Constante de Taxa de Dessorção

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654 19 PROCESSES AT SOLID SURFACES
E19B.7(b) �e rate constant for desorption is assumed to follow an Arrhenius law, kdes =
Ae−Ea,des/RT . Recall that for a �rst order process the half life is simply propor-
tional to the inverse of the rate constant, therefore the time needed for a certain
amount to desorb is also inversely proportional to the rate constant.�us
τ1/τ2 = e−(Ea,des/R)(1/T2−1/T1)
hence ln(τ1/τ2) = −
Ea,des
R
( 1
T2
− 1
T1
)
hence Ea,des =
−R ln(τ1/τ2)
(1/T2 − 1/T1)
With the data given
Ea,des =
−(8.3145 JK−1mol−1) ln[(1856 s)/(8.44 s)]
1/(1012 K) − 1/(873 K)
= 2.85... × 105 Jmol−1 = 285 kJmol−1
�e times for desorption at di�erent temperatures are computed using
τ1/τ2 = e−(Ea,des/R)(1/T2−1/T1) hence τ2 = τ1e(Ea,des/R)(1/T2−1/T1)
�e time needed at 298 K is related to that at 873 K
τ2 = (1856 s) e[(2.85...×10
5 J mol−1)/(8.3145 J K−1 mol−1)][1/(298 K)−1/(873 K)]
= 1.5 × 1036 s
E�ectively, the gas does not desorb at this temperature. Repeating the calcula-
tion at 1500 K
τ2 = (1856 s) e[(2.85...×10
5 J mol−1)/(8.3145 J K−1 mol−1)][1/(1500 K)−1/(873 K)]
= 0.14 ms
E19B.8(b) �e average time that a species remains adsorbed is proportional to its half-life,
given by [19B.14–839], t1/2 = τ0eEa,des/RT .�erefore, if the two times are τ1 and
τ2 at temperatures T1 and T2
τ2/τ1 = e(Ea,des/R)(1/T2−1/T1)
hence ln(τ2/τ1) =
Ea,des
R
( 1
T2
− 1
T1
)
hence Ea,des =
R ln(τ2/τ1)
1/T2 − 1/T1
�e data gives the lifetime at the higher temperature as τ2 = τ1(1 − 0.35) =
0.65 τ1
Ea,des =
(8.3145 JK−1mol−1) ln[(0.65τ1)/(τ1)]
1/(1000 K) − 1/(600 K)
= 5.4 kJmol−1