Problem 5

To remove sulfur from fuels, hydrogen-assisted hydrodesulfurization is used at refineries, typically over

\ce{MoS2}

supported on

\ce{SiO2}

. Isotope exchange at the gas–solid interface exchanges only the surface atoms. An experiment studies the exchange between an

\ce{MoS2/SiO2}

catalyst (

m_{\text{cat}} = 1.2350

g, Mo mass fraction

w_{\ce{Mo}} = 4.280\%

, initially containing only

^{32}

S) and gaseous

\ce{H2^{34}S}

in a flow reactor (

p = 1.00

bar, flow

v = 20.0

mL min

^{-1}

T = 23.0

°C,

\varphi(\ce{H2^{34}S}) = 1.00\%

, isotopic purity

\alpha = 99.95

mol%). After

t = 10.0

min, the fraction of

^{34}

S among sulfur atoms in the collected gas was

\gamma = 87.3

mol%. (a) Calculate the amount of exchanged (surface) sulfur atoms

n(\mathrm{S})_{\text{surface}}

in mol. (b) Assuming uniform spherical

\ce{MoS2}

particles of density

\rho = 5.06

g cm

^{-3}

, surface areas per atom

A_S = 3.00\times 10^{-19}

m² (S) and

A_{\ce{Mo}} = 5.00\times 10^{-19}

m² (Mo), and that only half of each

\ce{MoS2}

unit is exposed at the surface, calculate the particle radius

R

in nm (

M(\ce{MoS2}) = 160.07

M(\ce{Mo}) = 95.95

g mol

^{-1}

Step 3 of 4: Total surface and volume

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A_{\text{tot}} = \tfrac{1}{2}n(\mathrm{S})_{\text{surface}} N_A (2A_S + A_{\ce{Mo}}) = 3.38\ \text{m}^2,\quad V_{\text{tot}} = \dfrac{m_{\text{cat}}w_{\ce{Mo}}}{M(\ce{Mo})}\dfrac{M(\ce{MoS2})}{\rho} = 1.74\times 10^{-8}\ \text{m}^3

Analysis

Each surface $\ce{MoS2}$ unit exposes 2 S and 1 Mo, occupying $2A_S + A_{\ce{Mo}} = 1.10\times 10^{-18}$ m², but only half of the units lie at the surface, so $A_{\text{tot}} = \tfrac{1}{2}\times 1.02\times 10^{-5}\times 6.022\times 10^{23}\times 1.10\times 10^{-18} = 3.38$ m². The MoS2 mass is $1.2350\times 0.0428/95.95\times 160.07 = 0.0881$ g, giving $V_{\text{tot}} = 0.0881/5.06 = 1.74\times 10^{-8}$ m³.