Chemistry Labs

Problem 2

In interstellar space, HX2\ce{H2} forms on the surface of dust grains because two colliding H atoms cannot dissipate the bond energy. The rate constant for adsorption is ka=1.4×10−5 cm3 s−1k_a = 1.4 \times 10^{-5}\ \text{cm}^3\ \text{s}^{-1}, the H-atom number density is [H]=10 cm−3[\ce{H}] = 10\ \text{cm}^{-3}, the first-order desorption constant is kd=1.9×10−3 s−1k_d = 1.9 \times 10^{-3}\ \text{s}^{-1}, and the surface-reaction constant is kr=5.1×104 s−1k_r = 5.1 \times 10^{4}\ \text{s}^{-1} (surface atom numbers may be used like concentrations). Model A treats reaction as second order: dN/dt=ka[H]−kdN−krN2dN/dt = k_a[\ce{H}] - k_d N - k_r N^2. Model B limits each grain to 0, 1 or 2 atoms with fractions x0,x1,x2x_0, x_1, x_2 and rates: adsorption ka[H]xmk_a[\ce{H}]x_m, desorption kd m xmk_d\,m\,x_m, reaction 12kr m(m−1)xm\tfrac{1}{2}k_r\,m(m-1)x_m. (a) Find the steady-state NN and the rate of HX2\ce{H2} production per grain in each model. (b) Computer simulations give a true rate of 9.4×10−6 s−19.4 \times 10^{-6}\ \text{s}^{-1}; which model is consistent, and which step is rate-determining?
Step 4 of 4: Compare with simulation
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Intuition

With kr≫kdk_r \gg k_d, reaction is nearly instant once two atoms meet, so the slow step is waiting for a second H to adsorb.

vHX2=kr x2=5.1×104×1.8×10−10=9.0×10−6 s−1≈9.4×10−6 s−1v_{\ce{H2}} = k_r\,x_2 = 5.1 \times 10^{4} \times 1.8 \times 10^{-10} = 9.0 \times 10^{-6}\ \text{s}^{-1} \approx 9.4 \times 10^{-6}\ \text{s}^{-1}
Analysis

In model B only state 2 can react, so v=12kr⋅2⋅1⋅x2=krx2=9.0×10−6 s−1v = \tfrac{1}{2}k_r \cdot 2 \cdot 1 \cdot x_2 = k_r x_2 = 9.0 \times 10^{-6}\ \text{s}^{-1}, matching the simulated 9.4×10−6 s−19.4 \times 10^{-6}\ \text{s}^{-1}. Model A overestimates because it assumes reaction regardless of occupancy; the rate is governed by adsorption.