CHAP. 9: CHEMICAL KINETICS [CONTENTS] 291

two unknowns is solved as follows:

n= 1−

ln

[(

τ 1 / 2

)

1

/

(

τ 1 / 2

)

2

]

ln [(cA0) 1 /(cA0) 2 ]

, (9.84)

k=

2 n−^1 − 1

(n−1) (cA0)n 1 −^1

(

τ 1 / 2

)

1

. (9.85)

9.3.5 Generalized integral method

For the reaction
A→products,

we have a table of the measured valuesτi, i= 1, 2 ,... , M. The integrated form of the kinetic
equation for this reaction is [see equations (9.73)].

cA=cA0

[

1 + (n−1)cnA0−^1 kτ

] 1 − (^1) n
.
We determine the unknown constantsnandkusing the least squares method, see basic course
of mathematics. The method consists in finding the minimum of the function
S(n, k) =
∑M
i=1
(
cexpA,i−ccalcA,i
) 2

∑M
i=1
{
cexpA,i−cA0
[
1 + (n−1)cnA0−^1 kτ
] 1 −^1 n}^2

. (9.86)

The condition of the minimum is for the first derivatives to be zero. Hence we solve a system
of two equations for two unknowns
(
∂S
∂n

)

k

= 0,

(

∂S

∂k

)

n

= 0. (9.87)

The method is suitable when using a PC, but it is too laborious for “manual” calculation.