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11 The Rates of Chemical Reactions
We have
d[F]
dt
k 1 [A] k 1 [A] 0 e
−(k 1 +k 2 )t
(11.6-5)
The right-hand side of this equation does not contain [F], so we can multiply by dt and
integrate to obtain
[F] t − [F] 0 [F] t
−k 1 [A] 0
k 1 + k 2
(e
−(k 1 +k 2 )t − 1)
(11.6-6)
where we assume that [F] 0 0. A similar treatment for [G] gives the same result except
that k 1 is replaced by k 2 in the numerator:
[G] t − [G] 0 [G] t
−k 2 [A] 0
k 1 + k 2
(e
−(k 1 +k 2 )t − 1)
(11.6-7)
where we also assume that [G] 0 0. The ratio [F]/[G] has the same value at any time:
[F]
[G]
k 1
k 2
(reverse reactions negligible)
(11.6-8)
Exercise 11.22
For the reactions shown in Eq. (11.6-1) assume that [A] 0 0.500 mol L −1 , that k 1 0.100 s −1
and that k 2 0.0100 s −1 . Construct a graph showing [A], [F], and [G] for t ranging from
0 to 20 s.
If the reverse reactions cannot be neglected, the situation can be different. If the
system comes to equilibrium,
[F] eq
[A] eq
k 1
k
1
K 1
(11.6-9a)
[G] eq
[A] eq
k 2
k
2
K 2
(11.6-9b)
so that
[F] eq
[G] eq
k 1
k
1
k
2
k 2
k 1
k 2
k
2
k
1
K 1
K 2
(at equilibrium)
(11.6-10)
Depending on the values of the four rate constants, this ratio might differ significantly
from the ratio in Eq. (11.6-8). If F is a desired product and G is an undesired product,
the ratio of [F] to [G] might be optimized by allowing the system to come to equilibrium (using thermodynamic control) or by stopping the reaction before it comes to
equilibrium (using kinetic control).
11 The Rates of Chemical Reactions
We have
d[F]
dt
k 1 [A] k 1 [A] 0 e
−(k 1 +k 2 )t
(11.6-5)
The right-hand side of this equation does not contain [F], so we can multiply by dt and
integrate to obtain
[F] t − [F] 0 [F] t
−k 1 [A] 0
k 1 + k 2
(e
−(k 1 +k 2 )t − 1)
(11.6-6)
where we assume that [F] 0 0. A similar treatment for [G] gives the same result except
that k 1 is replaced by k 2 in the numerator:
[G] t − [G] 0 [G] t
−k 2 [A] 0
k 1 + k 2
(e
−(k 1 +k 2 )t − 1)
(11.6-7)
where we also assume that [G] 0 0. The ratio [F]/[G] has the same value at any time:
[F]
[G]
k 1
k 2
(reverse reactions negligible)
(11.6-8)
Exercise 11.22
For the reactions shown in Eq. (11.6-1) assume that [A] 0 0.500 mol L −1 , that k 1 0.100 s −1
and that k 2 0.0100 s −1 . Construct a graph showing [A], [F], and [G] for t ranging from
0 to 20 s.
If the reverse reactions cannot be neglected, the situation can be different. If the
system comes to equilibrium,
[F] eq
[A] eq
k 1
k
1
K 1
(11.6-9a)
[G] eq
[A] eq
k 2
k
2
K 2
(11.6-9b)
so that
[F] eq
[G] eq
k 1
k
1
k
2
k 2
k 1
k 2
k
2
k
1
K 1
K 2
(at equilibrium)
(11.6-10)
Depending on the values of the four rate constants, this ratio might differ significantly
from the ratio in Eq. (11.6-8). If F is a desired product and G is an undesired product,
the ratio of [F] to [G] might be optimized by allowing the system to come to equilibrium (using thermodynamic control) or by stopping the reaction before it comes to
equilibrium (using kinetic control).