Page 68 - Modern physical chemistry

3.15 The Collision Rate Density 51

In the last step, the result from example 10.5 has been introduced. Since the integrand
is even, we also have
f oo X e --- ) 112
2 -ux2 dx _ 1
[ 1r
-00 2a a
This fonnula was used in line (3.89).

Example 3. 11
Evaluate the integral

Again, integrate by parts:
00
00 3 _ax2 1 00 2 -ux2 1 2 _ax2 1 00_ux2 (-2ax dx)
1 x e dx = -- r x e (-2ax dx) = --x e 1 - - r e
o 2aJo 2a o 2a2Jo

I=O __ 1_e-ax21°O =_1_.
2a 2 0 2a 2

Example 3.12
Calculate the most probable speed of a molecule in an ideal gas.
The most probable speed is the speed at which the probability density j(v) is a
maximum. But the derivative of expression (3.94),

<if = 41r(~)3/2[2v_.E!:.v3)e-mv2/2kT

dv 2trkT kT '
vanishes when
2 2kT
v =--
m
and
112
[ 2kT )
v= --
m

3. 15 The Collision Rate Density
For two molecules to react, they must get together with the proper orientations and
with sufficient energy. From the theory developed so far, we can estimate the collision
rate per unit volume in a gas. The other factors will be considered later.
In Newtonian mechanics, the relative movement of two particles, of mass m A and
mass m B , respectively, is modeled by a single particle of mass

__ m....!.AO-m-=B_
J.l= [3.95]
mA+mB