Rates of reactions
Answer Section
MULTIPLE CHOICE
1. ANS: D
DIF: average
REF: K/U
LOC: EC 2.01
2. ANS: C
DIF: average
REF: K/U
LOC: EC 2.01, EC 1.03
3. ANS: A
DIF: average
REF: K/U
LOC: EC 2.01
4. ANS: C
DIF: average
REF: K/U
LOC: EC V.01, EC 2.06
5. ANS: D
DIF: average
REF: K/U
LOC: EC V.01, EC 2.06
6. ANS: E
DIF: average
REF: K/U
LOC: EC V.01, EC 2.06
7. ANS: B
DIF: easy
REF: K/U
LOC: EC V.01, EC 1.04
8. ANS: E
DIF: average
REF: I
LOC: EC 2.06, EC V.01
9. ANS: A
DIF: easy
REF: K/U
LOC: EC 2.06
10. ANS: E
DIF: easy
REF: K/U
LOC: EC 1.06
11. ANS: D
DIF: average
REF: K/U
LOC: EC 1.04
12. ANS: A
DIF: easy
REF: K/U, I
LOC: EC 1.06
13. ANS: A
DIF: average
REF: K/U, I
LOC: EC 1.06
14. ANS: D
DIF: easy
REF: K/U
LOC: EC 1.04
15. ANS: B
DIF: easy
REF: K/U
LOC: EC 1.04
16. ANS: A
DIF: average
REF: K/U
LOC: EC 1.04
17. ANS: D
DIF: average
REF: K/U
LOC: EC 2.06
18. ANS: D
DIF: easy
REF: K/U
LOC: EC 1.05
19. ANS: A
DIF: easy
REF: K/U
LOC: EC 1.05
20. ANS: B
DIF: easy
REF: K/U
LOC: EC 1.05
21. ANS: C
DIF: average
REF: I, K/U
LOC: EC 1.04
SHORT ANSWER
22. ANS:
An elementary reaction involves a single molecular event, such as a simple collision between atoms, molecules,
or ions. It may be one step in a series of steps that make up a reaction mechanism.
DIF:
23. ANS:
easy
REF: K/U
LOC: EC 1.06
The mechanism of a reaction is the series of steps that make up the overall reaction. Each step involves an
elementary reaction: a single molecular event, such as a simple collision between particles.
DIF:
easy
REF: K/U
LOC: EC 1.06
24. ANS:
A bimolecular elementary reaction is an elementary reaction in which two particles collide and react.
DIF:
easy
REF: K/U
LOC: EC 1.06
25. ANS:
A catalyst does not affect the enthalpy change of a chemical reaction. A catalyst increases the rate of a chemical
reaction by decreasing the activation energy required, but it does not affect the difference in enthalpy between
products and reactants.
DIF:
easy
REF: K/U, C
LOC: EC 1.04
26. ANS:
This is a first-order reaction. (The sum of the exponents is 1
DIF:
easy
REF: K/U
easy
REF: K/U
1.)
2
3.)
LOC: EC 1.03
27. ANS:
This is a third-order reaction. (The sum of the exponents is 1
DIF:
0
LOC: EC 1.03
28. ANS:
The rate of dissolution of the lump of sugar can be increased by
- crushing the lump of sugar in order to increase the surface area of sugar exposed to the tea
- stirring the tea in order to increase the amount of contact between undissolved sugar and tea that has a low
concentration of dissolved sugar molecules
DIF:
average
REF: K/U, C
LOC: EC 1.04
29. ANS:
Particles in the gaseous state move much faster than particles in the solid or liquid state. They have a higher
kinetic energy. This allows a greater percent of all collisions to be successful. Because gas reactants exist as
elementary molecules or atoms, the bonds that hold the molecules or atoms together in the liquid or solid state do
not have to be broken. Therefore, lower activation energy is required. The rapid movement of individual
molecules or atoms provides maximum surface area for interaction. All these factors combine to make gaseous
reactions extremely fast.
DIF:
difficult
REF: K/U, C
LOC: EC 1.04
30. ANS:
Ratio of moles NO to moles NO2 3:1
\Rate of loss of NO
3 Rate of production of NO2
3 [0.035 mol/(L s)]
0.105 mol/(L s)
The rate of loss of NO is
DIF:
average
0.105 mol/(L s).
REF: I
31. ANS:
Since this is a first order reaction,
LOC: EC V.01, EC 2.01
Reaction rate
k [X]1
(0.25 s 1)(0.05 mol/L)
1.25 10 2 mol/(L s)
-2
10 mol/(L s).
The rate of the reaction is 1.25
DIF:
average
REF: I
LOC: EC 1.03
32. ANS:
The following two requirements must be met for a reaction to occur:
- A collision must occur between reactant particles that have the correct orientation.
- Reactants must have sufficient collision energy to be able to overcome the activation energy barrier.
DIF:
easy
REF: K/U
LOC: EC 1.04
33. ANS:
An activated complex is a chemical species that exists at the transition state, or change-over point, of a reaction.
The activated complex is neither reactant nor product. It has partial bonds between aligned reactants and is highly
unstable.
DIF:
average
REF: K/U
LOC: EC 1.04
34. ANS:
a) mass of calcium carbonate, pH, volume of carbon dioxide gas if pressure constant, change in pressure if
volume constant, change in electrical conductivity
b) change in pressure if constant volume, change in volume if constant pressure
c) pH, mass of zinc, volume of hydrogen at constant pressure or pressure due to hydrogen at constant volume,
change in electrical conductivity
DIF:
average
REF: I
LOC: EC 2.06
35. ANS:
a) pH, volume of CO2(g) at constant pressure or pressure due to CO2(g) at constant volume, change in electrical
conductivity
b) change in colour due to the iodine in the water
c) change in volume at constant pressure or change in pressure at constant volume due to change in the number of
moles, colour change due to N2O4(g)
DIF:
average
REF: I
LOC: EC 2.06
36. ANS:
The equation has only CH3CHO(g) as a reactant, and the molar coefficient is one. Therefore, this is a first-order
reaction.
For a first-order reaction,
0.693 k
0.693 (8.6
8.06 s
The half-life is 8.06 s.
DIF:
difficult
10
REF: I
2
s 1)
LOC: EC 1.03
37. ANS:
The equation has only N2O5(g) as a reactant, and the molar coefficient is one. Therefore, this is a first-order
reaction.
For a first-order reaction,
0.693
k
0.693 (6.0
1.16 103 s
19.3 min
The half-life is 19.3 min.
DIF:
difficult
10
4
s 1)
REF: I
LOC: EC 1.03
38. ANS:
A homogeneous catalyst exists in the same phase as the reactants. An example is the use of ZnCl2(aq) as the
catalyst in the reaction between aqueous solutions of (CH3)2CHOH and HCl. A heterogeneous catalyst exists in a
phase that is different from the phase of the reaction it catalyses. An example is the use of platinum or palladium
to catalyse the hydrogenation of alkenes.
DIF:
easy
REF: K/U
LOC: EC 1.04
PROBLEM
39. ANS:
a)
b)
c)
The average rate is determined by drawing the secant to join the beginning of the required time interval to the
end of this time interval. The average rate is equal to the slope of the secant.
d)
The instantaneous rate is determined by drawing the tangent to the curve at the time for which the
instantaneous rate is required. The instantaneous rate is equal to the slope of the tangent.
DIF:
average
REF: K/U
LOC: EC 2.06
40. ANS:
a) The reaction is endothermic. The enthalpy of the products is higher than the enthalpy of the reactants. This
means that energy is absorbed as the reaction proceeds.
b) A
c) C
d) B
DIF:
easy
REF: K/U
LOC: EC 1.05
average
REF: K/U
LOC: EC 1.05
easy
REF: K/U
LOC: EC 1.05
41. ANS:
DIF:
42. ANS:
DIF:
43. ANS:
a)
b)
Slope of the tangent at 20 min
Slope of the tangent at 180 min
DIF:
average
REF: I
(0.31 mol/L 0.21 mol/L)
0.001 mol/L/min
(0.20 mol/L 0.14 mol/L)
0.0005 mol/L/min
92 min
130 min
LOC: EC 2.06
44. ANS:
a) Draw a tangent to the curve at A. Then calculate the slope of the tangent to the curve.
b) The rate of reaction at B is less than the rate of reaction at A. As the reaction proceeds, the concentration of
H2O2 decreases because it is used up in the reaction. This decrease in concentration results in fewer successful
collisions and hence a reduced reaction rate.
c)
DIF:
average
REF: C, K/U
LOC: EC 2.06
45. ANS:
a)
b) A 3B E G
c) Hr
H1
H2
H3
Hr
42 21 42 21 kJ/mol
d) The rate-determining step is the first step because it has the highest activation energy barrier.
DIF:
46. ANS:
average
REF: K/U, I
LOC: EC 1.05
a) B
b) A: A lighted match is needed to ignite the candle, hence a high activation energy barrier is needed. Since the
reaction is self-sustaining, however, it must be exothermic.
DIF:
average
REF: K/U
LOC: EC 1.05
47. ANS:
a) A, B, C
b) A: The rusting of iron takes place without the application of energy from outside the system, and yet the rusting
of iron is a fairly slow reaction. This is consistent with an activation energy barrier that is not too small (which
would be a fast reaction) and a barrier that is not too high (which would be a very slow or non-spontaneous
reaction). Thus, diagram A represents this reaction.
DIF:
average
REF: K/U
LOC: EC 1.05, EC 3.04
48. ANS:
a) C: It has the lowest activation energy barrier, therefore a greater percent of the particles have sufficient energy
for a successful collision.
b) B: Two solids have the highest activation energy barrier because more bonds have to be broken to form
products.
DIF:
average
REF: K/U
LOC: EC 1.05
49. ANS:
a) B: It has the lowest activation energy barrier therefore a greater percent of the particles have sufficient energy
for a successful collision.
b) C: The ionization of sulfuric acid takes place very quickly, with the release of a large amount of energy.
Diagram C has a low activation energy barrier, with a subsequent large negative enthalpy change. Thus,
diagram C best matches the ionization of sulfuric acid.
DIF:
average
REF: K/U
LOC: EC 1.05
50. ANS:
Students might include a diagram to show that breaking a solid into smaller pieces provides a greater surface area.
Students might also show, on the same diagram, that more particles of the solvent (or other reactants) collide
more frequently with the increased surface area. The increase in surface area results in an increase in the number
of collisions. According to collision theory, this should result in an increase in reaction rate.
DIF:
51. ANS:
average
REF: K/U, C
LOC: EC 1.04
As the Maxwell-Boltzmann distribution diagram shows, the number of particles with energy that is equal to or
greater than the activation energy is increased. This increases the number of successful collisions, thereby
increasing the reaction rate.
DIF:
average
REF: K/U, C
LOC: EC 1.04
52. ANS:
As shown in the diagram, an increase in the number of particles of one or more of the reactants increases the
probability of collision. The increase in the number of collisions per unit of time results in an increase in reaction
rate.
DIF:
average
REF: K/U, C
LOC: EC 1.04
53. ANS:
A catalyst provides an alternate reaction pathway, or reaction mechanism, that has a lower activation energy
barrier. Lowering the activation energy barrier results in an increase in the number of particles with sufficient
energy to have a successful collision. This results in an increase in the reaction rate.
DIF:
average
REF: K/U, C
LOC: EC 1.04
54. ANS:
a) Since doubling the concentration of hydroxide ions causes the reaction rate to double, this reaction is first order
with respect to hydroxide ions.
b) Doubling the concentration of both reactants causes the reaction rate to quadruple, and doubling the hydroxide
ion concentration causes the reaction rate to double. Thus, doubling the concentration of bromoethane would
also cause the reaction rate to double. Therefore, this reaction is first order with respect to the bromoethane.
c) The overall reaction order is 2(1
.
DIF:
55. ANS:
a) Rate
b) Rate
difficult
REF: I, C
LOC: EC 1.03
k[A][B]
k[A][B]
The rate constant is 3.33 10 1 s 1(mol/L) 1 for this reaction.
c) The rate constant is temperature specific because any change in temperature results in a change in the reaction
rate even though there has not been a change in reactant concentrations.
DIF:
average
REF: I
LOC: EC 1.03
56. ANS:
a) The overall reaction order is first order. (The sum of the exponents is 1 0 1.)
b) The reaction is first order with respect to (CH3)3CBr.
c) The reaction order is not dependent on the concentration of water.
d) Since the reaction order is not dependent on the concentration of water, doubling the concentration of water has
no effect on the reaction rate.
e) If the concentration of (CH3)3CBr is halved, the reaction rate is also halved, since the reaction rate is first order
with respect to the concentration of (CH3)3CBr.
DIF:
average
REF: K/U, I, C
LOC: EC 1.03
57. ANS:
a)
Rate of formation of H2 Initial rate for decomposition of PH3
PH3
2.0
10
4
mol/(L s)
3.0
10
4
mol/(L s)
Mole ratio of H2 to
b)
Rate of formation of P4
DIF:
58. ANS:
Step 1:
Step 2:
DIF:
average
Initial rate for decomposition of PH3
PH3
2.0
10
4
mol/(L s)
3.3
10
5
mol/(L s)
REF: I
LOC: EC 1.03
O3(g) O2(g) O(g)
O3(g) O(g) 2O2(g)
2O3(g) 3O2(g)
average
REF: I
LOC: EC 1.06
59. ANS:
Balanced equation: 2NO(g) O2(g) 2NO2(g)
Step 1: NO(g) O2(g) NO3(g)
Step 2: NO3(g) NO(g) 2NO2(g)
2NO(g) O2(g) 2NO2(g)
DIF:
60. ANS:
Step 1:
Step 2:
DIF:
average
REF: I
LOC: EC 1.06
SO2(g) O2(g) SO3(g) O(g)
SO2(g) O(g) SO3(g)
2SO2(g) O2(g) 2SO3(g)
difficult
REF: I
LOC: EC 1.06
Mole ratio of P4 to
61. ANS:
a)
b) and c)
Reaction rate
Slope of tangent to curve at 100 s
2.5 10 4 mol/(L•s)
The reaction rate at 100 s is 2.5 10 4 mol/(L•s).
DIF:
average
REF: I
LOC: EC 2.06
62. ANS:
Compare trials 1 and 2: Doubling the concentration of H2(g) causes the reaction rate to double. Therefore, the
reaction rate is first order with respect to the concentration of H2(g).
Compare trials 1 and 3: Doubling the concentration of NO(g) causes the reaction rate to quadruple. Therefore, the
reaction rate is second order with respect to NO(g).
Reaction rate k[H2(g)]1[NO(g)]2
DIF:
difficult
REF: I
LOC: EC 1.03
63. ANS:
Compare trials 1 and 2: Doubling the concentration of SO2Cl2(g) causes the reaction rate to double. Therefore, the
reaction rate is first order with respect to the concentration of SO2Cl2(g).
Comparing trials 1 and 3 verifies this conclusion. The tripling of the concentration of SO2Cl2(g) causes the reaction
rate to triple.
Reaction rate k[SO2Cl2 (g)]1
DIF:
average
REF: I
LOC: EC 1.03
64. ANS:
Compare trials 1 and 2: Doubling the concentration of cyclopropane causes the reaction rate to double. Therefore,
the reaction rate is first order with respect to the concentration of cyclopropane.
Comparing trials 1 and 3 verifies this conclusion. Tripling the concentration of cyclopropane causes the reaction
rate to triple.
Reaction rate k[cyclopropane]1
DIF:
average
REF: I, MC
LOC: EC 1.03
65. ANS:
Compare trials 1 and 2: Doubling the concentration of OCl causes the reaction rate to double. Therefore, the
reaction rate is first order with respect to the concentration of OCl .
Compare trials 1 and 3: Doubling the concentration of I also causes the reaction rate to double. Therefore, the
reaction rate is first order with respect to I .
Reaction rate k[OCl ]1[I ]1
DIF:
difficult
REF: I
LOC: EC 1.03
66. ANS:
Reaction rates play an important role in our everyday lives, from the decay of food to the way we start a fire.
Foods stored at room temperature decay quite quickly. Milk sours in a day, while fruit goes rotten after a few
days. The rates of both these reactions can be slowed down by a decrease in temperature. Hence, we store food in
a refrigerator or (for longer periods of time) in a freezer, in order to keep it fresh longer.
We start a fire using shavings and small twigs, because increased surface area results in an increased reaction
rate, which allows us to start the fire much more easily. A catalyst, such as palladium, is put in the fire heads of
portable heaters, which are used in confined spaces, such as tents. The catalyst allows the reaction to take place
more readily, thus reducing the need for an open flame to provide the energy to keep the reaction self-sustaining.
Students may provide a variety of examples, based on their personal experiences and the examples discussed
in class.
DIF:
average
REF: C, MC
LOC: EC 3.04, EC 1.04
67. ANS:
Catalysts are important in industries because they make some reactions, which might otherwise be too expensive,
efficient and financially possible. Catalysts also help to increase the rate at which products can be produced, and
subsequently sold, in order to be profitable. For example, platinum is used in the hydrogenation of unsaturated
fats and oils in the food industry. Oil is converted to a solid when the process of hydrogenation reduces the
number of unsaturated hydrocarbons (oil to margarine). Platinum or palladium is used in the industrial process of
hydrogenating ethene.
Students may provide other examples to illustrate their answers.
DIF:
average
REF: C, MC
LOC: EC 3.03, EC 1.04
68. ANS:
a) Enzymes are enormous protein molecules that catalyse biological processes. Enzymes are important because
they make possible many functions in the human body, such as digestion and metabolism.
b) One example of an enzyme that is important in the human body is lactase, which catalyses the breakdown of
lactose, a sugar in milk. People who are lactose-intolerant are usually missing lactase and have to take
supplements that contain lactase. Another example is amylase, an enzyme in saliva. Amylase is important in
the breakdown of carbohydrates into simple sugars in digestion.
DIF:
average
REF: C, MC
LOC: EC 3.04
69. ANS:
The rate of a chemical reaction involving a solid reactant is affected by the amount of surface area that is exposed
to the other reactant(s). Grain dust has a much larger surface area than grain. Thus, if a reaction is initiated, it will
proceed at a very fast rate, because a large number of collisions can quickly occur between the grain dust and
oxygen.
DIF:
easy
REF: C, MC
LOC: EC 1.04
70. ANS:
Fires can be started more easily using small twigs, because the increase in surface area results in an increased
reaction rate.
DIF:
easy
REF: C, MC
LOC: EC 1.04
71. ANS:
Food that is stored at room temperature decays quite quickly. Milk sours in a day, and fruit goes rotten after a few
days. The reaction rates of both these processes can be slowed down by a decrease in temperature. Hence, we
store food in a refrigerator or (for longer periods of time) in a freezer, in order to keep it fresh longer. Although a
decrease in temperature decreases the reaction rate, it does not stop the reaction. Thus, food that is left in a
refrigerator will eventually spoil.
DIF:
average
REF: C, MC
LOC: EC 3.04, EC 1.04