11.6 Competing Reactions
513
P R O B L E M S
Section 11.5: A Simple Reaction Mechanism: Two
Consecutive Steps
11.32 For the consecutive first-order reaction without reverse
reactions, assume that k 1 0.01000 s −1 and
k 2 0.1000 s −1 . If [A] 0 0.100 mo L −1 , find the value
of each concentration at t 10.00 s and at t 100.0 s.
11.33 a. For the consecutive first-order reactions without
reverse reactions, make a graph of [A], [B], and [F] in
the case that k 1 0.0100 s −1 and k 2 0.100 s −1 .
Assume that [A] 0 1.00 mol L −1 and that B and F
are initially absent.
b. For the values of part a, make a graph of d[A]/dt,
d[B]/dt, and d [F]/dt.
11.34 a. For the consecutive first-order reactions without
reverse reactions A − → B − → F, obtain the expression
for [B] in the case that k 1 k 2 , assuming that no B or
F is present at the beginning of the reaction. Proceed
by taking the limit of the expression of Eq. (11.5-6) as
k 1 − → k 2 , using l’Hôpital’s rule. 6
b. Construct a graph showing [A], [B], and [F] as
functions of time for the case that
k 1 k 2 0.100 s −1 and [A] 0 1.00 mol L −1 .
c. Draw a graph showing [A], [B], and [F] as functions of
time for the case that k 1 0.099 s −1 and
k 2 0.101 s −1 and [A] 0 1.00 mol L −1 . Compare it
with your graph from part b.
11.35 For the case of two consecutive reactions in which the
reverse reaction of the first reaction cannot be neglected,
sketch a rough graph showing the three concentrations as
functions of time. Compare your graph with those in
Figure 11.6.
11.6
Competing Reactions
In many syntheses a side reaction consumes part of the reactants but gives undesired
products. We consider the simplest case: that two competing reactions are first order
with negligible reverse reaction.
A
k 1
− → F
(11.6-1a)
A
k 2
− → G
(11.6-1b)
The rates of the two reactions combine to give
−
d[A]
dt
(k 1 + k 2 )[A]
(11.6-2)
This equation is the same as Eq. (11.2-2) except that k f is replaced by k 1 + k 2 , and its
solution is
[A] t [A] 0 e
−(k 1 +k 2 )t
(11.6-3)
The half-life for the disappearance of A is
t 1/2
ln(2)
k 1 + k 2
(11.6-4)
6 See Robert G. Mortimer, Mathematics for Physical Chemistry, 3rd ed., Elsevier/Academic Press, San Diego, CA, 2005, p. 113ff, or any calculus textbook.
513
P R O B L E M S
Section 11.5: A Simple Reaction Mechanism: Two
Consecutive Steps
11.32 For the consecutive first-order reaction without reverse
reactions, assume that k 1 0.01000 s −1 and
k 2 0.1000 s −1 . If [A] 0 0.100 mo L −1 , find the value
of each concentration at t 10.00 s and at t 100.0 s.
11.33 a. For the consecutive first-order reactions without
reverse reactions, make a graph of [A], [B], and [F] in
the case that k 1 0.0100 s −1 and k 2 0.100 s −1 .
Assume that [A] 0 1.00 mol L −1 and that B and F
are initially absent.
b. For the values of part a, make a graph of d[A]/dt,
d[B]/dt, and d [F]/dt.
11.34 a. For the consecutive first-order reactions without
reverse reactions A − → B − → F, obtain the expression
for [B] in the case that k 1 k 2 , assuming that no B or
F is present at the beginning of the reaction. Proceed
by taking the limit of the expression of Eq. (11.5-6) as
k 1 − → k 2 , using l’Hôpital’s rule. 6
b. Construct a graph showing [A], [B], and [F] as
functions of time for the case that
k 1 k 2 0.100 s −1 and [A] 0 1.00 mol L −1 .
c. Draw a graph showing [A], [B], and [F] as functions of
time for the case that k 1 0.099 s −1 and
k 2 0.101 s −1 and [A] 0 1.00 mol L −1 . Compare it
with your graph from part b.
11.35 For the case of two consecutive reactions in which the
reverse reaction of the first reaction cannot be neglected,
sketch a rough graph showing the three concentrations as
functions of time. Compare your graph with those in
Figure 11.6.
11.6
Competing Reactions
In many syntheses a side reaction consumes part of the reactants but gives undesired
products. We consider the simplest case: that two competing reactions are first order
with negligible reverse reaction.
A
k 1
− → F
(11.6-1a)
A
k 2
− → G
(11.6-1b)
The rates of the two reactions combine to give
−
d[A]
dt
(k 1 + k 2 )[A]
(11.6-2)
This equation is the same as Eq. (11.2-2) except that k f is replaced by k 1 + k 2 , and its
solution is
[A] t [A] 0 e
−(k 1 +k 2 )t
(11.6-3)
The half-life for the disappearance of A is
t 1/2
ln(2)
k 1 + k 2
(11.6-4)
6 See Robert G. Mortimer, Mathematics for Physical Chemistry, 3rd ed., Elsevier/Academic Press, San Diego, CA, 2005, p. 113ff, or any calculus textbook.