AP Chemistry – Chapter 12 Review Packet
Name______________________________
For questions 1 - 7, choose true or false.
1.
The rate of a given reaction is constant at any specified temperature. False
2.
A balanced equation does not indicate the mechanism by which a reaction is proceeding.
True
The rate constant is the rate of reaction at constant temperature. False
3.
4.
In the reaction H2 + I2  2 HI, only two molecules (H2 and I2) react so this reaction is a
bimolecular reaction. False
5.
Relatively few reaction mechanisms have been completely characterized. True
(Special ingenuity and techniques are needed to detect fast steps in a reaction mechanism.)
The rate-determining step of a reaction is the step that proceeds fastest. False
6.
7.
The half-life of a reaction is the time required for half of the original concentration of the
limiting reagent to be consumed in the reaction. True
8.
Factors that determine reaction rates include all of the following except:
a. size of reactant particles
b. magnitude of the equilibrium constant (Large values of k do not imply large rates.)
c. reaction temperature
d. concentration of reactants
9.
For the reaction 2A + B  C, the reaction rate is:
a. R = k[A]2 [B]
b. R = k[A] [B]
c. R = k[A]2
d. possibly not any of the above
(A valid reaction rate expression usually cannot be
written from the overall reaction.)
For a reaction of the type A + B  2C, it is found that doubling the amount of A causes
the reaction rate to be four times as great but that doubling the amount of B has no
apparent effect on the rate. The rate equation that is most correct is:
a. R = k[A]2 (second order in A and zero order in B)
b. R = k[A] [B]
c. R = k[A]
d. R = k[C]2
[A] [B]
10.
11.
Which of the following is the rate equation for a termolecular elementary reaction?
a. R = k [A]
b. R = k [A]2
c. R = k [A] [B]
d. R = k [A] [B] [C] (three component collision)
12.
Catalysts are characterized by each of the following except that:
a. they are not consumed in the overall reaction
b. they can accelerate or retard a reaction
Catalysts often undergo
c. they are stable and unreactive
reactions but are regenerated
d. they cause a change in the energy of activation
in a later step of the reaction.
13.
In first-order reactions, a decrease in t½ causes k to:
a. increase
b. remain the same
t½ = 0.693/k
c. decrease
d. can’t determine effect without more information
Questions 14-16
A+B C
The following are possible rate laws for the hypothetical reaction given above.
a. Rate = k[A]
b. Rate = k[A]2
c. Rate = k[A] [B]
d. Rate = k[A]2 [B]
e. Rate = k[A]2 [B]2
14.
This is the rate law for a first-order reaction.
a.
15.
This is the rate law for a reaction that is second order with respect to B.
16.
This is the rate law for a third order reaction. d.
17.
A multi-step reaction takes place by the following mechanism.
A+BC+D
A+CD+E
Which of the species shown above is an intermediate in the reaction?
a. A
b. B
c. C
d. D
e. E
18.
2 NOCl  2 NO + Cl2
The reaction above takes place with all of the reactants and products in the gaseous
phase. Which of the following is true of the relative rates of disappearance of the
reactants and appearance of products?
a. NO appears at twice the rate that NOCl disappears. 1Δ[NOCl] = 1Δ[NO] = Δ[Cl2]_
b. NO appears at the same rate that NOCl disappears. 2 Δt
2
Δt
Δt
c. NO appears at half the rate that NOCl disappears.
d. Cl2 appears at the same rate that NOCl disappears. Δ[NOCl] = Δ[NO] = 2Δ[Cl2]__
e. Cl2 appears at twice the rate that NOCl disappears.
Δt
Δt
Δt
e.
Questions 19-21
A+B C
The following are possible rate laws for the hypothetical reaction given above.
a. Rate = k[A]
b. Rate = k[B]2
c. Rate = k[A] [B]
d. Rate = k[A]2 [B]
e. Rate = k[A]2 [B]2
19.
When [A] and [B] are doubled, the initial rate of reaction will increase by a factor of
eight. d = If A and B are 1, then Rate = k, if A and B are doubled, Rate = k(2)2(2) = 8k
20.
When [A] and [B] are doubled, the initial rate of reaction will increase by a factor of two.
a = If A and B are 1, then Rate = k, if A and B are doubled, Rate = k(2) = 2k
When [A] is doubled and [B] is held constant, the initial rate of reaction will not change.
b = independent of changes in A is Rate = k[B]2
21.
22.
H2 (g) + I2 (g)  2 HI (g)
When the reaction given above takes place in a sealed isothermal container, the rate law
is:
Rate = k[H2] [I2]
If a mole of H2 gas is added to the reaction chamber, which of the following will be true?
a. The rate of reaction and the rate constant will increase.
b. The rate of reaction and the rate constant will not change.
c. The rate of reaction will increase and the rate constant will decrease.
d. The rate of reaction will increase and the rate constant will not change.
e. The rate of reaction will not change and the rate constant will increase.
23.
A+B C
When the reaction given above takes place, the rate law is: Rate = k[A]
If the temperature of the reaction chamber were increased, which of the following will be
true?
a. The rate of reaction and the rate constant will increase.
b. The rate of reaction and the rate constant will not change.
c. The rate of reaction will increase and the rate constant will decrease.
d. The rate of reaction will increase and the rate constant will not change.
e. The rate of reaction will not change and the rate constant will increase.
24.
A+B C
Based on the following experimental data, what is the rate law for the hypothetical
reaction given above?
Experiment
1
2
3
a.
b.
c.
d.
e.
25.
[B] (M)
0.10
0.20
0.20
Initial Rate of Formation of C (1/M·s)
3 x 102
6 x 102
6 x 102
Rate = k[A]
Rate = k[A]2
Rate = k[B]
Rate = k[B]2
Rate = k[A] [B]
A+B  C+D
The rate law for the hypothetical reaction shown above is as follows: Rate = k[A]
Which of the following changes to the system will increase the rate of reaction?
I An increase in the concentration of A
II. An increase in the concentration of B
III. An increase in the temperature
a.
b.
c.
d.
e.
26.
[A] (M)
0.20
0.20
0.40
I only
I and II only
I and III only
II and III only
I, II, and III
A+B C
Based on the following experimental data, what is the rate law for the hypothetical
reaction given above?
Experiment
1
2
3
a.
b.
c.
d.
e.
[A] (M)
0.20
0.20
0.40
Rate = k[A]
Rate = k[A]2
Rate = k[B]
Rate = k[B]2
Rate = k[A] [B]
[B] (M)
0.10
0.20
0.40
Initial Rate of Formation of C (1/M·s)
2.0 x 106
4.0 x 106
1.6 x 105
27.
A+B C
Based on the following experimental data, what is the rate law for the hypothetical
reaction given above?
Experiment
1
2
3
a.
b.
c.
d.
e.
[A] (M)
0.10
0.40
0.40
[B] (M)
0.10
0.10
0.20
Initial Rate of Formation of C (1/M·s)
1.5 x 103
6.0 x 103
2.4 x 102
Rate = k[A]
Rate = k[A]2
Rate = k[A] [B]2
Rate = k[B]2
Rate = k[A]2 [B]2
28.
Time (hours)
0
1
2
3
[A] (M)
0.40
0.02
0.10
0.05
[B] (M)
5.00
5.00
5.00
5.00
A reaction occurred in which A reacted with a large excess of B to form C. The
concentrations of the reactants were measured periodically and recorded in the chart
above. Based on the data in the chart, which of the following is the rate law for the
reaction?
a. Rate = k[A]
b. Rate = k[A]2
c. Rate = k[B]
d. Rate = k[B]2
e. Rate = k[A] [B]
29.
Consider the reaction: 4 NH3 + 7 O2  4 NO2 + 6 H2O
At a certain instant the initial rate of disappearance of the oxygen gas is X. What is the
value of the appearance of water at the same instant?
a. 1.2X
b. 1.1X
1Δ[O2] = 1Δ[H2O] so 6Δ[O2] = Δ[H2O]
c. 0.86X
7 Δt
6 Δt
7 Δt
Δt
d. 0.58X
e. cannot be determined from the data
6/7 = 0.86
30.
Consider the reaction X  Y + Z
Which of the following is a possible rate law?
a.
b.
c.
d.
e.
31.
Rate = k[X]
Rate = k[Y]
Rate = k[Y] [Z]
Rate = [X] [Y]
Rate = k[Z]
Rate laws do NOT include products!!
Consider the following rate law: Rate = k[A]n [B]m
How are the exponents n and m determined?
a.
b.
c.
d.
e.
By using the balanced chemical equation
By using the subscripts for the chemical formulas
By using the coefficients of the chemical formulas
By educated guess
By experiment
Questions 32-33
The following initial rate data were found for the reaction
2 MnO41 + 5 H2C2O4 + 6 H+1 → 2 Mn+2 + 10 CO2 + 8 H2O
[MnO41]0
1 x 103
2 x 103
2 x 103
2 x 103
[H+1]0
1.0
1.0
1.0
2.0
[H2C2O4]0
1 x 103
1 x 103
2 x 103
2 x 103
32.
Which of the following is the correct rate law?
a. Rate = k[MnO41]2 [H2C2O4]5 [H+1]6
b. Rate = k[MnO41]2 [H2C2O4] [H+1]
c. Rate = k[MnO41] [H2C2O4] [H+1]
d. Rate = k[MnO41]2 [H2C2O4]
e. Rate = k[MnO41]2 [H2C2O4]2
33.
What is the value of the rate constant?
a. 2 x 105 Ms1
b. 2 x 105 M2s1
c. 200 M1s1
d. 200 M2s1
e. 2 x 104 Ms1
Initial Rate (M/s)
2 x 104
8 x 104
1.6 x 103
1.6 x 103
Rate = k[MnO41]x [H2C2O4]y [H+1]z
x = 2 because as the concentration is
doubled, the rate is quadrupled.
y = 1 because as the concentration is
doubled, the rate is doubled.
z = 0 because as the concentration is
doubled, the rate does not change.
Rate = k[MnO41]2 [H2C2O4]
2 x 104 M/s = k(1 x103 M)2 (1 x103 M)
2 x 104 M/s = k
(1 x109 M3)
2 x 105 1/M2s = k
34.
Tabulated below are initial rate data for the reaction:
2 Fe(CN)63 + 2I1  2 Fe(CN)64 + I2
Run
1
2
3
4
5
[Fe(CN)63]0
0.01
0.01
0.02
0.02
0.02
[I1]0
0.01
0.02
0.02
0.02
0.02
[Fe(CN)64]0
0.01
0.01
0.01
0.02
0.02
[I2]0
0.01
0.01
0.01
0.01
0.02
Initial Rate (M/s)
1 x 105
2 x 105
8 x 105
8 x 105
8 x 105
The experimental rate law is:
a.
[I2] = k [Fe(CN)63]2 [I1]2 [Fe(CN)64]2 [I2]
t
35.
b.
[I2] = k [Fe(CN)63]2 [I1] [Fe(CN)64] [I2]
t
c.
[I2] = k [Fe(CN)63]2 [I1]
t
d.
[I2] = k [Fe(CN)63] [I1]2
t
e.
[I2] = k [Fe(CN)63] [I1] [Fe(CN)64]
t
The only two possibilities are c
and d because the others include
products. [Fe(CN)63] is second
order because when the
concentration is doubled, the
rate is quadrupled. [I1] is first
order because when the
concentration is doubled, the
rate doubles.
For a reaction in which A and B react to form C, the following initial rate data were
obtained:
[A] (M)
0.2
0.4
0.8
0.2
0.2
[B] (M)
0.2
0.2
0.2
0.4
0.8
Initial Rate of Formation of C (M/s)
0.50
2.00
80..
1.00
2.00
What is the rate law for the reaction?
a. Rate = k[A] [B]
b. Rate = k[A]2 [B]
c. Rate = k[A] [B]2
d. Rate = k[A]2 [B]2
e. Rate = k[A]3
[A] is second order because when [B] is
constant and [A] is doubled, the rate is
quadrupled. [B] is first order because
when [A] is constant and [B] is doubled,
the rate is doubled.
Questions 36-39
The following questions refer to the gas-phase decomposition of ethylene chloride.
C2H5Cl → products
Experiment shows that the decomposition is first order. The following data show kinetics
information for this reaction.
Time (s)
ln [C2H5Cl] (M)
1.0
1.625
2.0
1.735
36.
What is the rate constant for this decomposition?
a. 0.29 s1
Put the times in L1 and the ln [C2H5Cl] in L2.
b. 0.35 s1
1
Do a linear regression to find the slope.
c. 0.11 s
1
Slope =  k so  k =  0.11 so k = 0.11
d. 0.02 s
1
e. 0.22 s
37.
What was the initial concentration of the ethylene chloride?
a. 0.29 M
ln [A] =  kt + ln [A]0
b. 0.35 M
1.625 M =  (0.11s1) (1.0 s) + ln [A]0
c. 0.11 M
1.625 M =  0.11 + ln [A]0
d. 0.02 M
1.515 M = ln [A]0
e. 0.22 M
0.22 M = [A]0
38.
What would the concentration be after 5.0 seconds?
a. 0.13 M
ln [A] =  kt + ln [A]0
b. 0.08 M
ln [A] =  (0.11s1) (5.0 s) + ln [0.22 M]
c. 0.02 M
ln [A] =  0.55 +  1.5
d. 0.19 M
ln [A] =  2.05
e. 0.12 M
[A] = 0.13 M
What is the half life?
a. 0.7 s
b. 1.3 s
t½ = 0.693/k
c. 8.9 s
t½ = 0.693/0.11
d. 6.3 s
t½ = 6.3 s
e. 2.2 s
39.
40.
The decomposition of ozone may occur through the two-step mechanism shown:
step 1 O3 → O2 + O
step 2 O3 + O → 2O2
The oxygen atom is considered to be a(n)
a. reactant
b. product
c. catalyst
d. reaction intermediate
e. activated complex
Questions 41-43
The kinetics of the reaction A + 3 B → C + 2D were studied and the following results
obtained, where the rate law is
 Δ[A] = k [A]n [B]m
Δt
For a run where [A]0 = 1.0 x 103 M and [B]0 = 5.0 M, a plot of ln[A] versus t was found
to give a straight line with slope = 5.0 x 102 s1.
For a run where [A]0 = 1.0 x 103 M and [B]0 = 10.0 M, a plot of ln[A] versus t was
found to give a straight line with slope = 7.1 x 102 s1.
41.
42.
43.
What is the value of n?
a. 0
b. 0.5
Since the plots of ln[A] vs. t gives a straight line, then [A] is first order.
c. 1
d. 1.5
e. 2
Trial 2: k2 = k[B2]m
Trial 1: k1 = k[B1]m
7.1 x 102 = k[10.0]m
5.0 x 102 = k[5.0]m
What is the value of m?
k = 7.1 x 102
k = 5.0 x 102
a. 0
[10.0]m
[5.0]m
b. 0.5
c. 1
d. 1.5
5.0 x 102 = 7.1 x 102
e. 2
[5.0]m
[10.0]m
Calculate the value of k (ignore units).
a. 2.2 x 102
k1 = k[B1]½
b. 1.0 x 102
5.0 x 102 = k[5.0] ½
c. 5.0 x 102
k = 5.0 x 102
d. 1.1 x 102
[5.0] ½
e. none of these
(5.0 x 102)(10.0)m = (7.1 x 102)(5.0)m
(10.0)m = (7.1 x 102)
(5.0)m (5.0 x 102)
2m = 1.42 or 2m = 2½ so m = ½
k = 0.022
44.
Using the following information determine the activation energy for the reaction shown
here: 2NO → N2 + O2
Temperature (K)
1400
1500
a.
b.
c.
d.
e.
3.2 x 104 J/mol
9.5 x 106 J/mol
2.8 x 104 J/mol
6.8 x 105 J/mol
2.7x 105 J/mol
Rate Constant (1/Ms)
0.143
0.659
ln (k2/k1) = Ea/R (1/T1  1/T2)
ln (0.659/0.143) = Ea/8.31 J/molK (1/1400 K  1/1500 K)
ln (4.61) = Ea/8.31 J/molK {(7.14 x 104)  (6.67 x 104)}
1.53 = Ea/8.31 J/molK (4.7 x 105)
1.53 = (5.66 x 106)Ea
2.7 x 105 J/mol = Ea
Questions 45-46
The reaction
2NOBr → 2NO + Br2
exhibits the rate law
Rate = k [NOBr]2 =  Δ[NOBr]
Δt
where k = 1.0 x 105 M1s1 at 25C. This reaction is run where the initial concentration
of NOBr ([NOBr]0) = 1.00 x 101 M.
45.
What is one half-life for this experiment?
a. 5.0 x 101 s
t½ = 1/k[A]0
b. 6.9 x 104 s
5
t½ = 1/{(1.0 x 105 M1s1)(1.00 x 101 M)}
c. 1.0 x 10 s
t½ = 1.0 x 106 s
d. 1.0 x 106 s
e. none of these
46.
The [NO] after 1.00 hour has passed is
a. 3.5 x 104 M
[A] = [NOBr]
b. 9.9 x 103 M
c. 9.7 x 103 M
1/[A] = kt + 1/[A]0
3
d. 1.0 x 10 M
e. none of these
1/[A] = (1.0 x 105 M1s1)(3600 s) + 1/(1.00 x 101 M)
1/[A] = 0.036 + 10
1/[A] = 10.036 so [A] = 0.0996 M
Questions 47-48
The following questions refer to the reaction 2A2 + B2 → 2C. The following mechanism
has been proposed:
step 1 (very slow)
A2 + B2 → R + C
step 2 (slow)
A2 + R → C
47.
What is the molecularity of step 2?
a. unimolecular
b. bimolecular
c. termolecular
d. quadmolecular
e. the molecularity cannot be determined
48.
Which step is “rate determining”?
a. both steps
b. step 1
c. step 2
d. a step that is intermediate to step 1 and step 2
e. none of these
Questions 49-51
The questions below refer to the following diagram:
49.
Why is this reaction considered to be exothermic?
a. Because energy difference B is greater than energy difference C
b. Because energy difference B is greater than energy difference A
c. Because energy difference A is greater than energy difference C
d. Because energy difference B is greater than energy difference C plus energy
difference A
e. Because energy difference A and energy difference C are about equal
50.
At what point on the graph is the activated complex present?
a. point W
b. point X
c. point Y
d. point Z
e. none of these
51.
If the reaction were reversible, would the forward or the reverse reaction have a higher
activation energy?
a. The diagram shows no indication of any activation energy.
b. The forward and reverse activation energies are equal.
c. The forward activation energy.
d. The reverse activation energy.
The forward reaction goes from W to Y and
e. none of these
the reverse reaction goes from Z to Y.
52.
The reaction 2H2O2 → 2 H2O + O2 has the following proposed mechanism:
2H2O2 + I1 → H2O + IO1
H2O + IO1 → H2O + O2 + I1
The catalyst in the reaction is:
a. H2O
b. I1
c. H2O2
d. IO1
53.
When ethyl chloride, CH3CH2Cl, is dissolved in 1.0 M NaOH, it is converted into
ethanol, CH3CH2OH, by the reaction:
CH3CH2Cl + OH1 → CH3CH2OH + Cl1
At 25C the reaction is first order in CH3CH2Cl, and the rate constant is 1.0 x 101 s1. If
the activation parameters are A = 3.4 x 1014 s1 and Ea = 100.0 kJ/mol, what will the rate
constant be at 40C?
a.
b.
c.
d.
e.
6.9 x 103 s1
1.7 x 102 s1
5.0 x 103 s1
2.0 x 103 s1
5.0 x 1014 s1
100.0 kJ/mol = 1.000 x 105 J/mol
k = AeEa/RT
k = (3.4 x 1014 s1) e 1.000 x 105 J/mol/{(8.31 J/molK)(313 K)}
k = (3.4 x 1014 s1) e 38.4
k = (3.4 x 1014 s1) (2.10 x 1017)
k = 7.1 x 103 s1
If the entire problem is done without rounding and
using R = 8.3145, then k = 3.97 x 103 s1
Essay questions:
1.
The following results were obtained in experiments designed to study the rate of this
reaction.
A + 2B  2C
Experiment
1
2
3
4
[A]0 (M)
0.05
0.05
0.10
0.20
[B]0 (M)
0.05
0.10
0.10
0.10
Initial Rate of Disappearance of A (M/s)
3.0 x 103
6.0 x 103
1.2 x 102
2.4 x 102
a. Determine the order of the reaction with respect to each of the reactants and write the
rate law for the reaction.
When experiments 3 and 4 are compared, [A] doubles and the rate doubles so the reaction
is first order with respect to A.
When experiments 1 and 2 are compared, [B] doubles and the rate doubles so the reaction
is first order with respect to B.
Rate = k[A] [B]
b. Calculate the value of the rate constant, k, for the reaction. Include the units.
k = Rate_ = (1.2 x 102 M/s) = 1.2 M1s1
[A] [B]
(0.10 M)(0.10 M)
c. If another experiment is attempted with [A] and [B] both 0.02 M, what will be the
initial rate of disappearance of A?
Rate = k[A] [B]
Rate = (1.2 M1s1)(0.02 M)(0.02 M)
Rate = 4.8 x 104 M/s
d. The following reaction mechanism was proposed for the reaction above.
A + B  C +D
D + B  C
(i) Show that the mechanism is consistent with the balanced reaction.
A + B  C +D
D + B  C
A + 2B + D  2C + D
becomes A + 2B  2C
(ii) Show which step is the rate-determining step and explain your choice.
The first part of the mechanism is the slow, rate-determining step because its
reactants (A and B) are the ones present in the rate law.
2.
The following results were obtained in experiments designed to study the rate of this
reaction.
2 NO (g) + Br2 (g)  2 NOBr (g)
Experiment
1
2
3
[NO]0 (M)
0.02
0.04
0.02
[Br2]0 (M)
0.02
0.02
0.04
Initial Rate of Disappearance of A (M/s)
9.6 x 102
3.8 x 101
1.9 x 101
a. Write the rate law for the reaction.
When experiments 1 and 2 are compared, [NO] doubles and the rate quadruples so the
reaction is second order with respect to NO.
When experiments 1 and 3 are compared, [Br2] doubles and the rate doubles so the
reaction is first order with respect to Br2.
Rate = k[NO]2 [Br2]
b. Calculate the value of the rate constant, k, for the reaction. Include the units.
k=
Rate_ = (9.6 x 102 M/s) = 1.2 x 104 M2s1
[NO]2 [Br2]
(0.02 M)(0.02 M)
c. In experiment 2, what was the concentration of NO remaining when half of the
original amount of Br2 was consumed?
In experiment 2, we started with [Br2] = 0.02 M, so 0.01 M was consumed. From the
balanced equation, 2 moles of NO are consumed for every mole of Br2 consumed. So
0.02 M of NO are consumed.
[NO] remaining = 0.04 M  0.02 M = 0.02 M
d. Which of the following reaction mechanisms is consistent with the rate law
established in part a? Explain your choice.
I.
(fast) NO + NO  N2O2
(slow) N2O2 + Br2  2 NOBr
II.
(slow) Br2  Br + Br
(fast) 2(NO + Br  NOBr)
I.
(fast) NO + NO  N2O2
(slow) N2O2 + Br2  2 NOBr
The slow step is the rate-determining step, with the following rate law:
Rate = k[N2O2][Br2]
We can replace the intermediate (N2O2) by assuming that the first step reaches
equilibrium instantaneously and remembering that at equilibrium, the rates of the forward
and reverse reactions are equal.
kf[NO]2 = kr[N2O2]
kf [NO]2 = [N2O2]
kr
Rate = k kf [NO]2 [Br2]
kr
3.
The following results were obtained in experiments designed to study the rate of this
reaction.
2A + B  C + D
Experiment
1
2
3
[A]0 (M)
0.10
0.20
0.20
[B]0 (M)
0.10
0.20
0.40
Initial Rate of Formation of D (M/min)
1.5 x 103
3.0 x 103
6.0 x 103
a. Write the rate law for the reaction.
Experiments 2 and 3 prove that the rate must double when [B] doubles. Knowing
this, we can see that when the rate doubles from experiment 1 to 2, it must be
because of B, and the change in A has no effect. So the reaction is zero order
with respect to A.
b. Calculate the value of the rate constant, k, for the reaction. Include the units.
k=
Rate_ = (1.5 x 103 M/s) = 1.5 x 102 s1
[B]
(0.10 M)
c. If experiment 2 goes to completion, what will be the final concentration of D?
Assume that the volume is unchanged over the course of the reaction and that no D
was present at the start of the experiment.
0.2 M of A is consumed. According to the balanced equation, for every 2 moles
of A consumed, 1 mole of D is produced. So if 0.2 M of A is consumed, 0.1 M of
D is produced.
d. Which of the following reaction mechanisms is consistent with the rate law
established in part a? Explain your choice.
I.
(slow) A + B  C + E
(fast) A + E  D
II.
(slow) B  C + E
(fast) A + E  F
(fast) A + F  D
II.
The slow, rate-determining step gives the proper rate law: Rate = k[B]
A (g) + B (g)  C (g)
4.
The reaction above is second order with respect to A and zero order with respect to B.
Reactants A and B are present in a closed container. Predict how each of the following
changes to the reaction system will affect the rate and rate constant.
a. More gas A is added to the container.
The rate of the reaction will increase because the rate depends on the concentration of A
as given in the rate law: Rate = k[A]2. The rate constant is independent of the
concentration of the reactants and will not change.
b. More gas B is added to the container.
The rate of reaction will not change. If the reaction is zero order with respect to B, then
the rate is independent of the concentration of B. The rate constant is independent of the
concentration of the reactants and will not change.
c. The temperature is increased.
The rate of the reaction will increase with increasing temperature because the rate
constant increases with increasing temperature. The rate constant increases with
increasing temperature because at a higher temperature more gas molecules will collide
with enough energy to overcome the activation energy for the reaction.
d. An inert gas D is added to the container.
Neither the rate nor the rate constant will be affected by the addition of an inert gas.
e. The volume of the container is decreased.
The rate of the reaction will increase because decreasing the volume of the container will
increase the concentration of A. Rate = k[A]2 The rate constant is independent of the
concentration of the reactants and will not change.
5. Use your knowledge of kinetics to answer the following questions.
a.
Line 2 is the catalyzed
reaction. Adding a
catalyst lowers the
activation energy of the
reaction, making it easier
for the reaction to occur.
The two lines in the diagram above show different reaction pathways for the same
reaction. Which of the two lines shows the reaction when a catalyst has been added?
Explain.
b.
Line 2 shows the higher
temperature distribution. At
a higher temperature, more
of the molecules will be at
higher energies, causing the
distribution to flatten out and
shift to the right.
Which of the two lines in the energy distribution diagram shows the conditions at a
higher temperature?
5.
c.
Line 1 is correct. ln[reactant]
for a first order reaction changes
in a linear fashion over time, as
shown:
ln[A] = kt + ln[A]0
y = mx + b
Which of the two lines in the diagram above shows the relationship of ln[A] versus time
for a first order reaction with the following rate law: Rate = k[A]
d.
Graph 1 is correct,
showing a decrease
in the concentration
of A as it is
consumed in the
reaction, and a
corresponding
increase in the
concentration of B as
it is produced.
Which of the two graphs above shows the changes in concentration over time for the
following reaction: A  B
6.
Use your knowledge of kinetics to explain each of the following statements.
a. An increase in the temperature at which a reaction takes place causes an increase in
reaction rate.
An increase in temperature means an increase in the energy of the molecules present. If
the molecules have more energy, then more of them will collide more often with enough
energy to overcome the activation energy required for a reaction to take place, causing
the reaction to proceed more quickly.
b. The addition of a catalyst increases the rate at which a reaction will take place.
A catalyst offers a reaction an alternate pathway with a lower activation energy. If the
activation energy is lowered, then more molecular collisions will occur with enough
energy to overcome the activation energy, causing the reaction to go more quickly.
c. A catalyst that has been ground into powder will be more effective than a solid block
of the same catalyst.
The effectiveness of a solid catalyst depends on the surface area of the catalyst that is
exposed to the reactants. Grinding a solid into powder greatly increases its surface area.
d. Increasing the concentration of reactants increases the rate of a reaction.
Increasing the concentration of reactants crowds the reactants more closely together,
making it more likely that they will collide with each other. The more collisions that
occur, the more likely that collisions that will result in a reaction will occur.