Download Ch. 15 Study Guide

Chemistry 201: General Chemistry II - Lecture
Dr. Namphol Sinkaset
Chapter 15 Study Guide
Concepts
1. rate of reaction: the speed at which reactants disappear and products form.
2. There can only be one numerical rate for a reaction. The rate of formation of any
product is related to the rate of consumption of any reactant by the stoichiometry of
the equation.
3. Reaction rates depend on concentration of reactants.
4. A rate law describes the mathematical relationship between the concentration of
reactants and how fast the reaction occurs.
5. For a general reaction A + B −→ products, the rate law is expressed as Rate =
k[A]m [B]n .
6. A complete rate law for the reaction above has k, m, and n specified.
7. The units of k will depend on the individual rate law; k is called the rate constant.
8. The exponents m and n are known as reaction orders with respect to that particular
reactant.
9. The exponents m and n can only be found by experiment and are not related
to the stoichiometry of the reaction.
10. The overall order of a reaction is the sum of all the individual orders.
11. Although orders are ordinarily small whole numbers, fractional numbers are possible.
Additionally, a negative order means that a particular reactant retards the reaction.
12. Zero, first, and second order are terms used to describe when the exponents in a rate
law are equal to 0, 1, and 2, respectively.
13. If a reaction is zero order with respect to a reactant, that means the rate of reaction
is independent of the concentration of that reactant.
14. If a reaction is first order with respect to a reactant, that means the rate of reaction is
directly proportional to the concentration of that reactant.
15. If a reaction is second order with respect to a reactant, that means the rate of reaction
is proportional to the square of the concentration of that reactant.
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16. An integrated rate law is a relationship between the concentrations of reactants and
time.
17. The order of reaction with respect to a reactant can be found by graphical analysis,
i.e. plotting [A] vs. t, plotting ln [A] vs. t, and plotting 1/[A] vs. t in order to find
which relationship fits the data best.
18. Knowing the order of a reaction and its integrated rate law enable the prediction of
when the reaction will end.
19. half-life, t1/2 : time required for the concentration of a reactant to decrease to half its
initial value.
20. First-order reaction half-lives do not depend on the initial concentration.
21. Second-order and zero-order reaction half-lives do depend on initial concentration.
22. Generally, reaction rates are very sensitive to temperature. This temperature dependence is wrapped up in the rate constant k.
23. The Arrhenius equation describes what factors go into k. Its important aspects are
the activation energy, frequency factor, and exponential factor.
24. To get to the product state, reactants must go through a high-energy activated complex or transition state. To get there, reactants must overcome the activation
energy.
25. The frequency factor represents the number of approaches to the activation barrier
per unit time. Not all approaches result in a successful reaction.
26. The exponential factor is a number between 0 and 1 that represents the fraction of
molecules that successfully react upon approach to the activation barrier.
27. The influence of temperature on a reaction is seen in the exponential factor. Higher
temperatures result in more successful approaches because of the higher energy molecules
have when heated.
28. The activation energy and frequency factor can be found using an Arrhenius plot or
the 2-point form of the Arrhenius equation.
29. For reactions involving more than one reactant, the collision model breaks up the
frequency factor into an orientation factor and a collision frequency. In addition
to having enough energy to react, the collision must occur with the proper orientation.
30. Chemical equations tell us the net change, but they don’t tell us anything about how
the reaction happens.
31. Most reactions occur in a series of small steps that leads to the final products. The
series of steps is known as the mechanism of the reaction.
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32. A simple reaction that is part of a mechanism is called an elementary process.
The rate law for an elementary process can be written based on the coefficients of its
balanced equation.
33. An intermediate is a short-lived species that exists only during a reaction. It is
formed in one step and consumed in another.
34. molecularity: the number of reactant particles involved in a step.
35. The slow step in a mechanism determines its rate. It is called the rate-determining
step.
36. Valid mechanisms must satisfy two criteria: (1) elementary steps must add up to overall
reaction; (2) rate law predicted by mechanism must be consistent with experimentallydetermined rate law.
37. A catalyst is a substance that increases the rate of a reaction without getting used
up. Catalysts provide a lower-energy mechanism for the reaction.
38. There are homogeneous catalysts and heterogeneous catalysts. Homogeneous
catalysts are in the same phase as the reactants. Heterogeneous catalysts are in a
different phase than the reactants.
39. An enzyme is a biological catalyst. An enzyme has an active site into which a
specific substrate fits in a lock and key type of arrangement.
Equations
1. Rate =
∆[A]
∆t
(Rate of change of product A in a reaction)
2. Rate = k[A]m [B]n (Generic rate law for a reaction A + B)
3. ln[A]t − ln[A]0 = −kt (First order integrated rate law)
4. ln[A]t = −kt + ln[A]0 (Alternative first order integrated rate law)
5.
1
[A]t
= kt +
1
[A]0
(Second order integrated rate law)
6. [A]t = −kt + [A]0 (Zero order integrated rate law)
7. t1/2 =
ln 2
k
8. t1/2 =
1
k[A]0
9. t1/2 =
[A]0
2k
(First order reaction half life)
(Second order reaction half life)
(Zero order reaction half life)
10. k = Ae−Ea /RT (Arrhenius equation)
11. ln k =
12. ln kk12 =
−Ea
R
1
T
Ea 1
(
R T1
+ ln A (Arrhenius equation, linear form)
−
1
)
T2
(2-point form Arrhenius equation)
13. k = pze−Ea /RT (Arrhenius equation, collision model)
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Representative Problems
69. The formation of small amounts of nitric oxide, NO, in automobile engines
is the first step in the formation of smog. As noted earlier, nitric oxide is readily
oxidized to nitrogen dioxide by the reaction 2NO(g) + O2(g) −→ 2NO2(g) . The
following data were collected in a study of the rate of this reaction. What is the
rate law for the reaction? What is the rate constant with its correct units?
First, we look at the table of data they gave to us.
[O2 ]0 (mole/L)
0.0010
0.0040
0.0040
[NO]0 (mole/L)
0.0010
0.0010
0.0030
NO2 formation rate (mole/L·s)
7.10
28.4
255.6
This problem is a candidate for the inspection method, but we’ll use the more general
long method. First we write out the generic rate law for the reaction.
Rate = k[O2 ]x [NO]y
Now, we choose two runs where one reactant’s concentration is the same while the other
reactant’s concentration changes. For Trials 1 and 2, the [NO] remains constant while the
[O2 ] changes. We’ll use these to determine the order with respect to O2 .
7.10
28.4
1
4
1
4
1
4
1
log
4
x
=
=
=
=
=
=
[0.0010]x [0.0010]y
[0.0040]x [0.0010]y
[0.0010]x
[0.0040]x
0.0010 x
0.0040
x
1
4
1
x log
4
1
Now, we do the same thing with Trials 2 and 3.
28.4
255.6
1
9
1
9
1
9
[0.0040]x [0.0010]y
[0.0040]x [0.0030]y
[0.0010]y
=
[0.0030]y
0.0010 y
=
0.0030
y
1
=
3
=
4
1
9
log
1
3
= y log
y = 2
Therefore, the rate law is:
Rate = k[O2 ][NO]2
To solve for k, we can use any Trial. Let’s use Trial 1.
Rate
7.10 M/s
7.10 M/s
k
=
=
=
=
k[O2 ][NO]2
k(0.0010 M)(0.0010 M)2
k(1.0 × 10−9 M3 )
7.1 × 109 M−2 s−1
75. If it takes 75.0 min for the concentration of a reactant to drop to 20.0% of its
initial value in a first-order reaction, what is the rate constant for the reaction
in the units min−1 ?
This is a concentration-time problem. We need to use the first-order integrated rate
law. First, some mathematically manipulation using the properties of the ln function to
ensure there are no rounding errors.
ln[A]t − ln[A]0 = −kt
[A]t
ln
= −kt
[A]0
We’re not given concrete concentrations, just a percentage change. It’s easiest to imagine
you have 100 units. If the concentration drops by 20%, that means we must have 80 units
left. We use these numbers as our initial and final.
ln
80.0
= −k(75.0 min)
100.0
k = 0.00298 min−1
91. If a reaction has k = 3.0 × 10−4 s−1 at 25 ◦ C and an activation energy of
100.0 kJ/mole, what will the value of k be at 50 ◦ C?
The problem gives us both the rate constant and the activation energy. This should
be a clue that it’s an Arrhenius equation problem. Since it’s asking for the k at another
temperature, we know we must use the 2-point form.
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k2
Ea
ln
=
k1
R
1
1
−
T1 T2
1
1
100.0 kJ/mole
−
=
0.008314 kJ/mole·K 298 K 323 K
k2
3.0 × 10−4 s−1
k2
ln
= 3.124
3.0 × 10−4 s−1
k2
= 22.737
3.0 × 10−4 s−1
k2 = 6.8 × 10−3 s−1
ln
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