Porous silicates are important as ion exchangers, molecular sieves and catalysts in petrochemistry. — Organic Chemistry Chemistry Question
Silica Nanostructures
Porous silicates are important as ion exchangers, molecular sieves and catalysts in petrochemistry. Classic zeolites are silicates having defined channels with diameters of 0.4 nm to 1.5 nm. Materials with larger pore diameters are needed to accept larger molecules and make them react. Surfactants or block copolymers are used as ”templates” for the production of amorphous silicates with pore sizes of 1.6 nm to 50 nm.
The production of SiO2 is carried out according to the following equation:
[VISUAL]
When this reaction is carried out in the presence of surfactants, silica-organic hybrid materials form. The organic component can be removed by combustion or dissolution and pure minerals with pores can be obtained. In the following example, X-ray scattering detects hexagonal pore structures.
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The table contains the scattering angles 2θ of the first diffraction peaks (wavelength λ = 0.15 nm) and the relative mass loss w after the removal of the surfactants.
| surfactant | 2θ | w |
| :--- | :--- | :--- |
| C12H25N(CH3)3Cl | 2.262° | 37.2 % |
| C14H29N(CH3)3Cl | 2.046° | 47.6 % |
| C16H33N(CH3)3Cl | 1.829° | 54.4 % |
| C18H37N(CH3)3Cl | 1.719° | 60.0 % |
ρ(SiO2) = 2.2 g·cm-3, ρ(surfact.) = 1 g cm-3
Write down the formulas of A and B.
Model Answer
A = Si(OCH3)4 , B = Si(OH)4
a) Calculate the pore distance d using Bragg's law for the diffraction peaks.
b) What are the radii r of the pores? Calculate.
(Disregard possible end caps of cylindrical pores.)
Model Answer
a) Bragg: n λ = 2 d∙sinθ, with n = 1, results see table below
b) w = V(surfact.) * ρ(surfact.) / (V(SiO2) * ρ(SiO2) + V(surfact.) * ρ(surfact.))
V(surfact.) = V(pore) and V(SiO2) can be calculated for a structure of height l:
V(pore) = π r^2 l
V(SiO2) = 6∙A(triangle)∙l – π r^2 l
with A(triangle) = a∙(d/2) and d/2 = a∙sqrt(3)/2, which gives a = d/sqrt(3)
V(SiO2) = 0.5 * sqrt(3) * d^2 * l – π r^2 l
This leads to:
r = d * sqrt((3^0.5 * w * ρ(SiO2)) / (2 * π * w * ρ(SiO2) + 2 * π * (1 - w) * ρ(surfact.)))
Results:
- C12H25N(CH3)3Cl: d = 3.80 nm, r = 1.50 nm
- C14H25N(CH3)3Cl: d = 4.20 nm, r = 1.80 nm
- C16H25N(CH3)3Cl: d = 4.70 nm, r = 2.10 nm
- C18H25N(CH3)3Cl: d = 5.00 nm, r = 2.30 nm
In another experiment, hexagonal pore structures form by using surfactants of different chain lengths but the same surfactant mass concentrations.
How do a) pore diameter and b) pore distance depend on the tail lengths of the surfactants? Answer qualitatively and explain.
Model Answer
Increasing tail length leads to a) an increase in diameter and b) an increase in pore distance (the same total volume of surfactants but more surfactant molecules per pore, i.e. fewer pores and larger pore distances).
The specific surface Asp (surface area per mass) of porous materials can be determined by gas adsorption experiments. The Langmuir adsorption isotherm can be derived from a kinetic consideration of adsorption and desorption in a monolayer.
Show that the relation between pressure p, volume of adsorbed gas Vads and maximum adsorbable volume V* can be expressed as
1/Vads = 1/(KV*p) + 1/V* (K = constant)
Model Answer
In equilibrium the rate of adsorption (kads(n*– nads)p) is equal to the rate of desorption (kdes nads):
kdes nads = (kads(n*– nads)p)
where n* = maximum adsorbable amount (in mol dm^-3) and nads = adsorbed amount (in mol dm^-3)
nads / (n* - nads) = (kads / kdes) * p or
Vads / (V* - Vads) = (kads / kdes) * p
and with kads / kdes = K:
1/Vads = 1/(KV*p) + 1/V*
Concerning the adsorption of N2 to 1 g silica material at 77 K, the following volumes as functions of pressure are adsorbed. The volumes have been normalized to standard pressure. The area of one adsorbed N2 molecule is A(N2) = 0.16 nm^2.
| p \ surfactant | 1.30×10^5 Pa | 2.60×10^5 Pa | 4.00×10^5 Pa | 5.30×10^5 Pa | 6.60×10^5 Pa | 8.00×10^5 Pa |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| C12H25N(CH3)3Cl | 4.6 | 8.2 | 11.9 | 14.5 | 16.7 | 19.0 |
| C14H29N(CH3)3Cl | 6.0 | 11.5 | 16.0 | 19.0 | 23.1 | 25.5 |
| C16H33N(CH3)3Cl | 7.8 | 14.0 | 19.0 | 24.0 | 28.0 | 31.3 |
| C18H37N(CH3)3Cl | 8.1 | 14.7 | 20.8 | 25.5 | 29.0 | 34.0 |
(volumes Vads in cm^3)
Calculate the specific surfaces Asp (m^2 g^-1) of the materials.
Model Answer
Linear regression of p * Vads^-1 versus p yields the slope (V*)^-1.
Asp = V*_std * p_std / (R * T * m) * A(N2) * N_A
with m(SiO2) = 1 g and T = 77 K.
Results:
- C12H25N(CH3)3Cl: V* = 49.0 cm3, Asp = 747.1 m2 g^-1
- C14H29N(CH3)3Cl: V* = 67.5 cm3, Asp = 1029.1 m2 g^-1
- C16H33N(CH3)3Cl: V* = 77.3 cm3, Asp = 1178.6 m2 g^-1
- C18H37N(CH3)3Cl: V* = 86.5 cm3, Asp = 1318.8 m2 g^-1
Imagine you don’t have an x-ray machine to measure the pore distances in 34.2.
Calculate the pore radii from mass loss (in 34.2) and the specific surfaces Asp determined in 34.5 without using the pore distance d.
Model Answer
Using:
w = V(surfact.) * ρ(surfact.) / (V(SiO2) * ρ(SiO2) + V(surfact.) * ρ(surfact.))
Which gives V(pore) / m(SiO2) = ρ(surfact.)^-1 * w / (1 - w)
And specific surface:
Asp = S(pore) / m(SiO2) = 2 * π * r * l / m(SiO2)
Since V(pore) / m(SiO2) = π * r^2 * l / m(SiO2), we find:
V(pore) / m(SiO2) = Asp * r / 2
Combining these yields:
r = 2 * ρ(surfact.) * w / ((1 - w) * ρ(SiO2) * Asp)
Results:
- C12H25N(CH3)3Cl: r = 1.6 nm
- C14H29N(CH3)3Cl: r = 1.8 nm
- C16H33N(CH3)3Cl: r = 2.0 nm
- C18H37N(CH3)3Cl: r = 2.3 nm
(Note: X-ray scattering is still necessary to detect the hexagonal structure.)