , θH* ,θCO2* and θOH*by equation A2-A5 and A19. Whereas θO* is given by:

θO*=K4θOH* θ*

θH*(A27)

By putting the values of θ OH*andθ H*from equation A19 and A4 in equation A27:

θO*=K2K3K4 K7 PH 2O

PH 2

θ* (A28)

Putting values of θCO*,θH2O* , θH* ,θCO2* , θOH*, and θO* from A2- A5, A19 and A28 in site balance equation A10 and taking θ*common:


K 6+√ PH 2

K 7+K2K3 K4 K7 PH 2O

PH 2

+K 3K2 √K 7PH 2O

√PH 2)=1 (A29)

Therefore,

θ* 2= 1


K 6+√ PH 2

K 7+K2K 3K 4K 7 PH 2O

PH 2

+K3 K2 √K7 PH 2O

√PH 2)

2

(A30)

By putting the values of θCO* andθO*from A2and A28 in equation A26:

r5=k5K 1K2K 3K 4K 7PCOPH 2OPH 2

−1θ* 2(A31)


r5=k5K1 K2K3 K4 K7 PCOPH2O PH 2

−1


K6+√ PH 2

K 7+K 2K 3K 4 K 7PH 2O

PH 2

+K 3 K 2√K7 PH 2O

√PH 2)

2

(A32)

We assume the surface coverage of θCO2* andθH* species is negligible as they are involved after rds:

r5=K IIIaPCOPH 2O PH 2

−1

(1+K1PCO+K2PH 2O+K 2K 3K 4K 7PH 2O

PH2

+K3 K2 √K7 PH 2O

√PH 2)

2

(A33)

Where,

K IIIa=k5K 1K2 K3K 4K 7 (A34)

2.0Carboxyl(b)

2.1 Assuming H2O dissociation (3) as rds:

If step 3 is rds, the rate of H2O* dissociation via sequence of elementary steps shown in mechanism b, and other elementary steps are at equilibrium, is given by:

r3=k3θH2O*θ* (B1)


θCO*=K 1PCOθ* (B2)

θH2O*=K2PH 2Oθ* (B3)

θH*=√ PH 2

K7θ* (B4)

θCO2*=PCO2

K6θ* (B5)

θCOOH*=θCO2* θH*K5θ* (B6)

θOH*=θCOOH* θ*K 4θCO* (B7)

By putting the values of θ CO2* and θ H* from equation B5 and B4 in equation B6:

θCOOH*=PCO2 √PH 2

K 5K 6 √K 7

θ*

(B8)

By putting the values of θ COOH* and θ CO*from equation B8 and B2 in equation B7:

θOH*=PCO2√PH2

K 1K 4K 5K 6√K7 PCOθ* (B9)

A site balance gives:

θ* + θCO* +θH2O* + θH*+θCO2* + θCOOH* +θOH* = 1(B10)

Putting values of θCO*,θH2O* , θH* ,θCO2* , θCOOH*, and θOH* from B2- B5, B8 and B9 in equation B10 and taking θ*common:


K 6+√ PH 2

K7+PCO2

√PH 2

K 5K 6√K7

+PCO2

√PH2

K1K4 K5K 6√K7 PCO )=1 (B11)

Therefore,

θ* 2= 1


K 6+√ PH 2

K 7+PCO2

√PH 2

K5K 6 √K7

+PCO2

√PH2

K1 K4 K5K 6√K7 PCO )2

(B12)

By putting the values of θH2O *from B3 in equation B1:

r3=k3K 2PH 2Oθ*2 (B13)

Substituting the value of θ*2from equation B12 in equation B13:

r3=K3K 2PH2O


K 6+√ PH 2

K7+PCO2 √PH 2

K5K6 √K 7

+PCO2 √PH 2

K1K 4K 5K6 √K7PCO )2

(B14)

We assume the surface coverage of θCO2*,θCO*, θOH* and θCOOH* species is negligible as they are involved after rds:

r3=K Ib PH 2O

(1+√ PH 2

K7+K

2

PH 2O)2

(B15)

Where,

K Ib=k3K 2 (B16)

2.2. Assuming COOH formation (4) as rds:

If step 4 is rds, the rate of COOH* formation via sequence of elementary steps shown in mechanism b, and other elementary steps are at equilibrium, is given by:

r 4=k 4θOH*θCO* (B17)

The equilibrium assumption for other elementary steps givesθCO*,θH2O* , θH* ,θCO2* and θCOOH*

by equation B2-B5 and B8. Whereas θOH* is given by:

θOH*=K3θH2O* θ*

θH*(B18)

By putting the values of θ H2O*andθ H*from equation B3 and B4 in equation B18:


√PH 2

θ*

(B19)

Putting values of θCO*,θH2O* , θH* ,θCO2* , θCOOH*, and θOH* from B2- B5, B8 and B19 in site balance equation B10 and taking θ*common:


K 6+√ PH 2

K7+PCO2

√PH 2

K 5K 6√K7

+K3K2 √K7 PH 2O

√PH 2)=1 (B20)

Therefore,

θ* 2= 1


K 6+√ PH 2

K 7+PCO2

√PH 2

K5K 6 √K7

+K3 K2 √K7 PH 2O

√PH 2)

2(B21)

By putting the values of θOH *∧θCO*from B19 and B2in equation B17:

r 4=k 4K 1K3 K2 √K7 PCOPH2O PH2

−0.5θ* 2(B22)


r 4=k 4K1 K 3K 2√K7PCOPH 2O PH 2

−0.5

(1+K 1 PCO+K2 PH 2O+PCO2

K6+√ PH 2

K 7+PCO2

√PH 2

K5K 6√K 7

+K3 K 2 √K 7 PH2O

√PH 2)

2

(B23)

We assume the surface coverage of θCO2*, θH* and θCOOH* species is negligible as they are involved after rds:

r 4=K IIbPCOPH 2OPH 2

−0.5

(1+K1 PCO+K2 PH 2O+K3 K2√K7PH 2O

√PH 2)

2

(B24)

Where,

K IIb=k4K 1K 3K2 √K 7 (B25)

2.3. Assuming COOH dissociation (5) as rds:

If step 5 is rds, the rate of COOH* dissociation via sequence of elementary steps shown in mechanism b, and other elementary steps are at equilibrium, is given by:

r5=k4θCOOH*θ* (B26)

The equilibrium assumption for other elementary steps givesθCO*,θH2O* , θH* ,θCO2* and θOH*

by equation B2-B5 and B19. Whereas θCOOH* is given by:

θCOOH*=K 4θCO* θOH*

θ*(B27)

By putting the values of θ CO* and θ OH*from equation B2 and B19 in equation B27:

θCOOH*=K 1K 2K 3K4 √K7 PH 2O PCO

√PH 2

θ*

(B28)

Putting values of θCO*,θH2O* , θH* ,θCO2* , θCOOH*, and θOH* from B2- B5, B28 and B19 in site balance equation B10 and taking θ*common:


K 6+√ PH 2

K7+K 1K2 K3K 4 √K 7 PH 2O PCO

√PH2

+K3K 2√K7PH 2O

√PH 2)=1 (B29)

Therefore,

θ* 2= 1


K 6+√ PH 2

K 7+K1K 2K 3K 4 √K7 PH2O

PCO√PH 2

+K 3K 2√K7PH 2O

√PH 2)

2

(B30)

By putting the values of θCOOH*from B28in equation B26:

r5=k5K 1K3 K2K 4 √K7 PCOPH 2O PH 2

−0.5θ* 2 (B31)


r5=k5K 1K 3 K2K 4 √K 7 PCOPH2O PH2

−0.5


K 6+√ PH 2

K7+K 1 K 2K 3K 4 √K 7 PH 2O PCO

√PH 2

+K 3K 2 √K 7PH 2O

√PH 2)

2

(B32)

We assume the surface coverage of θCO2* species is negligible as that is involved after rds:

r5=K IIIbPCOPH 2OPH 2

−0.5

(1+K 1PCO+K2PH 2O+√ PH 2

K7+K 1K2K3K 4 √K7 PH2O PCO

√PH 2

+K3K 2√K7PH 2O

√PH 2)

2

(B33)

Where,

K IIIb=k5 K1K 3K 2K 4 √K7 (B34)

3.0Mechanism (c) involving COH (c)

3.1. Assuming OH dissociation (4) as rds:

If step 4 is rds, the rate of OH* dissociation via sequence of elementary steps shown in mechanism c, and other elementary steps are at equilibrium, is given by:

r 4=k 4θOH*θ*(C1)


θCO*=K 1PCOθ*(C2)

θH2O*=K2PH 2Oθ* (C3)

θH*=√ PH 2

K9θ* (C4)

θCO2*=PCO2

K8θ* (C5)

θOH*=K3θH2O* θ*

θH* (C6)

θCOOH*=θCO2* θH*K7θ* (C7)

θCOH*=K5θCO* θH*

θ* (C8)

θO*=θCOOH* θ*K6θCOH* (C9)

By putting the values of θ CO2* and θ H*from equationC5 and C4 in equation C7:

θCOOH*=PCO2 √PH 2

K 8K 7 √K 9

θ* (C10)

By putting the values of θ CO* and θ H*from equationC2 and C4 in equation C8:

θCOH*=K1K5 PCO√PH 2

√K 9

θ* (C11)

By putting the values of θH2O* and θ H*from equationC3 and C4 in equation C6:


√PH 2

θ* (C12)

By putting the values of θCOOH*and θ COH*from equation C10 and C11 in equation C9:

θO*=PCO2

K1K6 K7K 8K 5PCOθ* (C13)

A site balance gives

θ* + θCO* +θH2O* + θH*+θCO2* + θOH* +θHCOOH* +θCOH* +θO*= 1(C15)

Putting values of θCO*,θH2O* , θH* ,θCO2* , θOH* , θCOOH* ,θO* and θCOH* from C2- C5 and C10 -C13, in equation C15 and taking θ*common:


K 8+√ PH 2

K 9+

PCO2

K 1K6 K7 K8K 5PCO+K3 K2√K 9PH 2O

√PH 2

+K1 K5PCO √PH 2

√K9

+PCO2

√PH 2

K8K 7 √K9)=1

(C16)

Therefore,

θ* 2= 1


K 8+√ PH 2

K 9+

PCO2

K1K 6K 7K 8K 5PCO+K 3K 2√K9PH 2O

√PH 2

+K1K5PCO √PH 2

√K9

+PCO2

√PH 2

K8 K7 √K9)

2

(C17)

By putting the values of θO H*fromC12inequation C1:

r 4=k 4K 3K2 √K 9PH 2OPH 2

−0.5θ* 2(C18)

Substituting the value of θ*2from equation C17 in equation C18:

r 4=k 4 K 3K 2 √K 9PH 2OPH 2

−0.5

(1+K 1 PCO+K 2 PH 2O+PCO2

K 8+√ PH 2

K 9+

PCO2

K1K 6 K 7 K 8 K 5PCO+K3 K2√K9 PH 2O

√PH 2

+K1 K 5PCO√PH 2

√K 9

+PCO2

√PH 2

K 8 K7 √K9)

2

(C19)

We assume the surface coverage of species θCO2*,θCOOH*,θCOH* andθO* is negligible as they are involved after rds:

r 4=K Ic PH 2O PH 2

−0.5

(1+K 1 PCO+K 2 PH 2O+√ PH 2

K9+K3 K2 √K9 PH 2O

√PH 2)

2

(C20)

Where,

K Ic=k 4K 3 K2 √K 9 (C21)

3.2. Assuming COH formation (5) as rds:

If step 5 is rds, the rate of COH* formation via sequence of elementary steps shown in mechanism c, and other elementary steps are at equilibrium, is given by:

r5=k5θCO*θH* (C22)

The equilibrium assumption for other elementary steps givesθCO* θH2O* θH* θCO2* θCOOH* θOH* by equation C2-C5, C10 and C12.

θCOH*=θCOOH* θ*K 6θO* (C23)

θO*=K4θOH* θ*

θH*(C24)

By putting the values of θOH* and θ H*from equation C12 and C4 in equation C24:

θO*=K2K3 K9K 4 PH 2O

PH 2

θ* (C25)

By putting the values of θ COOH* andθ O*from equation C10 and C25 in equation C23:

θCOH*=PCO2

PH 2 √PH 2

K2K3 K 4K 6K 7 K8K 9√K9 PH 2O

θ*(C26)

Putting values of θCO*,θH2O* , θH* ,θCO2* , θOH* , θCOOH* ,θO* and θCOH* from C2- C5,C12, C10and C25 in site balance equation C15 and taking θ*common:


K 8+√ PH 2

K 9+K 2K3 K9 K4 PH 2O

PH 2

+K 3K2 √K 9PH 2O

√PH 2

+PCO2

PH 2√PH 2

K2K 3K4 K6 K7K 8K 9√K9 PH 2O

+PCO2

√PH 2

K 8K7 √K9)=1

(C27)

Therefore,

θ* 2= 1


K 8+√ PH 2

K 9+K2K 3K 9K 4 PH 2O

PH 2

+K 3K2 √K 9PH 2O

√PH 2

+PCO2

PH 2√PH 2

K 2K3 K4 K6K7K8K 9√K9 PH 2O

+PCO2

√PH 2

K 8K 7 √K 9)

2

(C28)

By putting the values of θCO*∧θH*from C2 and C4 in equation C22:

r5=k5

K1

√K 9

PCOPH 2

0.5θ* 2

(C29)


r5=k5

K1

√K 9

PCOPH 2

0.5


K 8+√ PH 2

K9+K2K3K 9K 4PH 2O

PH 2

+K3K2 √K9 PH 2O

√PH 2

+PCO2

PH 2√PH 2

K 2K 3K 4K 6K 7K 8K9 √K9PH 2O

+PCO2√PH 2

K8 K7 √K9 )2

(C30)

We assume the surface coverage of species θCO2*,θCOOH*,θCOH* andθO* is negligible as they are involved after rds:

r5=K IIcPCOPH 2

0.5

(1+K 1PCO+K2PH 2O+√ PH 2

K 9+K 3K 2 √K9PH 2O

√PH2)

2

(C31)

Where,

K IIc=k5 K1

√K 9(C32)

3.3. Assuming COOH formation (6) as rds:

If step 6 is rds, the rate of COOH* formation via sequence of elementary steps shown in mechanism c, and other elementary steps are at equilibrium, is given by:

r6=k6θCOH*θO* (C33)

The equilibrium assumption for other elementary steps givesθCO* θH2O* θH* θCO2* θCOOH* θCOH*θOH* and θO* by equation C2-C5, C10-C12 and C25.Putting their values in site balance equation C15 and taking θ*common:


K 8+√ PH 2

K 9+K 2K3 K9 K4 PH 2O

PH 2

+K 3K2 √K 9PH 2O

√PH 2

+K1K 5 PCO√PH2

√K 9

+PCO2

√PH 2

K 8K7 √K 9)=1

(C34)

Therefore,

θ* 2= 1


K 8+√ PH 2

K 9+K2K 3K 9K 4 PH 2O

PH 2

+K 3K2 √K 9PH 2O

√PH 2

+K 1K5 PCO√PH 2

√K 9

+PCO2

√PH 2

K 8K7 √K9)

2

(C35)

By putting the values of θCO*∧θO *from C2 and C25in equation C33:

r6=k6K1K5K 4K 3K 2 √K9 PCOPH2O PH 2

−0.5θ* 2(C36)


r6=k6 K1K5 K4 K3 K2√K 9PCOPH 2O

PH 2

−0.5


K 8+√ PH2

K9+K 2K3 K9K 4PH 2O

PH2

+K3K2 √K9 PH 2O

√PH 2

+K 1K 5 PCO√PH 2

√K9

+PCO2

√PH 2

K 8K 7 √K 9)

2

(C37)

We assume the surface coverage of species θCO2*,θCOOH* andθH* is negligible as they are involved after rds:

r6=K IIIc PCOPH 2OPH 2

−0.5

(1+K 1PCO+K2PH 2O+K 2K3K9K 4 PH 2O

PH 2

+K3 K2 √K 9PH2O

√PH 2

+K1K5 PCO√PH 2

√K9 )2

(C38)

Where,

K IIIc=k6K 1K 5K 4K3K2 √K9 (C39)

3.4. Assuming COOH dissociation (7) as rds:

If step 7 is rds, the rate of COOH* dissociation via sequence of elementary steps shown in mechanism c, and other elementary steps are at equilibrium, is given by:

r7=k7θCOOH*θ* (C40)

The equilibrium assumption for other elementary steps givesθCO* θH2O* θH* θCO2* θCOH*θOH* and θO* by equation C2-C5, C11, C12 and C25, whereas θCOOH* is given by:

θCOOH*=K 6θCOH*θO*

θ*(C41)

By putting the values of θCOH* and θO* from equation C11and C25in equation C41

θCOOH*=K 6K 1 K5K 4K3 K2 √K 9PCOPH 2O

√PH 2

θ*

(C42)

Putting the values of θCO* θH2O* θH* θCO2* θCOH*θOH* θO* and θCOOH* from equation C2-C5, C11, C12, C25 and C42 in site balance equation C15 and taking θ*common:


K 8+√ PH 2

K 9+K 2K3 K9 K4 PH 2O

PH 2

+K 3K2 √K 9PH 2O

√PH 2

+K1K 5 PCO√PH2

√K 9

+K 6K1 K5K 4K 3K2 √K 9PCOPH 2O

√PH 2)=1

(C43)

Therefore,

θ* 2= 1


K 8+√ PH 2

K 9+K2K 3K 9K 4 PH 2O

PH 2

+K 3K2 √K 9PH 2O

√PH 2

+K 1K5 PCO√PH 2

√K 9

+K 6K1K5K 4K 3K 2 √K 9PCOPH 2O

√PH 2)

2

(C44)

By putting the values of θCOOH*fromC42in equation C40:

r7=k7 K6K 1K5 K4 K3K 2√K9PCOPH 2O PH 2

−0.5θ*2(C45)


r7=k 7K 6K1K 5K 4K3 K2 √K 9PCOPH 2O PH 2

−0.5


K 8+√ PH 2

K9+K 2K3 K9K 4PH 2O

PH2

+K3K2 √K9 PH 2O

√PH 2

+K 1K 5 PCO√PH 2

√K9

+K 6K 1K 5K4K3 K2 √K9 PCOPH 2O

√PH2)

2

(C46)

We assume the surface coverage of species θCO2* and θH* is negligible as they are involved after rds:

r7=K IVcPCOPH 2OPH 2

−0.5

(1+K1PCO+K2PH 2O+K 2K 3K 9K 4PH 2O

PH 2

+K3 K2 √K 9PH2O

√PH 2

+K1K5 PCO√PH 2

√K9

+K6 K 1K 5K 4K3K2 √K 9PCOPH 2O

√PH 2)

2

(C47)

Where,

K IVc=k7K 6K 1K5K4K3K 2√K9 (C48)

The rate expression with the formate mechanism(d) is identical to the mechanism with COH(c) since the isomers (i.e., HCO and COH, as well as HCOO and COOH) cannot be discriminated in the kinetic expression.

Documents

ars.els-cdn.com · Web viewThe diffraction pattern of samples calcined at 600oC showed no residual hydrotalcite or hydroxide phase is present (Figure 1b), which is due to loss of