Chemistry Labs

Problem 5

To remove sulfur from fuels, hydrogen-assisted hydrodesulfurization is used at refineries, typically over MoSX2\ce{MoS2} supported on SiOX2\ce{SiO2}. Isotope exchange at the gas–solid interface exchanges only the surface atoms. An experiment studies the exchange between an MoSX2/SiOX2\ce{MoS2/SiO2} catalyst (mcat=1.2350m_{\text{cat}} = 1.2350 g, Mo mass fraction wMo=4.280%w_{\ce{Mo}} = 4.280\%, initially containing only 32^{32}S) and gaseous HX2X34X2234S\ce{H2^{34}S} in a flow reactor (p=1.00p = 1.00 bar, flow v=20.0v = 20.0 mL min−1^{-1}, T=23.0T = 23.0 °C, φ(HX2X34X2234S)=1.00%\varphi(\ce{H2^{34}S}) = 1.00\%, isotopic purity α=99.95\alpha = 99.95 mol%). After t=10.0t = 10.0 min, the fraction of 34^{34}S among sulfur atoms in the collected gas was γ=87.3\gamma = 87.3 mol%. (a) Calculate the amount of exchanged (surface) sulfur atoms n(S)surfacen(\mathrm{S})_{\text{surface}} in mol. (b) Assuming uniform spherical MoSX2\ce{MoS2} particles of density ρ=5.06\rho = 5.06 g cm−3^{-3}, surface areas per atom AS=3.00×10−19A_S = 3.00\times 10^{-19} m² (S) and AMo=5.00×10−19A_{\ce{Mo}} = 5.00\times 10^{-19} m² (Mo), and that only half of each MoSX2\ce{MoS2} unit is exposed at the surface, calculate the particle radius RR in nm (M(MoSX2)=160.07M(\ce{MoS2}) = 160.07, M(Mo)=95.95M(\ce{Mo}) = 95.95 g mol−1^{-1}).
Step 4 of 4: Particle radius
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Intuition

The volume-to-area ratio of a sphere is R/3; both totals share the factor N so the radius comes out without knowing N.

R=3VtotAtot=3×1.74×10−83.38=1.55×10−8 m=15.5 nmR = \dfrac{3V_{\text{tot}}}{A_{\text{tot}}} = \dfrac{3\times 1.74\times 10^{-8}}{3.38} = 1.55\times 10^{-8}\ \text{m} = 15.5\ \text{nm}
Analysis

For NN identical spheres Vtot=N43πR3V_{\text{tot}} = N\tfrac{4}{3}\pi R^3 and Atot=N4πR2A_{\text{tot}} = N 4\pi R^2, so Vtot/Atot=R/3V_{\text{tot}}/A_{\text{tot}} = R/3 and R=3Vtot/Atot=15.5R = 3V_{\text{tot}}/A_{\text{tot}} = 15.5 nm — independent of the unknown number of particles.