Problem 1

Hydrogen molecules reaching a metal surface dissociate and adsorb as H atoms,

\ce{H2(g) -> 2H(ad)}

. The adsorbed atoms are either absorbed into the bulk,

\ce{H(ad) -> H(ab)}

, or recombine and desorb. The rates per surface site are

r_1 = k_1 P_{\ce{H2}}(1-\theta)^2

(adsorption),

r_2 = k_2\theta^2

(desorption) and

r_3 = k_3\theta

(absorption), where

\theta

is the fraction of sites occupied by H atoms. Adsorption and desorption are fast compared with absorption (

r_1, r_2 \gg r_3

) and

\theta

is constant. (a) Express

r_3

as a function of

P_{\ce{H2}}

and show that it can be written

r_3 = k_3\dfrac{\sqrt{C P_{\ce{H2}}}}{1+\sqrt{C P_{\ce{H2}}}}

; express

C

through

k_1

and

k_2

. (b) A metal sample of area

S = 1.0\times 10^{-3}

^2

is placed in a 1.0 L container with

\ce{H2}

P_{\ce{H2}} = 1.0\times 10^{4}

Pa; the density of adsorption sites is

N = 1.3\times 10^{18}

^{-2}

and

T = 400

K. As absorption proceeds,

P_{\ce{H2}}

decreases at the constant rate

v = 4.0\times 10^{-4}

Pa s

^{-1}

. Calculate the amount of H atoms absorbed per unit area per unit time,

A

[mol s

^{-1}

^{-2}

]. (c) At 400 K,

C = 1.0\times 10^{2}

^{-1}

; calculate

k_3

Step 2 of 3: Hydrogen absorption flux

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A = \dfrac{2\,v\,V}{RT\,S} = \dfrac{2\times 4.0\times 10^{-4}\times 1.0\times 10^{-3}}{8.314\times 400\times 1.0\times 10^{-3}} = 2.4\times 10^{-7}\ \text{mol s}^{-1}\text{m}^{-2}

Analysis

Each $\ce{H2}$ lost from the gas phase yields 2 absorbed H atoms, so the molar absorption rate is $2\,vV/RT = 2.4\times 10^{-10}$ mol s $^{-1}$ ; dividing by the surface area gives $A = 2.4\times 10^{-7}$ mol s $^{-1}$ m $^{-2}$ .

Common pitfall. Remember the factor 2: one H2 molecule gives two H atoms; forgetting it halves A.