Chemistry Labs

Problem 1

Hydrogen molecules reaching a metal surface dissociate and adsorb as H atoms, HX2(g)→2 H(ad)\ce{H2(g) -> 2H(ad)}. The adsorbed atoms are either absorbed into the bulk, H(ad)→H(ab)\ce{H(ad) -> H(ab)}, or recombine and desorb. The rates per surface site are r1=k1PHX2(1−θ)2r_1 = k_1 P_{\ce{H2}}(1-\theta)^2 (adsorption), r2=k2θ2r_2 = k_2\theta^2 (desorption) and r3=k3θr_3 = k_3\theta (absorption), where θ\theta is the fraction of sites occupied by H atoms. Adsorption and desorption are fast compared with absorption (r1,r2≫r3r_1, r_2 \gg r_3) and θ\theta is constant. (a) Express r3r_3 as a function of PHX2P_{\ce{H2}} and show that it can be written r3=k3CPHX21+CPHX2r_3 = k_3\dfrac{\sqrt{C P_{\ce{H2}}}}{1+\sqrt{C P_{\ce{H2}}}}; express CC through k1k_1 and k2k_2. (b) A metal sample of area S=1.0×10−3S = 1.0\times 10^{-3} m2^2 is placed in a 1.0 L container with HX2\ce{H2} at PHX2=1.0×104P_{\ce{H2}} = 1.0\times 10^{4} Pa; the density of adsorption sites is N=1.3×1018N = 1.3\times 10^{18} m−2^{-2} and T=400T = 400 K. As absorption proceeds, PHX2P_{\ce{H2}} decreases at the constant rate v=4.0×10−4v = 4.0\times 10^{-4} Pa s−1^{-1}. Calculate the amount of H atoms absorbed per unit area per unit time, AA [mol s−1^{-1} m−2^{-2}]. (c) At 400 K, C=1.0×102C = 1.0\times 10^{2} Pa−1^{-1}; calculate k3k_3.
Step 3 of 3: Rate constant k3
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r3=A NAN=2.4×10−7×6.02×10231.3×1018=1.1×10−1 s−1,k3=r3/θ=1.1×10−1 s−1r_3 = \dfrac{A\,N_A}{N} = \dfrac{2.4\times 10^{-7}\times 6.02\times 10^{23}}{1.3\times 10^{18}} = 1.1\times 10^{-1}\ \text{s}^{-1},\quad k_3 = r_3/\theta = 1.1\times 10^{-1}\ \text{s}^{-1}
Analysis

Per site, the absorption rate is r3=ANA/N=0.11r_3 = A N_A/N = 0.11 s−1^{-1}. Here CPHX2=102×104=106≫1CP_{\ce{H2}} = 10^{2}\times 10^{4} = 10^{6} \gg 1, so θ≈1\theta \approx 1 and k3=r3/θ≈r3=1.1×10−1k_3 = r_3/\theta \approx r_3 = 1.1\times 10^{-1} s−1^{-1}.