On the Integrated Velocity Equations of Chemical Reactions. 227 
It may be worth while pointing out some special simplifications of (18) 
that occur in practice. 
a — \ the two summations fall away, leaving 
/.a^S = (-1)^-1 ^i^,;^^-i(.0 . . . (21) 
If a = 2 the integral may be written 
7,0^+1$ = [^(1) - ^{n) - (l-.O^'X/O - ... - 
^^jrrr-^^-K'O] .... (22) 
If = 1 we get 
Jca"^ = (-1)"+' ^— ^j^i [-^('O - ^(1) - («- l)f (1) - ... - 
■ • • • (23) 
§ 4. The Determination of the Molecular Order. 
The results already given seem sufficient for the determination of 
molecular order with a minimum of experimental labour. Suppose, for 
instance, three reactants, A, B and C, are involved. The total number, N, 
of molecules, taking part may be determined in the usual way by performing- 
two different experiments with all reactants in the same initial concentration, 
say «j and a^. Then from (14), 
The same fraction, X, must be transformed in each case. These determine 
N and Jc. Now repeat the experiment with two concentrations, say those of 
B and C, initially equal, and the other lower, the ratio being n. Suppose N 
has been found to be 5. Of A, either one, or two, or three molecules must 
be involved. The appropriate solutions would be 
ka'^^ r= - , ■4^"'(n) (25) 
ka^^ = j^y, [^(1) - a-nyV(n) - ^Aj-^' V(n)^ (26) 
and 
Jca^^ = -^^^^^[2{Un)-M^(l)} - (/.-l){^'(l) + -fW}] . (27) 
Knowing Ic, a, S and n, a little arithmetic will show which of these equations 
is satisfied, and thus determine the molecular order of A. If purely chemical 
considerations do not fix the orders of B and C, another reaction would have 
to be timed, using A and C in initially equal concentrations, and B lower. 
The same process would then give the molecular order of B. 
