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 3 of 4: Model B: discrete steady state
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x2x1=ka[H]2kd+kr=2.7×10−9;x1x0=ka[H]kd+kr x2/x1≈6.9×10−2;x0=0.94, x1=0.064, x2=1.8×10−10\dfrac{x_2}{x_1} = \dfrac{k_a[\ce{H}]}{2k_d + k_r} = 2.7 \times 10^{-9};\quad \dfrac{x_1}{x_0} = \dfrac{k_a[\ce{H}]}{k_d + k_r\,x_2/x_1} \approx 6.9 \times 10^{-2};\quad x_0 = 0.94,\ x_1 = 0.064,\ x_2 = 1.8 \times 10^{-10}
Analysis

Setting dx1/dt=0dx_1/dt = 0 and dx2/dt=0dx_2/dt = 0 gives the ratios x2/x1=ka[H]/(2kd+kr)=2.7×10−9x_2/x_1 = k_a[\ce{H}]/(2k_d+k_r) = 2.7 \times 10^{-9} and x1/x0≈ka[H]/kd=6.9×10−2x_1/x_0 \approx k_a[\ce{H}]/k_d = 6.9 \times 10^{-2}; normalising x0+x1+x2=1x_0+x_1+x_2=1 yields x0=0.94x_0=0.94, x1=0.064x_1=0.064, x2=1.8×10−10x_2=1.8\times10^{-10}.