11.3 Forward Reactions with More Than One Reactant
501
This is the same as
1
[A] t
−
1
[A] 0
k f bt
(11.3-8)
where we drop the prime symbol ( ) on t. Equation (11.3-8) is the same as Eq. (11.2-11)
except for the appearance of the factor b in the right-hand side.
E X A M P L E 11.8
Some of the damage to the ozone layer of the upper atmosphere might involve the reaction
NO + O 3 − → NO 2 + O 2
The reaction is first order in each reactant and the rate constant is equal to 1.3 × 10 6 L
mol −1 s −1 at 298 K. If the initial concentrations of NO and O 3 are both equal to 1.00 ×
10 −6 mol L −1 , find the concentrations of NO and O 3 at time t 2.00 s.
Solution
1
[NO]
1
NO 0
+ k f t
1
1.00 × 10 −6 mol L −1 + (1.3 × 10 6 L mol −1 s −1 )(2.00 s)
3.6 × 10 6 L mol −1
[NO] [O 3 ]
1
3.6 × 10 6 L mol −1 2.8 × 10 −7 mol L −1
Exercise 11.12
a. Find the half-life of the reaction in the previous example with the given initial concentrations.
b. Find the half-life of the reaction in the previous example with initial concentrations both
equal to 5.00 × 10 −5 mol L −1 .
In the case that the reactants are not mixed in the stoichiometric ratio we separate
the variables in Eq. (11.3-6a):
1
([A] 0 − ax)([B] 0 − bx)
dx k f dt
(11.3-9)
This equation can be integrated by the method of partial fractions. If we write
1
([A] 0 − ax)([B] 0 − bx)
G
[A] 0 − ax
+
H
[B] 0 − bx
(11.3-10)
then G and H are guaranteed by a theorem of algebra to be constants. These constants
are found to be
G
1
[B] 0 − b[A] 0 /a
and H
1
[A] 0 − a[B] 0 /b
Exercise 11.13
Verify the expressions for G and H.