Problem 2

In interstellar space,

\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

k_a = 1.4 \times 10^{-5}\ \text{cm}^3\ \text{s}^{-1}

, the H-atom number density is

[\ce{H}] = 10\ \text{cm}^{-3}

, the first-order desorption constant is

k_d = 1.9 \times 10^{-3}\ \text{s}^{-1}

, and the surface-reaction constant is

k_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 = k_a[\ce{H}] - k_d N - k_r N^2

. Model B limits each grain to 0, 1 or 2 atoms with fractions

x_0, x_1, x_2

and rates: adsorption

k_a[\ce{H}]x_m

, desorption

k_d\,m\,x_m

, reaction

\tfrac{1}{2}k_r\,m(m-1)x_m

. (a) Find the steady-state

N

and the rate of

\ce{H2}

production per grain in each model. (b) Computer simulations give a true rate of

9.4 \times 10^{-6}\ \text{s}^{-1}

; which model is consistent, and which step is rate-determining?

Step 2 of 4: Model A: quadratic steady state

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k_a[\ce{H}] = k_d N + k_r N^2 \Rightarrow N = \dfrac{-k_d + \sqrt{k_d^2 + 4k_r k_a[\ce{H}]}}{2k_r} = 5.2 \times 10^{-5};\quad v_{\ce{H2}} = \tfrac{1}{2}k_r N^2 = 7.0 \times 10^{-5}\ \text{s}^{-1}

Analysis

Adding the reaction term makes $N$ the positive root of $k_r N^2 + k_d N - k_a[\ce{H}] = 0$ , giving $N = 5.2 \times 10^{-5}$ . The $\ce{H2}$ formation rate is $\tfrac{1}{2}k_r N^2 = 7.0 \times 10^{-5}\ \text{s}^{-1}$ per grain — about 7 times the simulated value.

Common pitfall. The factor 1/2 in

\tfrac{1}{2}k_rN^2

prevents double counting of the two H atoms consumed per

\ce{H2}