Physical Chemistry Third Edition

11.6 Competing Reactions 513

PROBLEMS

Section 11.5: A Simple Reaction Mechanism: Two

Consecutive Steps

11.32For the consecutive first-order reaction without reverse
reactions, assume thatk 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 att 10 .00 s and att 100 .0s.

11.33 a.For the consecutive first-order reactions without
reverse reactions, make a graph of [A], [B], and [F] in
the case thatk 1  0 .0100 s−^1 andk 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 ofd[A]/dt,

d[B]/dt, andd[F]/dt.

11.34a. For the consecutive first-order reactions without
reverse reactions A−→B−→F, obtain the expression

for [B] in the case thatk 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 thatk 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.

11.35For 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 thatkfis replaced byk 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.