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 1 of 4: Adsorption and desorption balance
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ka[H]=1.4×10−4 s−1;Nss=ka[H]kd=7.4×10−2k_a[\ce{H}] = 1.4 \times 10^{-4}\ \text{s}^{-1};\quad N_{\text{ss}} = \dfrac{k_a[\ce{H}]}{k_d} = 7.4 \times 10^{-2}
Analysis

The adsorption rate per grain is ka[H]=1.4×10−5×10=1.4×10−4 s−1k_a[\ce{H}] = 1.4 \times 10^{-5} \times 10 = 1.4 \times 10^{-4}\ \text{s}^{-1}. Ignoring reaction, the steady state of dN/dt=ka[H]−kdNdN/dt = k_a[\ce{H}] - k_d N gives N=7.4×10−2N = 7.4 \times 10^{-2} atoms per grain — far below one atom on average.