Physical Chemistry 3: — Chemical Kinetics — - Christian-Albrechts ...

Physical Chemistry 3: — Chemical Kinetics — - Christian-Albrechts ... Physical Chemistry 3: — Chemical Kinetics — - Christian-Albrechts ...

21.10.2014 Views

5.4 Advanced collision theory 127 (4) Arrhenius activation energy: y = 2 ln ln ∝ 1 µ 2 ln +ln 1+ 0 − 0 ln = 1 µ 1 2 + 1 − µ 0 0 + 1+ 0 2 2 = 0 + 1 2 − 0 1+ 0 (5.172) (5.173) (5.174) (5.175) b) Line-of-Centers model (Figs. 5.21 - 5.22) Even for hard spheres, because of angular momentum conservation (see below), not the entire collision energy, but only the fraction “directed” along the line of centers () is effective. In particular, we have • for a collision with =0: = (5.176) y central collisions are most effective, • for a collision with = : =0 (5.177) y glancing collisions are ineffective because =0, • general expression (see below): = µ 1 − 2 2 (5.178) y decreases with increasing .

5.4 Advanced collision theory 128 I Figure 5.21: Collision energy along line-of-centers Functional expression for ( ): (1) We assume that a reaction can occur only if ≥ 0 . Thus, we first determine (see Fig. 5.21): y = cos (5.179) 2 = 2 cos 2 (5.180) = ¡ 2 1 − sin 2 ¢ (5.181) µ = 2 1 − 2 (5.182) 2 = µ 1 − 2 2 (5.183) (2) We see from this expression that decreases with increasing . Thus, for a given ,wehaveamaximalvalueof (i.e., a value max ), for which ≥ 0 . We determine max from the condition that µ = 1 − 2 ≥ 2 0 (5.184) y y 2 ≤ 2 µ 1 − 0 = 2 max (5.185) ( )= 2 max = 2 µ 1 − 0 (5.186)

5.4 Advanced collision theory 128<br />

I<br />

Figure 5.21: Collision energy along line-of-centers<br />

Functional expression for ( ):<br />

(1) We assume that a reaction can occur only if ≥ 0 . Thus, we first determine<br />

(see Fig. 5.21):<br />

y<br />

= cos (5.179)<br />

2 = 2 cos 2 (5.180)<br />

= ¡ 2 1 − sin 2 ¢ (5.181)<br />

µ <br />

= 2 1 − 2<br />

(5.182)<br />

2<br />

= <br />

µ<br />

1 − 2<br />

2 <br />

(5.183)<br />

(2) We see from this expression that decreases with increasing . Thus, for a<br />

given ,wehaveamaximalvalueof (i.e., a value max ), for which ≥ 0 .<br />

We determine max from the condition that<br />

µ <br />

= 1 − 2<br />

≥ <br />

2 0 (5.184)<br />

y<br />

y<br />

2 ≤ 2 µ<br />

1 − 0<br />

<br />

<br />

= 2 max (5.185)<br />

( )= 2 max = 2 µ<br />

1 − 0<br />

<br />

<br />

(5.186)

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