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11 The Rates of Chemical Reactions
When the expressions for G and H are substituted into Eq. (11.3-10) and the resulting
expression is substituted into Eq. (11.3-9), a definite integration gives
1
a[B] 0 − b[A] 0
ln
[B] t [A] 0
[A] t [B] 0
k f t
(11.3-11)
This integrated rate law can now be compared with experimental data.
Exercise 11.14
Verify Eq. (11.3-11) by carrying out the integration.
E X A M P L E 11.9
Assume that for the reaction
NO + O 3 − → NO 2 + O 2
[NO] 0 1.00 × 10 −6 mol L −1 and [O 3 ] 0 5.00 × 10 −7 mol L −1 . Find the concentration
of O 3 after 3.50 s.
Solution
Let O 3 be denoted by A and NO be denoted by B. Use of Eq. (11.3-11) with a b 1 gives,
with some manipulation:
[B] 0 − x
[A] 0 − x
[B] 0
[A] 0
exp[kt([B] 0 − [A] 0 )]
Using the values given, solution of the equation gives x 4.73 × 10 −7 and [O 3 ] 0.27 ×
10 −7 mol L −1 .
Exercise 11.15
Carry out the mathematical steps to verify the solution of the preceding example.
The Method of Initial Rates
Consider the hypothetical reaction
aA + bB + f F − → products
(11.3-12)
We assume the reaction has definite orders so that the initial rate is
r initial k f [A]
a
0 [B]
β
0 [F]
φ
0 ≈ −
1
a
∆[A]
∆t
(11.3-13)
Several experiments are carried out with the same values of [B] 0 and [F] 0 but
with different values of [A] 0 . The initial rate is determined for each experiment. We
write
ln(r initial ) ln(k f [B]
β
0 [F]
φ
0 ) + αln([A] 0 )
(11.3-14)