IChO 2023 · Theory problem 5

Hydrodesulfurization

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The production of sulfur-free fuels is the general trend towards lowering the emission of sulfur-containing compounds that are toxic to the environment. To remove sulfur, the hydrogen-assisted hydrodesulfurization process is used at the refineries.

5.1 Draw the structure of products A to E of thiophene hydrodesulfurization, knowing that A and B are cyclic regioisomers and C is cyclic.

Figure 1

Figure 1. Thiophene hydrodesulfurization process

Sulfur has two most common natural stable isotopes, \(\ce{^{32}S}\) and \(\ce{^{34}S}\), with a relative molar abundance of \(x(\ce{^{32}S})=94.8\,\%\) and \(x(\ce{^{34}S})=4.37\,\%\), respectively. For hydrogen, the stable natural isotopes are \(\ce{^{1}H}\) and \(\ce{^{2}H}\) (\(\ce{D}\)), with a relative molar abundance of \(x(\ce{^{1}H})=99.986\,\%\) and \(x(\ce{^{2}H})=0.014\,\%\), respectively.

5.2 Considering only the isotopes listed above, list all isotopologues of \(\ce{H2S}\).

5.3 Considering only the isotopes listed above, list all isotopologues of \(\ce{H2S}\) containing simultaneously \(\ce{D}\) and \(\ce{^{34}S}\) nuclei and for each calculate the respective relative molar abundance in %.

The desulfurization is a catalytic process typically carried out over \(\ce{MoS2}\) supported on \(\ce{SiO2}\) (\(\ce{MoS2}/\ce{SiO2}\)) catalyst. To study the surface of the catalyst, isotope exchange methods can be employed. The isotope exchange reaction takes place at the gas-solid interface, resulting in the exchange of the surface atoms exclusively. In the first approximation, the bulk atoms do not participate in the exchange (Figure 2).

In the experiment, the isotope exchange between the \(\ce{MoS2}/\ce{SiO2}\) catalyst (Mo mass fraction \(w_{\mathrm{Mo}}=4.280\,\%\), initially containing only \(\ce{^{32}S}\)) and gaseous isotopically-labeled \(\ce{H2^{34}S}\) was studied in a flow reactor (Figure 2). The \(\ce{MoS2}/\ce{SiO2}\) catalyst (\(m_{\mathrm{cat}}=1.2350\,\mathrm{g}\)) was kept in a flow (\(p=1.00\,\mathrm{bar}\), \(v=20.0\,\mathrm{mL\,min^{-1}}\), \(T=23.0\,^\circ\mathrm{C}\)) of gas mixture containing \(\ce{H2^{34}S}\) balanced with \(\ce{Ar}\) (volume fraction \(\varphi(\ce{H2^{34}S})=1.00\,\%\), \(\ce{^{34}S}\) isotopic purity \(\alpha=99.95\ \text{mol.}\,\%\)). The experiment duration was \(t=10.0\,\mathrm{min}\) and gas from the outlet was collected during the entire experiment. The measured fraction of \(\ce{^{34}S}\) isotope among the sulfur atoms (\(\gamma\)) in the collected gas phase was \(\gamma=87.3\ \text{mol.}\,\%\). Assume ideal gas behavior, and that the elemental (not isotopic!) composition of \(\ce{MoS2}\) on the surface and in the bulk are identical, and by the end of the experiment all sulfur atoms from the surface are exchanged with the gas phase.

Figure 2

Figure 2. Schematic representation of the experiment in the course (A) and at the final stage (B). Sulfur atoms on the surface are shown in red, sulfur atoms in the bulk are shown in blue. Molybdenum atoms are not shown.

5.4 Calculate the amount of exchanged sulfur atoms \(n(\ce{S})_{\mathrm{surface}}\) in mol.

If you have been unable to calculate the amount of exchanged sulfur atoms, use the value \(1.53\times10^{-5}\,\mathrm{mol}\) in all the following calculations.

Assume that the \(\ce{MoS2}\) phase consists of uniform spherical particles, and at the end of the experiment all sulfur atoms located on the surface are exchanged, while the bulk atoms did not participate in the exchange reaction. The density of \(\ce{MoS2}\) is \(\rho=5.06\,\mathrm{g\,cm^{-3}}\), the average area occupied by \(\ce{S}\) and \(\ce{Mo}\) atoms on the surface is equal to \(A_{\mathrm{S}}=3.00\times10^{-19}\,\mathrm{m^2}\) and \(A_{\mathrm{Mo}}=5.00\times10^{-19}\,\mathrm{m^2}\), respectively. The area of a sphere with radius \(R\) can be calculated as \(S=4\pi R^2\), and its volume as \(V=(4/3)\pi R^3\). Assume that isotopic composition does not affect the density of \(\ce{MoS2}\).

5.5 Calculate the particle radius \(R\) of the \(\ce{MoS2}\) particles. Give your answer in nm.

In reality, the isotopically-labeled atoms from the surface diffuse into the bulk and the non-labeled atoms from the bulk travel to the surface, undergoing a gradual exchange (Figure 3A). Therefore, for a given moment, the fraction of the labeled atoms inside the particle decreases from the surface of the particle to its center. Simultaneously, with an increase in time of exchange, the involvement of bulk atoms to the exchange reaction increases, as sketched in Figure 3B.

Figure 3

Figure 3. A) Schematic representation of the diffusion of sulfur isotopes from the surface to the bulk in a \(\ce{MoS2}\) particle. Sulfur atoms on the surface are shown in red, sulfur atoms in the bulk are shown in blue. Molybdenum atoms are not shown. B) The fraction of \(\ce{^{34}S}\) atoms in the bulk as function of time and distance from center of the particle; \(r_{\mathrm{edge}}\) corresponds to the radius of \(\ce{MoS2}\) particle.

At the end of the experiment, the surface atoms are completely exchanged, and additionally a fraction of the bulk is exchanged via diffusion. The fraction of the exchanged bulk atoms, \(n(\ce{S})_{\mathrm{bulk}}^{\mathrm{ex}}\), of the total bulk atoms of sulfur, \(n(\ce{S})_{\mathrm{bulk}}^{\mathrm{total}}\), can be calculated as:

\[ F=n(\ce{S})_{\mathrm{bulk}}^{\mathrm{ex}}/ n(\ce{S})_{\mathrm{bulk}}^{\mathrm{total}} =1-\exp(-Dt/R^2) \]

Where \(t\) is the time of exchange experiment (described above), \(R\) is the particle radius, and \(D\) is the diffusion coefficient. The catalyst described above was independently studied by means of electron microscopy, which showed that the \(\ce{MoS2}\) particles are uniformly distributed spheres with a radius of \(35.0\,\mathrm{nm}\).

5.6 Using \(R=35.0\,\mathrm{nm}\) and the data of the exchange experiment described above, calculate the diffusion coefficient \(D\) (in \(\mathrm{m^2\,s^{-1}}\)) for the diffusion of sulfur atoms in \(\ce{MoS2}\). In your calculations, use the approximation \(e^x\approx1+x\) for \(x\ll1\).

Official English text of the problem.

Answers

  • 5.1: A: 2,3-dihydrothiophene S1CCC=C1; B: 2,5-dihydrothiophene S1CC=CC1; C: thiolane (tetrahydrothiophene) S1CCCC1; D: butane-1-thiol CCCCS; E: butane CCCC. A and B may be interchanged.
  • 5.2: Six isotopologues: sulfur-32 hydrogen sulfide [1H][32S][1H], its monodeuterated form [1H][32S][2H] and dideuterated form [2H][32S][2H]; sulfur-34 hydrogen sulfide [1H][34S][1H], its monodeuterated form [1H][34S][2H] and dideuterated form [2H][34S][2H].
  • 5.3: Monodeuterated sulfur-34 hydrogen sulfide, \(\ce{HD^{34}S}\), [1H][34S][2H]: \(1.2\times10^{-3}\,\%\). Dideuterated sulfur-34 hydrogen sulfide, \(\ce{D2^{34}S}\), [2H][34S][2H]: \(8.6\times10^{-8}\,\%\).
  • 5.4: \(n(\ce{S})_{\mathrm{surface}}=1.03\times10^{-5}\,\mathrm{mol}\).
  • 5.5: \(R\approx15.4\,\mathrm{nm}\).
  • 5.6: \(D\approx1.07\times10^{-20}\,\mathrm{m^2\,s^{-1}}\).

Solution

5.1 — Products and balanced reactions

The starting material is thiophene, c1ccsc1. The other small molecules in the reaction scheme are hydrogen, [H][H], and hydrogen sulfide, [SH2].

Addition of one molecule of hydrogen leaves one double bond in the five-membered sulfur-containing ring. There are two distinct positions for this double bond:

Product Name and SMILES Precise description of the requested drawing
A 2,3-Dihydrothiophene S1CCC=C1 A five-membered ring with sequence \(\ce{S-CH2-CH2-CH=CH}\), closed back to sulfur. The double bond includes a carbon adjacent to sulfur.
B 2,5-Dihydrothiophene S1CC=CC1 A five-membered ring with sequence \(\ce{S-CH2-CH=CH-CH2}\), closed back to sulfur. The double bond joins the two carbons not directly bonded to sulfur.
C Thiolane (tetrahydrothiophene) S1CCCC1 A saturated five-membered ring containing one sulfur atom and four \(\ce{CH2}\) groups.
D Butane-1-thiol CCCCS An open, unbranched chain, \(\ce{CH3CH2CH2CH2SH}\).
E Butane CCCC An open, unbranched chain, \(\ce{CH3CH2CH2CH3}\).

The diagram does not distinguish the two branches, so assigning A and B the other way round is equally valid.

Hydrogenation of either remaining double bond gives the same saturated ring C. Hydrogenolysis of one ring carbon–sulfur bond then opens the ring to give D. Hydrogenolysis of the remaining carbon–sulfur bond produces E and hydrogen sulfide.

The balanced reactions for both branches are:

\[ \ce{C4H4S + H2 -> C4H6S} \qquad\text{(thiophene to A or B)} \]

\[ \ce{C4H6S + H2 -> C4H8S} \qquad\text{(A or B to C)} \]

\[ \ce{C4H8S + H2 -> CH3CH2CH2CH2SH} \qquad\text{(C to D)} \]

\[ \ce{CH3CH2CH2CH2SH + H2 -> CH3CH2CH2CH3 + H2S} \qquad\text{(D to E)} \]

Thus the overall hydrodesulfurization is:

\[ \boxed{\ce{C4H4S + 4H2 -> C4H10 + H2S}} \]

5.2 — All isotopologues

The two hydrogen positions are equivalent. Consequently, the hydrogen isotope combinations are two protium atoms, one protium and one deuterium atom, or two deuterium atoms. Combining these three possibilities with either sulfur isotope gives \(3\times2=6\) isotopologues.

Here and below, \(\ce{H}\) in an isotope-specific formula denotes \(\ce{^{1}H}\).

Formula Name Isotope-explicit SMILES
\(\ce{H2^{32}S}\) Sulfur-32 hydrogen sulfide [1H][32S][1H]
\(\ce{HD^{32}S}\) Monodeuterated sulfur-32 hydrogen sulfide [1H][32S][2H]
\(\ce{D2^{32}S}\) Dideuterated sulfur-32 hydrogen sulfide [2H][32S][2H]
\(\ce{H2^{34}S}\) Sulfur-34 hydrogen sulfide [1H][34S][1H]
\(\ce{HD^{34}S}\) Monodeuterated sulfur-34 hydrogen sulfide [1H][34S][2H]
\(\ce{D2^{34}S}\) Dideuterated sulfur-34 hydrogen sulfide [2H][34S][2H]

Interchanging the two hydrogen positions does not create another isotopologue.

5.3 — Abundances of isotopologues containing both deuterium and sulfur-34

Use the stated natural isotope fractions:

\[ p_{\mathrm H}=0.99986,\qquad p_{\mathrm D}=0.00014,\qquad p_{34}=0.0437. \]

The instruction to consider only the listed isotopes restricts the species being enumerated; the supplied natural sulfur abundances are used without renormalization.

For monodeuterated sulfur-34 hydrogen sulfide, \(\ce{HD^{34}S}\), [1H][34S][2H], either hydrogen position can contain deuterium. Hence the factor of two:

\[ \begin{aligned} x(\ce{HD^{34}S}) &=2p_{\mathrm H}p_{\mathrm D}p_{34}\\ &=2(0.99986)(0.00014)(0.0437)\\ &=1.223428696\times10^{-5}. \end{aligned} \]

Converting the mole fraction to percent,

\[ x(\ce{HD^{34}S}) =0.001223428696\,\% \approx\boxed{1.2\times10^{-3}\,\%}. \]

For dideuterated sulfur-34 hydrogen sulfide, \(\ce{D2^{34}S}\), [2H][34S][2H],

\[ \begin{aligned} x(\ce{D2^{34}S}) &=p_{\mathrm D}^{\,2}p_{34}\\ &=(0.00014)^2(0.0437)\\ &=8.5652\times10^{-10}, \end{aligned} \]

so

\[ x(\ce{D2^{34}S}) =8.5652\times10^{-8}\,\% \approx\boxed{8.6\times10^{-8}\,\%}. \]

The final percentages are rounded to two significant figures, consistent with the supplied deuterium abundance.

5.4 — Amount of exchanged sulfur

For the numerical calculations, use the standard constants

\[ R_{\mathrm g}=8.31446\,\mathrm{J\,mol^{-1}\,K^{-1}}, \qquad N_{\mathrm A}=6.02214\times10^{23}\,\mathrm{mol^{-1}}. \]

The total gas volume passed through the reactor is

\[ V_{\mathrm{gas}}=vt =(20.0\,\mathrm{mL\,min^{-1}})(10.0\,\mathrm{min}) =200\,\mathrm{mL} =2.00\times10^{-4}\,\mathrm{m^3}. \]

Each hydrogen sulfide molecule contains one sulfur atom. Its volume fraction equals its mole fraction for an ideal gas, so the amount of sulfur passed through the reactor is

\[ \begin{aligned} n_{\mathrm{gas}}(\ce{S}) &=\frac{pV_{\mathrm{gas}}\varphi}{R_{\mathrm g}T}\\ &=\frac{(1.00\times10^5\,\mathrm{Pa}) (2.00\times10^{-4}\,\mathrm{m^3})(0.0100)} {(8.31446\,\mathrm{J\,mol^{-1}\,K^{-1}})(296.15\,\mathrm K)}\\ &=8.1224\times10^{-5}\,\mathrm{mol}. \end{aligned} \]

Exchange replaces one sulfur atom in the solid with one from the gas, leaving the total amount of gas-phase sulfur unchanged. Since the catalyst initially contains no sulfur-34, the net amount of sulfur-34 retained is

\[ \begin{aligned} n_{\mathrm{retained}}(\ce{^{34}S}) &=(\alpha-\gamma)n_{\mathrm{gas}}(\ce{S})\\ &=(0.9995-0.873)(8.1224\times10^{-5})\\ &=1.02748\times10^{-5}\,\mathrm{mol}. \end{aligned} \]

Under the completely labelled surface approximation shown in Figure 2, this is the amount of surface sulfur:

\[ \boxed{n(\ce{S})_{\mathrm{surface}} \approx1.03\times10^{-5}\,\mathrm{mol}}. \]

If the final surface isotope fraction is taken as \(0.9995\) rather than unity, this amount is divided by \(0.9995\), a correction of only \(0.050\,\%\). The calculations below use the completely labelled surface approximation.

5.5 — Radius inferred when exchange occurs only at the surface

Use \(M(\ce{Mo})=95.95\,\mathrm{g\,mol^{-1}}\) and \(M(\ce{^{32}S})\approx32.00\,\mathrm{g\,mol^{-1}}\). The mass and amount of molybdenum are

\[ m(\ce{Mo})=m_{\mathrm{cat}}w_{\mathrm{Mo}} =(1.2350)(0.04280)=0.052858\,\mathrm g, \]

\[ n(\ce{Mo})=\frac{0.052858}{95.95} =5.50891\times10^{-4}\,\mathrm{mol}. \]

Thus the mass of the supported molybdenum disulfide phase is approximately

\[ \begin{aligned} m(\ce{MoS2}) &=n(\ce{Mo})\left[M(\ce{Mo})+2M(\ce{^{32}S})\right]\\ &=(5.50891\times10^{-4})(159.95)\\ &=0.0881150\,\mathrm g. \end{aligned} \]

Its total volume is therefore

\[ V_{\mathrm{solid}} =\frac{0.0881150\,\mathrm g}{5.06\,\mathrm{g\,cm^{-3}}} =0.0174140\,\mathrm{cm^3} =1.74140\times10^{-8}\,\mathrm{m^3}. \]

The small mass increase from isotope substitution is negligible at the experimental precision.

The surface has the same elemental composition as the bulk, so

\[ n(\ce{Mo})_{\mathrm{surface}} =\frac12n(\ce{S})_{\mathrm{surface}}. \]

Both sulfur and molybdenum occupy surface area. Consequently,

\[ \begin{aligned} S_{\mathrm{total}} &=N_{\mathrm A}\left[ n(\ce{S})_{\mathrm{surface}}A_{\mathrm S} +n(\ce{Mo})_{\mathrm{surface}}A_{\mathrm{Mo}} \right]\\ &=N_{\mathrm A}n(\ce{S})_{\mathrm{surface}} \left(A_{\mathrm S}+\frac12A_{\mathrm{Mo}}\right). \end{aligned} \]

The area per surface sulfur atom, including its corresponding half molybdenum atom, is

\[ A_{\mathrm S}+\frac12A_{\mathrm{Mo}} =3.00\times10^{-19}+\frac12(5.00\times10^{-19}) =5.50\times10^{-19}\,\mathrm{m^2}. \]

Hence

\[ S_{\mathrm{total}} =(6.02214\times10^{23})(1.02748\times10^{-5}) (5.50\times10^{-19}) =3.4032\,\mathrm{m^2}. \]

For any number of identical spheres,

\[ \frac{S_{\mathrm{total}}}{V_{\mathrm{solid}}} =\frac{4\pi R^2}{(4/3)\pi R^3} =\frac3R. \]

Therefore,

\[ R=\frac{3V_{\mathrm{solid}}}{S_{\mathrm{total}}} =\frac{3(1.74140\times10^{-8})}{3.4032} =1.5351\times10^{-8}\,\mathrm m, \]

\[ \boxed{R\approx15.4\,\mathrm{nm}}. \]

5.6 — Diffusion coefficient using the measured radius

Now use the independently measured radius

\[ R=35.0\,\mathrm{nm}=3.50\times10^{-8}\,\mathrm m. \]

The observed exchange from part 5.4 must now be divided into surface exchange and bulk exchange.

The actual total surface area is

\[ S_{\mathrm{actual}} =\frac{3V_{\mathrm{solid}}}{R} =\frac{3(1.74140\times10^{-8})}{3.50\times10^{-8}} =1.49263\,\mathrm{m^2}. \]

Thus the actual amount of surface sulfur is

\[ \begin{aligned} n(\ce{S})_{\mathrm{surface,actual}} &=\frac{S_{\mathrm{actual}}} {N_{\mathrm A}(A_{\mathrm S}+A_{\mathrm{Mo}}/2)}\\ &=\frac{1.49263} {(6.02214\times10^{23})(5.50\times10^{-19})}\\ &=4.50650\times10^{-6}\,\mathrm{mol}. \end{aligned} \]

The amount of exchanged bulk sulfur is the observed exchange minus this surface contribution:

\[ \begin{aligned} n(\ce{S})_{\mathrm{bulk}}^{\mathrm{ex}} &=1.02748\times10^{-5}-4.50650\times10^{-6}\\ &=5.76833\times10^{-6}\,\mathrm{mol}. \end{aligned} \]

The total amount of sulfur in the particles is

\[ n(\ce{S})_{\mathrm{total}} =2n(\ce{Mo}) =1.10178\times10^{-3}\,\mathrm{mol}. \]

The denominator in the definition of \(F\) is bulk sulfur, so surface sulfur must also be subtracted here:

\[ \begin{aligned} n(\ce{S})_{\mathrm{bulk}}^{\mathrm{total}} &=1.10178\times10^{-3}-4.50650\times10^{-6}\\ &=1.09728\times10^{-3}\,\mathrm{mol}. \end{aligned} \]

Therefore,

\[ F=\frac{5.76833\times10^{-6}}{1.09728\times10^{-3}} =5.25696\times10^{-3}. \]

Using the requested approximation,

\[ \exp\left(-\frac{Dt}{R^2}\right) \approx1-\frac{Dt}{R^2}, \]

so

\[ F=1-\exp\left(-\frac{Dt}{R^2}\right) \approx\frac{Dt}{R^2}. \]

With \(t=10.0\,\mathrm{min}=600\,\mathrm s\),

\[ \begin{aligned} D&\approx\frac{FR^2}{t}\\ &=\frac{(5.25696\times10^{-3})(3.50\times10^{-8}\,\mathrm m)^2} {600\,\mathrm s}\\ &=1.0733\times10^{-20}\,\mathrm{m^2\,s^{-1}}. \end{aligned} \]

\[ \boxed{D\approx1.07\times10^{-20}\,\mathrm{m^2\,s^{-1}}}. \]

The inferred exponent \(Dt/R^2\approx0.00526\) is much smaller than unity, consistent with the approximation. Bulk exchange also explains why the surface-only calculation in part 5.5 gives a smaller radius: it attributes too much of the observed exchange to surface area.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.