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 2 of 4: Amount of exchanged sulfur
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n(S)surface=p ΔVRT=105×0.25×10−68.314×296.15=1.0×10−5 moln(\mathrm{S})_{\text{surface}} = \dfrac{p\,\Delta V}{RT} = \dfrac{10^{5}\times 0.25\times 10^{-6}}{8.314\times 296.15} = 1.0\times 10^{-5}\ \text{mol}
Analysis

Ideal gas law on the 32^{32}S volume difference: n=pΔV/RT=105×0.25×10−6/(8.314×296.15)=1.02×10−5n = p\Delta V/RT = 10^{5}\times 0.25\times 10^{-6}/(8.314\times 296.15) = 1.02\times 10^{-5} mol.