Page 61 - Introduction to chemical reaction engineering and kinetics

3.1 Features of a Rate Law: Introduction 43
raised to a simple power, such as 1 or 2. This power or exponent is the order of reaction
with respect to that reactant.
Thus, for a reaction represented by

jvAjA + lvsjB + (v&Z ---, products (4

the rate of disappearance of A may be found to be of the form:

(-r*) = k*c;& (3.1-2)

where (Y is the order of reaction with respect to reactant A, p is the order with respect
to B, and y is the order with respect to C. The overall order of reaction, n, is the sum of
these exponents:

n=a+/?+y (3.1-3)

and we may refer to an nth-order reaction in this sense. There is no necessary connection
between a stoichiometric coeficient such as VA in reaction (A) and the corresponding
exponent a! in the rate law.
The proportionality “constant” kA in equation 3.1-2 is called the “rate constant,” but
it actually includes the effects of all the parameters in equation 3.1-1 other than con-
centration. Thus, its value usually depends on temperature, and we consider this in the
next section.
For reaction (A), the rate may be written in terms of ( -rg) or ( -rc) instead of ( -rA).
These rates are related to each other through the stoichiometry, as described in Section
1.4.4. Corresponding rate constants kB or k, may be introduced instead of kA, and these
rate constants are similarly related through the stoichiometry. Such changes do not alter
the form of equation 3.1-2 or values of (Y, p, and y; it is a matter of convenience which
species is chosen. In any case, it should clearly be specified. Establishing the form of
equation 3.1-2, including the values of the various parameters, is a matter for experi-
ment.

Repeat problem l-2(a) in light of the above discussion.

SOLUTION

The reaction in problem l-2(a) is represented by A + 3B + products. The rate law in terms
of A iS (-t-A) = kAcAcB, and in terms of B is ( -rB) = kBcAcB. We wish to determine
the value of kB given the value of kA. From equation 1.4-8,

(-rA)/(-1) = (-rB)/(-3)) or(-rg) = 3(-rA)

Thus,

kBCACB = 3k,cAc,
and

kB = 3kA = 3(1.34) = 4.02 Lmol-’ h-’