145 
easily calculate the velocity constant Kk. To find this we have to 
know which fraction of thé molecules dissociates per unit of time 
or of how many of the molecules one constituent leaves the sphere 
of attraction of the other one per unit of time. 
Consider a surface element do of the sphere of attraction. 
The number of atoms that passes this surface element of the 
sphere of attraction of the other atom per unit of time in outward 
direction is 
1 Are—2hm(? AS dy do dX. dY, dZ, fo e—thmV*—2ht = da dB dy do 
where v, represents the component of the relative velocity in the 
direction perpendicular to the surface. 
Let us take for the a-direction the direction of the normal, then 
we can write for the number in question 
4 Ate—2hmC? JS dy d§ dX. dY, dZ, fe eahmV*—2hb dedsdydo. (2) 
The integration limits for @, 8 and y are defined in the following 
way. 
The equations of motion of the two atoms are: 
d'r ( dE ; 
Mm zp (r etc. 
"ae : 
= v'—x 
Ss Sa (oe elc. 
mm €) 
When w(r) represents the potential energy of the two atoms, 
when their distance is 7, we have 
By subtraction we find the equation of the relative motion 
du HE Dn 
m—_=agpi{r)—  . ELC. 
dt ; r 
Multiplying these 3 equations respectively by «, 8 and y and 
taking their sum, we find the equation ot energy 
dee, En yee 
7 a 
wh 
dt Or \r dt 
so that 
im(?+ 2+ y)+2p=—C. 
As to the action of the sphere of attraction we shall use the 
image of a hard layer against the inside of which the atoms can 
impinge. As long as the velocity is small, the impulses are elastic. 
10 
Proceedings Royal Acad. Amsterdam. Vol. XXIII. 
