16
Chapter
Kinetics: Rates and Mechanisms of
Chemical Reactions
16.1
Changes in concentrations of reactants (or products) as a function of time are measured to determine
the reaction rate.
16.2
An increase in pressure increases the reactant concentrations and would result in an increased reaction
rate.
16.3
Additional solvent would decrease the reactant concentrations and would result in a decreased reaction
rate.
16.4
An increase in solid surface area would allow more gaseous components to react per unit time and thus
would increase the reaction rate.
16.5
An increase in temperature increases both the number and the energy of the collisions and thus would
increase the reaction rate.
16.6
The second experiment proceeds at the higher rate. I
2
in the gaseous state would experience more
collisions with gaseous H
2
.
16.7
The reaction rate is the change in the concentration of reactants or products per unit time. Reaction
rates change with time because reactant concentrations decrease, while product concentrations increase
with time.
16.8
a)
The slope of the line joining any two points on the graph of concentration vs. time provides the
average rate over that time period. The shorter this time period, the closer the average rate will
equal the instantaneous rate, the rate at a particular instant during the reaction.
b)
The initial rate is the instantaneous rate at the moment the reactants are mixed (at t
=
0).
16.9
The calculation of the overall rate is the difference between the forward and reverse rates. This
complication may be avoided by measuring the initial rate, where product concentrations are
negligible, so the reverse rate is negligible. Additionally, the calculations are simplified as the reactant
concentrations can easily be determined from the volumes and concentrations of the solutions mixed.
16.10
16.11
a)
Calculate the slope of the line connecting (0, [C]
o
) and (t
f
, [C]
f
) (final time and concentration). The
negative of this slope is the average rate.
b)
Calculate the negative of the slope of the line tangent to the curve at t
=
x.
c)
Calculate the negative of the slope of the line tangent to the curve at t
=
0.
d)
If you plotted [D]
vs.
time, you would not need to take the negative of the slopes in (a)–(c).
31
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a)
Rate
=

2
1 [AX ]
2
t
∆
∆
=

(
29
(
29
(
29
0.0088 mol/L
0.0500 mol/L
2 20.0 0 s


=
1.0
x
10

3
mol/L · s
b)
The initial rate is determined by the slope of the line tangent to the curve at the point t
=
0. The
initial rate is higher than the rate in part (a).
Initial rate
≈
4.0
x
10
6 3
mol/L · s
16.13
a)
Rate
=

2
[AX ]
∆
∆
t
=

(0.0088 mol/ L)
(0.0249 mol/ L)
(20.0 8.0) s


=
1.3
x
10

3
mol/L · s
b)
The rate at exactly 5.0 s will be higher than the rate in part (a).
Rate
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 Spring '07
 Kobiashi
 Kinetics, mol

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