Table 2 shows the integration of the peaks from Figure 7. This integration is performed by an automated software provided with the instrument. It should be noted that the first pulse was completely consumed by the sample, the pulse was injected between time 0 and 5 minutes. From Figure 7 we observe that during the first four pulses, hydrogen is consumed by the sample. After the fourth pulse, it appears the sample is not consuming hydrogen. The experiment continues for a total of seven pulses, at this point the software determines that no consumption is occurring and stops the experiment. Pulse eight is denominated the "saturation peak", meaning the pulse at which no hydrogen was consumed.
Table 2: Hydrogen pulse chemisorption data.
| Pulse n |
Area |
| 1 |
0 |
| 2 |
0.000471772 |
| 3 |
0.00247767 |
| 4 |
0.009846683 |
| 5 |
0.010348201 |
| 6 |
0.010030243 |
| 7 |
0.009967717 |
| 8 |
0.010580979 |
Using Equation 3 the change in area (Δarean) is calculated for each peak pulse area (arean)and compared to that of the saturation pulse area (areasaturation = 0.010580979). Each of these changes in area is proportional to an amount of hydrogen consumed by the sample in each pulse. Table 3 Shows the calculated change in area.
Δ
Area
n
=Area
saturation
-Area
n
Δ
Area
n
=Area
saturation
-Area
n
MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaaeyqaiaabkhacaqGLbGaaeyyamaaBaaaleaacaqGWbGaaeyzaiaabggacaqGRbGaaeOtaaqabaGccaqG9aGaaeyqaiaabkhacaqGLbGaaeyyamaaBaaaleaacaqGZbGaaeyyaiaabshacaqG1bGaaeOCaiaabggacaqG0bGaaeyAaiaab+gacaqGUbaabeaakiaab2cacaqGbbGaaeOCaiaabwgacaqGHbWaaSbaaSqaaiaabchacaqGLbGaaeyyaiaabUgacaqGobaabeaaaaa@4E7A@
(3)Table 3: Hydrogen pulse chemisorption data with ΔArea.
| Pulse n |
Arean |
ΔArean |
| 1 |
0 |
0.010580979 |
| 2 |
0.000471772 |
0.0105338018 |
| 3 |
0.00247767 |
0.008103309 |
| 4 |
0.009846683 |
0.000734296 |
| 5 |
0.010348201 |
0.000232778 |
| 6 |
0.010030243 |
0.000550736 |
| 7 |
0.009967717 |
0.000613262 |
| 8 |
0.010580979 |
0 |
The Δarean values are then converted into hydrogen gas consumption using Equation 4, where Fc is the area-to-volume conversion factor for hydrogen and SW is the weight of the sample. Fc is equal to 2.6465 cm3/peak area. Table 4 shows the results of the volume adsorbed and the cumulative volume adsorbed. Using the data on Table 4, a series of calculations can now be performed in order to have a better understanding of our catalyst properties.
V
adsorbed
=
ΔArea
n
×F
c
SW
V
adsorbed
=
ΔArea
n
×F
c
SW
(4)Table 4: Includes the volume adsorbed per pulse and the cumulative volume adsorbed.
| Pulse n |
arean |
Δarean |
Vadsorbed (cm3/g STP) |
Cumulative quantity (cm3/g STP) |
| 1 |
0 |
0.0105809790 |
0.2800256 |
0.2800256 |
| 2 |
0.000471772 |
0.0105338018 |
0.2787771 |
0.5588027 |
| 3 |
0.00247767 |
0.0081033090 |
0.2144541 |
0.7732567 |
| 4 |
0.009846683 |
0.0007342960 |
0.0194331 |
0.7926899 |
| 5 |
0.010348201 |
0.0002327780 |
0.0061605 |
0.7988504 |
| 6 |
0.010030243 |
0.0005507360 |
0.0145752 |
0.8134256 |
| 7 |
0.009967717 |
0.0006132620 |
0.0162300 |
0.8296556 |
| 8 |
0.010580979 |
0 |
0.0000000 |
0.8296556 |
Gram molecular weight is the weighted average of the number of moles of each active metal in the catalyst. Since this is a monometallic catalyst, the gram molecular weight is equal to the molecular weight of palladium (106.42 [g/mol]). The GMCCalc is calculated using Equation 5, where F is the fraction of sample weight for metal N and WatomicN is the gram molecular weight of metal N (g/g-mole). Equation 6 shows the calculation for this experiment.
GMW
Calc
=
1
(
F
1
W
atomic1
)+(
F
2
W
atomic2
)+...+(
F
N
W
atomicN
)
GMW
Calc
=
1
(
F
1
W
atomic1
)+(
F
2
W
atomic2
)+...+(
F
N
W
atomicN
)
MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5ADE@
(5)
GMW
Calc
=
1
(
F
1
W
atomicPd
)
=
W
atomicPd
F
1
=
106.42
g
g-mole
1
=106.42
g
g-mole
GMW
Calc
=
1
(
F
1
W
atomicPd
)
=
W
atomicPd
F
1
=
106.42
g
g-mole
1
=106.42
g
g-mole
(6)The metal dispersion is calculated using Equation 7, where PD is the percent metal dispersion, Vs is the volume adsorbed (cm3 at STP), SFCalc is the calculated stoichiometry factor (equal to 2 for a palladium-hydrogen system), SW is the sample weight and GMWCalc is the calculated gram molecular weight of the sample [g/g-mole]. Therefore, in Equation 8 we obtain a metal dispersion of 6.03%.
PD
=100×(
V
s
×
SF
Calc
SW×22414
)×
GMW
Calc
PD
=100×(
V
s
×
SF
Calc
SW×22414
)×
GMW
Calc
MathType@MTEF@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeiuaiaabseadaWfqaqaaaWcbaaabeaakiabg2da9iaabgdacaqGWaGaaeimaiabgEna0oaabmaabaWaaSaaaeaacaqGwbWaaSbaaSqaaiaabohaaeqaaOGaey41aqRaae4uaiaabAeadaWgaaWcbaGaae4qaiaabggacaqGSbGaae4yaaqabaaakeaacaqGtbGaae4vaiabgEna0kaabkdacaqGYaGaaeinaiaabgdacaqG0aaaaaGaayjkaiaawMcaaiabgEna0kaabEeacaqGnbGaae4vamaaBaaaleaacaqGdbGaaeyyaiaabYgacaqGJbaabeaaaaa@4FBF@
(7)
PD=100×(
0.8296556[
cm
3
]×2
0.1289[ g ]×22414[
cm
3
mol
]
)×106.42[
g
g-mol
]=6.03%
PD=100×(
0.8296556[
cm
3
]×2
0.1289[ g ]×22414[
cm
3
mol
]
)×106.42[
g
g-mol
]=6.03%
(8)The metallic surface area per gram of metal is calculated using Equation 9, where SAMetallic is the metallic surface area (m2/g of metal), SWMetal is the active metal weight, SFCalc is the calculated stoichiometric factor and SAPd is the cross sectional area of one palladium atom (nm2). Thus, in Equation 8 we obtain a metallic surface area of 2420.99 m2/g-metal.
SA
Metallic
=(
V
S
SW
Metal
×22414
)×(
SF
Calc
)×(
6
.022×10
23
)
×SA
Pd
SA
Metallic
=(
V
S
SW
Metal
×22414
)×(
SF
Calc
)×(
6
.022×10
23
)
×SA
Pd
(9)
SA
Metallic
=(
0.8296556[
cm
3
]
0.001289[
g
metal
]×22414[
cm
3
mol
]
)×(
2
)×(
6
.022×10
23
[
atoms
mol
]
)×0.07[
nm
2
atom
]=2420.99[
m
2
g-metal
]
SA
Metallic
=(
0.8296556[
cm
3
]
0.001289[
g
metal
]×22414[
cm
3
mol
]
)×(
2
)×(
6
.022×10
23
[
atoms
mol
]
)×0.07[
nm
2
atom
]=2420.99[
m
2
g-metal
]
(10)The active particle size is estimated using Equation 11, where DCalc is palladium metal density (g/cm3), SWMetal is the active metal weight, GMWCalc is the calculated gram molecular weight (g/g-mole), and SAPd is the cross sectional area of one palladium atom (nm2). As seen in Equation 12 we obtain an optical particle size of 2.88 nm.
APS=
6
D
Calc
×(
W
s
GMW
Calc
)×(
6
.022×10
23
)
×SA
Metallic
APS=
6
D
Calc
×(
W
s
GMW
Calc
)×(
6
.022×10
23
)
×SA
Metallic
(11)
APS=
600
(
1
.202×10
-20
[
g
Pd
nm
3
]
)×(
0.001289[ g ]
106.42[
g
Pd
mol
]
)×(
6
.022×10
23
[
atoms
mol
]
)×(
2420.99[
m
2
g
Pd
]
)
=2.88nm
APS=
600
(
1
.202×10
-20
[
g
Pd
nm
3
]
)×(
0.001289[ g ]
106.42[
g
Pd
mol
]
)×(
6
.022×10
23
[
atoms
mol
]
)×(
2420.99[
m
2
g
Pd
]
)
=2.88nm
(12)In a commercial instrument, a summary report will be provided which summarizes the properties of our catalytic material. All the equations used during this example were extracted from the AutoChem 2920-User's Manual.
Table 5: Summary report provided by Micromeritics AuthoChem 2920.
| Properties |
Value |
| Palladium atomic weight |
106.4 g/mol |
| Atomic cross-sectional area |
0.0787 nm² |
| Metal density |
12.02 g/cm³ |
| Palladium loading |
1 wt% |
| Metal dispersion |
6.03% |
| Metallic surface area |
2420.99 m²/g-metal |
| Active particle diameter (hemisphere) |
2.88 nm |