Unit 3 Review, pages 406–413
Knowledge
1. (a)
2. (c)
3. (c)
4. (d)
5. (c)
6. (d)
7. (d)
8. (b)
9. (d)
10. (d)
11. (d)
12. (d)
13. False. The terms “thermal energy” and “temperature” do not mean the same thing.
14. False. Much of the energy emitted from the Sun is released during the nuclear process
called fusion.
15. True
16. False. The quantity of energy required to break a chemical bond is known as bond
dissociation energy.
17. True
18. True
19. False. Electricity generated by wind turbines is slightly more expensive than
electricity generated by other sources.
20. True
21. False. Reactant concentration(s) can be used to express the rate law equation.
Understanding
22. Kinetic energy is the energy of motion. Potential energy is energy based on position
or composition.
23. (a) Combustion of methane gas to heat a home is an open system.
(b) Combustion of methane gas in a sealed bomb calorimeter is a closed system.
(c) Water boiling in a kettle with a closed lid is a closed system.
(d) An acid-base neutralization reaction in a sealed flask is a closed system.
24. (a) Splitting a large gas molecule into smaller gas molecules is an endothermic
reaction.
(b) Forming a cation from an atom in the gas phase is an endothermic reaction.
(c) Mixing elemental sodium and chlorine to form table salt is an exothermic reaction.
(d) Nuclear fission is an exothermic reaction.
25. Nuclear fusion is the process of building large atoms by combining nuclei; nuclear
fission is the process of splitting nuclei of large atoms to form smaller atoms.
26. Answers may vary. Sample answer: Three examples of endothermic reactions are
cooking an egg, photosynthesis, and the synthesis of NO(g) from N2(g) and O2(g).
27. Answers may vary. Sample answer: Three examples of exothermic reactions are the
combustion of methane, cellular respiration, and nuclear fission.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-4
28. The change in enthalpy is the difference in chemical potential energy between the
reactants and the products. It is consistent with the law of conservation of energy because
the amount of energy released or absorbed by the reaction is exactly equal to the change
in potential energy, so no energy is lost or gained.
29. Bond energies are used to find the approximate enthalpy change of a reaction. The
sum of the energies required to break the bonds in the reactants plus the sum of the
energies released by forming the bonds in the products (expressed as a negative quantity)
equals the approximate enthalpy change of the reaction.
30. Forming any bond releases energy. Breaking a bond requires energy. So breaking a
P−Cl bond releases more energy than forming a N–O bond does.
31. More energy needed to form and break triple bonds than single or double bonds
because the attractions between atoms are stronger in triple bonds than they are in single
or double bonds.
32. The two rules you need to apply when calculating enthalpy changes using Hess’s law
are:
1) If you reverse a chemical reaction, you must also reverse the sign of ΔH; and,
2) The magnitude of ΔH is directly proportional to the number of moles of reactants and
products in a reaction. If the coefficients in a balanced equation are multiplied by a
factor, the value of ΔH is multiplied by the same factor.
33. The standard enthalpy of formation of a compound is the change in enthalpy that
accompanies the formation of 1 mol of the compound from its elements, with all
elements in their standard states.
34. The three main forms of fossil fuels used for energy are coal, petroleum, and natural
gas. The primary concern with using fossil fuels as energy sources is that they are a
nonrenewable resource.
35. Answers may vary. Sample answer: Three physical properties that can change during
a reaction and that may be used to measure the rate of a reaction are pH of a solution,
volume of gas, and colour.
36. (a) The chemical nature of the reactants (in this case the differing reactivities) is the
factor that causes copper but not gold jewellery to turn green over time.
(b) Temperature is the factor that affects how quickly milk left out on the counter will
turn sour.
(c) Papain is sometimes added to meat to make it more tender because is a catalyst (it
accelerates the breaking down of peptide bonds).
(d) The greater surface area of the dust in grain silos makes it explosive, whereas kernels
of grain are almost non-flammable.
(e) The different concentrations is what makes pure acetic acid burn skin on contact but
vinegar safe to add to food and consume.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-5
37. Chart formats and examples may vary. Sample answer:
Factors That Affect Reaction Rate
Factor
nature of
reactants
temperature
of the
reaction
system
concentration
of reactants
surface area
of reactants
presence of a
catalyst
Effect
Reactivity of substances differs depending on state of
matter, bond type, bond strength, and number of
bonds (molecular size).
Higher temperature increases the kinetic energy of
the reactant entities so entities move faster and with
more energy; the probability of collisions increases
and more entities have enough energy to break their
bonds and form an activation complex.
Higher concentration increases the probability of
collisions between reactant molecules, and thus of
effective collisions.
Example
Sodium reacts with
water but gold does
not.
Chemical changes
occur more rapidly
at higher
temperatures when
cooking.
Concentrated acids
react with metal
faster than dilute
acids.
Increasing surface area increases the number of sites Powdered sugar
where reactants can collide, and thus the probability
dissolves faster
of effective collisions.
than sugar cubes.
A catalyst provides an alternative pathway for the
Sulfur and nitrogen
reaction that has a lower activation energy, so a much oxides break down
greater fraction of the collisions are successful at a
in a catalytic
given temperature.
converter.
38. Biocatalysts support the tenets of green chemistry because biocatalysts are generally
more efficient and usually less toxic than other catalysts.
39. Answers may vary. Sample answer: The order of reaction with respect to a reactant is
the exponent to which the concentration of that reactant is raised in the rate law equation
for the reaction. It tells us how the reaction rate is proportional to the initial concentration
of the reactant. The orders of reaction in the rate law equation tell you which reactants are
taking part in the rate-determining step, which can help you to work out a possible
mechanism. If more than one reactant is present in a reaction, the sum of the orders of
reaction of the reactants (the exponents in the rate law equation) is called the overall
reaction order; for example, in the reaction A + B → C, if the reaction is first order with
respect to A and second order with respect to B, the overall reaction order is 1 + 2 = 3.
40. An elementary step of a reaction mechanism is a single reaction that occurs during the
overall reaction. An elementary step involves a one-, two-, or three-entity collision that
cannot be explained by simpler reactions. The rate-determining step is the slowest
elementary step and is the step in a reaction mechanism that determines the rate of the
overall reaction.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-6
Analysis and Application
41. Answers may vary. Sample answer: Photosynthesis is an endothermic reaction that
we encounter in everyday life. The source of energy for the enthalpy change is sunlight.
42. Answers may vary. Sample answer:
Transfer of Energy when a Pan of Water Is Placed over a Hot Campfire
As the wood burns,
chemical potential
energy stored in the
wood is transformed
into thermal energy
and light energy.
Thermal energy is transferred
to the metal pan by
conduction, convection, and
radiation. The kinetic energy of
the vibrating particles in the
metal increases and the
temperature of the metal rises.
Thermal energy is transferred from the pan to
the water by conduction. The kinetic energy of
the water molecules increases. When the
boiling point is reached, continued heating
causes the bonds between particles to begin
to break. Potential energy increases as the
water evaporates.
43. Given: m = 1.5 kg = 1500 g; c = 4.18 J/(g·°C); Tinitial = 20 º C; Tfinal = 75 º C
Required: q
Analysis: q = mc!T
Solution: q = mc!T
# 4.18 J &
q = (1500 g) %
(75 °C ) 20 °C)
$ g " °C ('
# 4.18 J &
= (1500 g ) %
( (55 °C )
$ g " °C '
= 3.4 * 105 J
q = 340 kJ
Statement: The amount of thermal energy that is required to heat 1.5 kg of water from
20 °C to 75 °C is 3.4 ! 105 J or 340 kJ.
44. Given: !H vap = 20.7 kJ/mol ; mCl = 2.25 kg = 2250 g
2
Required: ΔH
Analysis: !H = n!H vap ; n =
m
M
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-7
" 1 mol Cl
2
Solution: !H = (2250 g Cl2 ) $
$# 70.90 g Cl2
% " 20.7 kJ
'$
'& $# mol Cl2
%
'
'&
!H = 657 kJ
Statement: The enthalpy change during the vaporization of 2.25 kg of elemental chlorine
is 657 kJ.
45. Given: VH O(l) = 200.0 mL ; d H O(l) = 1.00 g/mL ; c = 4.18 J/(g·°C); Tinitial 1 = 22.1 °C;
2
2
Tfinal 1 = 26.8 °C; mmetal = 5.1 g; Tinitial 2 = 48.6 °C
Required: c of metal
Analysis: Use q = mc!T to determine the amount of thermal energy required to heat the
q
and substitute the calculated value for q and the given
m!T
values to determine the specific heat capacity of the metal.
Solution: q = mc!T
metal. Rearrange as c =
# 4.18 J &
= (200.0 g) %
(26.8 °C ) 22.1 °C)
$ g " °C ('
# 4.18 J &
= (200.0 g ) %
( (4.7 °C )
$ g " °C '
q = 3929 J (two extra digits carried)
c=
=
q
m!T
3929 J
(5.1 g)(48.6 °C " 26.8 °C)
c = 35 J/(g # °C)
Statement: The specific heat capacity of the metal is 35 J/(g ⋅ °C) .
46. Given: mNaOH(s) = 0.40 g; VH O(l) = 100 mL ; d H O(l) = 1.00 g/mL ; c = 4.18 J/(g·°C);
2
2
Tinitial = 20.02 °C; Tfinal = 21.12 °C
Required: !H sol
Analysis: q = mc!T ; n =
m
; !H = n!H sol
M
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-8
Solution: qsurroundings = mc!T
# 4.18 J &
= (100 g H 2O(l)) %
(21.12 °C ) 20.02 °C)
$ g " °C ('
# 4.18 J &
= (100 g ) %
( (1.10 °C )
g
"
°C
$
'
qsurroundings = 459.8 J (two extra digits carried)
Since the final temperature is higher than the initial temperature, the system transfers
thermal energy to the surroundings and so the reaction is exothermic.
!H system = –459.8 J
Convert enthalpy change to molar enthalpy change using the molar mass, M.
mNaOH(s) = 0.40 g
M NaOH(s) = M Na + M O + M H
= (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol)
M NaOH(s) = 40.00 g/mol
nNaOH(s) =
=
mNaOH(s)
M NaOH(s)
0.40 g
40.00 g/mol
nNaOH(s) = 0.010 mol
!H
n
–459.8 J
=
0.010 mol
!H sol = 46 kJ/mol
Statement: The molar enthalpy change of the dissolution reaction of sodium hydroxide
in water
is –46 kJ/mol.
47. (a) Given: 60.0 mL of 0.700 mol/L NaOH(aq); 40.0 mL of H2SO4(aq); ΔT = 5.6 °C;
c = 4.18 J/(g·°C)
Required: !H neut NaOH
!H sol =
Analysis: q = mc!T ; !H = n!H neut
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-9
Solution: qsurroundings = mc!T
# 4.18 J &
= (100.0 g ) %
( (5.6 °C )
$ g " °C '
qsurroundings = 2341 J (two extra digits carried)
Since the temperature increased, the system transferred thermal energy to the
surroundings and so the reaction is exothermic.
!H system = –2341 J
Convert enthalpy change to molar enthalpy change.
! 0.700 mol NaOH $
n = (60 mL NaOH(aq) ) #
&
#" 1000 mL NaOH(aq) &%
n = 0.042 mol NaOH
!H
!H neut =
n
–2341 J
=
0.042 mol
!H neut = –56 kJ/mol
Statement: The molar enthalpy of neutralization for sodium hydroxide is –56 kJ/mol.
(b) The assumptions made in the answer to (a) are:
• Any thermal energy transferred from the calorimeter to the outside environment is
negligible.
• Any thermal energy absorbed by the calorimeter itself is negligible.
• All dilute, aqueous solutions have the same density (1.00 g/mL) and specific heat
capacity (4.18 J/(g·°C)) as water.
48. Given: VH O(l) = 2.00 L = 2000 mL ; d H O(l) = 1.00 g/mL ; c = 4.18 J/(g·°C);
2
2
Tinitial = 26.5 °C; Tfinal = 100.0 °C;
Required: q
Analysis: q = mc!T
Solution: q = mc!T
# 4.18 J &
= (2000 g) %
(100 º C ) 26.5 º C)
$ g"ºC ('
# 4.18 J &
= (2000 g ) %
( (73.5 °C )
$ g " °C '
q = 614 kJ
Statement: The amount of energy that was transferred to the water from sunlight was
614 J.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-10
49. Given: mH O(l) = 255 g ; mC H O(l) = 1.01 g ; cH O(l) = 4.18 J/(g!ºC) ;
2
3
6
2
cCu(s) = 0.385 J/(g!ºC) ; ΔT = +28.8 °C; mcalorimeter = 305 g
Required: !H c C3H 6O (l)
Analysis: q = mc!T ; n =
m
; !H = n!H c
M
Solution:
qsurroundings = mc!T
# 4.18 J &
# 0.385 J &
= (255 g H 2O(l)) %
( (28.8 °C ) + (305 g ) %
( (28.8 °C )
$ g " °C '
$ g " °C '
qsurroundings = 34 080 J (two extra digits carried)
Since the final temperature is higher than the initial temperature, the system transfers
thermal energy to the surroundings and so the reaction is exothermic.
!H system = –34 080 J
Convert enthalpy change to molar enthalpy change.
! 1 mol propanol $
n = (1.01 g propanol ) #
&
" 58 g propanol %
n = 0.0174 mol (two extra digits carried)
!H
n
–34 080 J
=
0.0174 mol
!H c = "1.96 # 103 kJ/mol
!H c =
Statement: The enthalpy of combustion of propanol is −1.96 ×103 kJ/mol .
50. Given: H2(g) + Cl2(g) → 2 HCl(g)
for H2(g): nH−H = 1 mol; DH−H = 432 kJ/mol;
for Cl2(g): nCl−Cl = 1 mol; DCl−Cl = 239 kJ/mol;
for: HCl(g): nH−Cl = 2 mol; DH−Cl = 427 kJ/mol
Required: ΔH
Analysis: !H = "n # Dbonds broken $ "n # Dbonds formed
!H = (nH $ H DH $ H + nCl$Cl DCl$Cl ) $ nH $Cl DH $Cl
Solution: !H = (nH " H DH " H + nCl"Cl DCl"Cl ) " nH "Cl DH "Cl
= (1 mol # DH " H + 1 mol # DCl"Cl ) " 2 mol # DH "Cl
*$
kJ ' $
kJ ' - $
kJ '
= ,& 1 mol # 432
+
1
mol
#
239
) &
) / " & 2 mol # 427
)
mol ( %
mol ( /. %
mol (
,+%
!H = " 183 kJ
Statement: The enthalpy change for the reaction is –183 kJ.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-11
51. (a) Given: H2O2(g) → H2O(l) + O2(g)
Required: ΔH
Solution: Step 1. Balance the chemical equation.
1
H2O2(g) → H2O(l) + O2(g)
2
Step 2. For each reactant and product, identify the number of bonds per mole, the amount
of bonds in the reaction, and the bond energy per mole.
reactants
products
Substance
H2O2(g)
H2O(l)
O2(g)
Number of bonds per
mole (nsubstance)
1 mol O–O bonds
2 mol O–H bonds
2 mol O–H bonds
1 mol O=O bonds
Amount of
bonds in
reaction
1 mol
2 mol
2 mol
0.5 mol
Bond
energy per
mole
146 kJ/mol
467 kJ/mol
467 kJ/mol
495 kJ/mol
Step 3: Calculate the enthalpy change, ΔH, of the reaction.
!H = "n # Dbonds broken $ "n # Dbonds formed
!H = (nDO$O + nDO$ H ) $ (nDO$ H + nDO=O )
Add the total energy absorbed to break the bonds in the reactants.
!n " Dbonds broken = (nDO#O + nDO# H )
$
kJ ' $
kJ '
= & 1 mol " 146
) + & 2 mol " 467
)
%
mol ( %
mol (
!n " Dbonds broken = 1080 kJ
Add the total energy released when the bonds of products form.
!n " Dbonds formed = (nDO=O + nDO# H )
$
kJ ' $
kJ '
= & 0.5 mol " 495
) + & 2 mol " 467
)
%
mol ( %
mol (
!n " Dbonds formed = 1181.5 kJ
Subtract the energy released when the bonds of the products form from the energy
absorbed to break the bonds of the reactants.
!H = "n # Dbonds broken $ "n # Dbonds formed
= 1080 kJ $ 1181.5 kJ
!H = –102 kJ
The sign of the enthalpy change is negative.
This is an exothermic reaction and energy is released.
Statement: The enthalpy change is –102 kJ.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-12
(b) Given: CH3OH(aq) + O2(g) → CO2(g) + H2O(l)
Required: ΔH
Solution: Step 1. Balance the chemical equation.
3
CH3OH(aq) + O2(g) → CO2(g) + 2 H2O(l)
2
Step 2. For each reactant and product, identify the number of bonds per mole, the amount
of bonds in the reaction, and the bond energy per mole.
reactants
products
Substance
CH3OH(aq)
O2(g)
CO2(g)
H2O(l)
Number of bonds per
mole (nsubstance)
3 mol C–H bonds
1 mol C–O bonds
1 mol O–H bonds
1 mol O=O bonds
2 mol C=O bonds
2 mol O–H bonds
Amount of bonds
in reaction
3 mol
1 mol
1 mol
1.5 mol
2 mol
4 mol
Bond energy
per mole
413 kJ/mol
358 kJ/mol
467 kJ/mol
495 kJ/mol
799 kJ/mol
467 kJ/mol
Step 3: Calculate the enthalpy change, ΔH, of the reaction.
!H = "n # Dbonds broken $ "n # Dbonds formed
!H = (nDC$ H + nDC$O + nDO$ H + nDO=O ) $ (nDC=O + nDO$ H )
Add the total energy absorbed to break the bonds in the reactants.
!n " Dbonds broken = (nDC# H + nDC#O + nDO# H + nDO=O )
$
kJ ' $
kJ '
= & 3 mol " 413
) + & 1 mol " 358
)
%
mol ( %
mol (
$
kJ ' $
kJ '
+ & 1 mol " 467
) + & 1.5 mol " 495
)
%
mol ( %
mol (
!n " Dbonds broken = 2806.5 kJ
Add the total energy released when the bonds of products form.
!n " Dbonds formed = (nDC=O + nDO# H )
$
kJ ' $
kJ '
= & 2 mol " 799
+
4
mol
"
467
) &
)
%
mol ( %
mol (
!n " Dbonds formed = 3466 kJ
Subtract the energy released when the bonds of the products form from the energy
absorbed to break the bonds of the reactants.
ΔH = Σn × Dbonds broken − Σn × Dbonds formed
= 2806.5 kJ − 3466 kJ
ΔH = –660 kJ
The sign of the enthalpy change is negative.
This is an exothermic reaction and energy is released.
Statement: The enthalpy change is –660 kJ.
(c) Given: CHF2CHF2(g) → CHCH(g) + 2 F2(g)
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-13
Required: ΔH
Solution: Step 1. Balance the chemical equation.
CHF2CHF2(g) → CHCH(g) + 2 F2(g)
Step 2. For each reactant and product, identify the number of bonds per mole, the amount
of bonds in the reaction, and the bond energy per mole.
reactants
Substance
CHF2CHF2(g)
products
CHCH(g)
F2(g)
Number of bonds per
mole (nsubstance)
1 mol C–C bonds
4 mol C–F bonds
2 mol C–H bonds
2 mol C–H bonds
1 mol C≡C bonds
1 mol F–F bonds
Amount of
bonds in
reaction
1 mol
4 mol
2 mol
2 mol
1 mol
2 mol
Bond
energy per
mole
347 kJ/mol
485 kJ/mol
413 kJ/mol
413 kJ/mol
839 kJ/mol
154 kJ/mol
Step 3: Calculate the enthalpy change, ΔH, of the reaction.
!H = "n # Dbonds broken $ "n # Dbonds formed
!H = (nDC$C + nDC$ F + nDC$ H ) $ (nDC$ H + nDC%C + nDF$ F )
Add the total energy absorbed to break the bonds in the reactants.
!n " Dbonds broken = (nDC#C + nDC# F + nDC# H )
$
kJ ' $
kJ ' $
kJ '
= & 1 mol " 347
) + & 4 mol " 485
) + & 2 mol " 413
)
%
mol ( %
mol ( %
mol (
!n " Dbonds broken = 3113 kJ
Add the total energy released when the bonds of products form.
!n " Dbonds formed = (nDC–H + nDC#C + nDF$ F )
%
kJ ( %
kJ ( %
kJ (
= ' 2 mol " 413
* + ' 1 mol " 839
* + ' 2 mol " 154
*
&
mol ) &
mol ) &
mol )
!n " Dbonds formed = 1973 kJ
Subtract the energy released when the bonds of the products form from the energy
absorbed to break the bonds of the reactants.
!H = "n # Dbonds broken $ "n # Dbonds formed
= 3113 kJ $ 1973 kJ
!H = 1140 kJ
The sign of the enthalpy change is positive.
This is an endothermic reaction and energy is absorbed.
Statement: The enthalpy change is 1140 kJ.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-14
52. Given: 4 CH 3NO 2 (g) + 3 O 2 (g) ! 4 CO 2 (g) + 2 N 2 (g) + 6 H 2O(g) and
(1) C(g) + O 2 (g) ! CO 2 (g)
"H = #393.5 kJ
(2) 2 H 2 (g) + O 2 (g) ! 2 H 2O(g)
"H = #483.6 kJ
(3) 2 C(g) + 3 H 2 (g) + 2 O 2 (g) + N 2 (g) ! 2 CH 3NO 2 (g) "H = –226.2 kJ
Required: !H
Solution: Step 1. Since this is a combustion reaction, rewrite the equation for the
combustion of nitromethane so that nitromethane has a coefficient of 1:
3
1
3
CH 3NO 2 (g) + O 2 (g) ! CO 2 (g) + N 2 (g) + H 2O(g)
4
2
2
1
Step 2. Reverse equation 3 and multiply it and its ΔH by .
2
3
Multiply equation 2 and its ΔH by .
4
Then, add the three equations and their changes in enthalpies.
C(g) + O 2 (g) ! CO 2 (g)
"H = #393.5 kJ
3
3
3
H 2 (g) + O 2 (g) ! H 2O(g)
2
4
2
"H = #362.7 kJ
3
1
H 2 (g) + O 2 (g) + N 2 (g) "H = 113.1 kJ
2
2
3
1
3
CH 3NO 2 (g) + O 2 (g) ! CO 2 (g) + N 2 (g) + H 2O(g)
"H = #643.1 kJ
4
2
2
CH 3NO 2 (g) ! C(g) +
Statement: The enthalpy change for the combustion of nitromethane is –643.1 kJ per
mole of nitromethane.
53. (a) The balanced chemical equation for the combustion of glucose is
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
(b) Given: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l); from Table 1,
!H fo C H O (s) = "1273.1 kJ/mol; !H fo CO (g) = "393.5 kJ/mol; ΔH fo H2O(l) = −285.8 kJ/mol ;
6
12
6
2
Required: !H for the combustion reaction
o
r
Analysis: !H r! = " nproducts !H !products # " nreactants !H !reactants
Since O2(g) is in its standard state, you can rewrite the equation as
ΔH ro = ⎡⎣nCO2 (g) ΔHfo CO2 (g) + nH2O(l) ΔHfo H2O(l) ⎤⎦ − nC6H12O6 (s) ΔHfo C6H12O6 (s)
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-15
Solution: Glucose has a coefficient of 1 in the balanced equation so you can use the
equation as is.
Insert the appropriate values into the equation for standard enthalpy of formation and
solve.
ΔH ro = ⎡⎣ nCO2 (g) ΔH fo CO2 (g) + nH2O(l) ΔH fo H2O(l) ⎤⎦ − nC6 H12O6 (s) ΔH fo C6 H12O6 (s)
= 6(−393.5 kJ) + 6(−285.8 kJ) − (−1273.1 kJ)
ΔH ro = − 2802.7 kJ/mol
Statement: The enthalpy change of the combustion of glucose is –2802.7 kJ.
(c) Given: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
Required: ΔH
Solution: Step 1. For each reactant and product in the balanced equation, identify the
number of bonds per mole, the amount of bonds in the reaction, and the bond energy per
mole.
Amount of
Bond
Number of bonds per
bonds in
energy per
Substance
mole (nsubstance)
reaction
mole
reactants C6H12O6(s)
5 mol C–C bonds
5 mol
347 kJ/mol
7 mol C–O bonds
7 mol
358 kJ/mol
5 mol O–H bonds
5 mol
467 kJ/mol
7 mol C–H bonds
7 mol
413 kJ/mol
O2(g)
1 mol O=O bonds
6 mol
495 kJ/mol
products
CO2(g)
2 mol C=O bonds
12 mol
799 kJ/mol
H2O(g)
2 mol O–H bonds
12 mol
467 kJ/mol
Step 2: Calculate the enthalpy change, ΔH, of the reaction.
!H = "n # Dbonds broken $ "n # Dbonds formed
!H = (nDC$C + nDC$O + nDO$ H + nDC$ H + nDO%O ) $ (nDC=O + nDO$ H )
Add the total energy absorbed to break the bonds in the reactants.
!n " Dbonds broken = (nDC#C + nDC#O + nDO# H + nDC# H + nDC=O + nDO$O )
%
kJ ( %
kJ ( %
kJ (
= ' 5 mol " 347
* + ' 7 mol " 358
* + ' 5 mol " 467
*
&
mol ) &
mol ) &
mol )
%
kJ ( %
kJ (
+ ' 7 mol " 413
* + ' 6 mol " 495
*
&
mol ) &
mol )
!n " Dbonds broken = 12 437 kJ
Add the total energy released when the bonds of products form.
!n " Dbonds formed = (nDC=O + nDO# H )
$
kJ ' $
kJ '
= & 12 mol " 799
) + & 12 mol " 467
)
%
mol ( %
mol (
!n " Dbonds formed = 15 192 kJ
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-16
Subtract the energy released when the bonds of the products form from the energy
absorbed to break the bonds of the reactants.
!H = "n # Dbonds broken $ "n # Dbonds formed
= 12 437 kJ $ 15 192 kJ
!H = –2755 kJ
The sign of the enthalpy change is negative.
This is an exothermic reaction and energy is released.
Statement: The enthalpy change of the combustion of glucose is –2755 kJ.
(d) Given: VH O(l) = 100.0 mL ; c = 4.18 J/(g·°C); ΔT = 37.0 °C; mC H O (s) = 1.00 g
2
6
12
6
Required: ΔH
m
; !H = n!H sol
M
= mc!T
Analysis: q = mc!T ; n =
Solution: qsurroundings
# 4.18 J &
= (100.0 g H 2O(l) ) %
( (37.0 °C )
$ g " °C '
qsurroundings = 15.466 kJ (two extra digits carried)
Since the temperature increased, the system transfers thermal energy to the surroundings
and so the reaction is exothermic.
!H system = –15.466 kJ
Convert enthalpy change to molar enthalpy change using the molar mass, M.
mC H O (s) = 1.00 g
6
12
MC H
6
6
12 O6 (s)
= 6 M C + 12 M H + 6 M O
= 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol)
MC H
6
nC H
6
12 O6 (s)
12 O6 (s)
=
= 180.18 g/mol
mC H
6
MC H
6
=
nC H
6
12 O6 (s)
12 O6 (s)
12 O6 (s)
1.00 g
180.18 g/mol
= 0.00555 mol (two extra digits carried)
!H
n
–15.466 kJ
=
0.00555 mol
!H sol = "2790 kJ/mol
Statement: The enthalpy change of the combustion of glucose is !2790 kJ/mol .
!H sol =
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-17
(e) The answers to (b), (c), and (d) are different because the calculations use average
values and because there is some experimental error in the calorimetry experiment.
54. Three different ways to express the standard molar enthalpy of formation for vinyl
chloride gas are:
2C(s) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g) + 37.3 kJ
2 C(s) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g) !H o = –37.3 kJ
55. (a) A thermochemical equation for the formation of sulfur dioxide gas is
S(s) + O2(g) → SO2(g) ΔH = –298.6 kJ
(b)
(c) Given: !H fo = "296.8 kJ/mol ; mSO
2 (g)
= 9.63 g
Required: ΔH
Analysis: !H = n!H fo ; n =
m
M
" 1 mol SO (g) % " (296.8 kJ %
2
Solution: !H = (9.63 g SO 2 (g) ) $
'$
'
$# 64.07 g SO 2 (g) '& $# mol SO 2 (g) '&
!H = 44.6 kJ
Statement: The reaction releases 44.6 kJ of thermal energy.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-18
56. (a) An equation for the combustion of 1 mol of pentane gas is
C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l)
(b) Given: C5H12(g) + 8 O2 → 5 CO2(g) + 6 H2O(l); from Table 1,
!H fo C H (g) = "146 kJ/mol; !H fo CO (g) = "393.5 kJ/mol; !H fo H O(l) = "285.8 kJ/mol ;
5
12
2
2
Required: !H for the combustion reaction
o
r
Analysis: !H r! = " nproducts !H !products # " nreactants !H !reactants
Since O2(g) is in its standard state, you can rewrite the equation as
!H ro = " nCO (g) !H fo CO (g) + nH O(l) !H fo H O(l) $ & nC H (g) !H fo C H (g)
# 2
2
2
2
%
5 12
5 12
Solution: Pentane gas has a coefficient of 1 in the balanced equation so you can use the
equation as is.
Insert the appropriate values into the equation for standard enthalpy of formation and
solve.
!H ro = " nCO (g) !H fo CO (g) + nH O(l) !H fo H O(l) $ & nC H (g) !H fo C H (g)
# 2
2
2
2
%
5 12
5 12
= 5(&393.5 kJ) + 6(&285.8 kJ) & (&146 kJ)
!H ro = & 3536.3 kJ/mol
Statement: The enthalpy change of the combustion of pentane is –3536.3 kJ.
(c) Given: !H c = "3534.5 kJ/mol ; mSO (g) = 2.0 g
2
Required: ΔH
Analysis: !H = n!H c ; n =
m
M
" 1 mol C H (g) % " (3530 kJ %
5 12
Solution: !H = (20 g C5 H12 (g) ) $
'$
'
$# 72.17 g C5 H12 (g) '& $# mol C5 H12 (g) '&
!H = 1000 kJ
Statement: The combustion of 20 g of pentane would release 1000 kJ of thermal energy.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-19
57. Answers may vary. Sample answer:
Comparison of Five Different Uses of Fossil Fuels
Fossil
fuel
petroleum
coal
natural
gas
natural
gas
petroleum
Possible
alternative fuel
syngas
Use
gasoline for
transportation,
small motors
electricity
wind energy
electricity
hydro power
heating and
cooling
diesel fuel for
transportation
Advantage(s) of
alternative
more available
than petroleum
Disadvantage(s) of
alternative
emits greenhouse
gases
no emissions
no emissions;
renewable
limited availability
limited availability;
environmental
concerns
limited availability
at times
impact on food
supply
solar energy
renewable
biodiesel
renewable;
reduced
greenhouse gases
58. Answers may vary. Sample answer:
Advantages and Disadvantages of Wind Energy
Advantages
Social factors:
– can offer energy independence to
individuals or small towns
Economic factors:
– can be located close to where it is needed
– in the long term, may cost less than fossil
fuels because damage to environment may
be less
Environmental factors:
– renewable
– no greenhouse gases are created
– no pollution
Disadvantages
Social factors:
– distracting noise and motion
Economic factors:
– less efficient than fossil fuels
– in the short term, more expensive
per unit of energy than fossil fuels
– requires large land or water areas
– needs consistent wind
Environmental factors:
– can interfere with flying or
swimming animals
59. For the reaction represented by the balanced equation
CO(g) + NO2(g) → CO2(g) + NO(g)
(a) The sum of the two concentrations must equal 0.10 mol/L, since CO(g) becomes
CO2(g) in the reaction, so the missing values are:
Time (s)
0
20
40
60
80
100
[CO(g)] (mol/L)
0.100
0.050
0.033
0.026
0.020
0.017
Copyright © 2012 Nelson Education Ltd.
[CO2(g)] (mol/L)
0.000
0.050
0.067
0.074
0.080
0.083
Unit 3: Energy Changes and Rates of Reaction
U3-20
(b) The rate of NO2(g) consumption is the same as the rate of CO(g) consumption. If the
initial concentration of nitrogen dioxide gas is 0.250 mol/L, [NO2(g)] after 80 s is
0.250 mol/L – 0.080 mol/L = 0.170 mol/L.
(c)
(d) Yes, the graph reflects the stoichiometry of the balanced equation, because [CO(g)]
decreases at the same rate that [CO2(g)] increases.
60. For the reaction represented by the balanced equation 2 X2O5(g) → 4 XO2(g) + O2(g):
(a)
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-21
(b)
[X2O5(g)]
Time (h) (mol/L)
0.0
1.20
2.0
0.80
4.0
0.55
7.0
0.30
12.0
0.10
[XO2(g)]
(mol/L)
0
0.80
1.30
1.68
2.20
[O2(g)]
(mol/L)
0
0.20
0.325
0.42
0.55
![X 2O5 (g)] [X 2O5 (g)]t =12.0 h " [X 2O5 (g)]t =0 h
=
!t
!t
0.10 mol/L " 1.20 mol/L
=
12.0 h " 0 h
![X 2O5 (g)]
= "0.092 mol/(L # h)
!t
![O 2 (g)] [O 2 (g)]t =12.0 h " [O 2 (g)]t =0 h
=
(ii) Rate of O2(g) formation:
!t
!t
0.55 mol/L " 0.00 mol/L
=
12.0 h " 0 h
![O 2 (g)]
= 0.046 mol/(L # h)
!t
![XO 2 (g)] [XO 2 (g)]t =12.0 h " [XO 2 (g)]t =0 h
=
(iii) Rate of XO2(g) formation:
!t
!t
2.20 mol/L " 0.00 mol/L
=
12.0 h " 0 h
![XO 2 (g)]
= 0.183 mol/(L # h)
!t
(c) (i) Rate of X2O5(g) consumption:
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-22
(d) Answers may vary, as graphs in (a) may vary. Sample answer:
To calculate the instantaneous rate of consumption of X2O5(g) at t = 2.0 h, use the points
(0 h, 1.08 mol/L) and (4.0 h, 0.52 mol/L) on the tangent to the [X2O5(g)] curve at 2.0 h.
rate instantaneous at t = 2.0 h = slope of the tangent at 2.0 h
!y
at 2.0 h
!x
1.08 mol/L " 0.52 mol/L
=
0 h " 4.0 h
= "0.14 mol/(L # h)
=
rate instantaneous at t = 2.0 h
The instantaneous rate of consumption of X2O5(g) at t = 2.0 h is 0.14 mol/L ⋅ h .
To calculate the instantaneous rate of consumption of X2O5(g) at t = 7.0 h, use the points
(3.0 h, 0.55 mol/L) and (10.0 h, 0.12 mol/L) on the tangent to the [X2O5(g)] curve at
7.0 h.
rate instantaneous at t =7.0 h = slope of the tangent at 7.0 h
!y
at 7.0 h
!x
0.55 mol/L " 0.12 mol/L
=
3.0 h " 10.0 h
= "0.061 mol/(L # h)
=
rate instantaneous at t =7.0 h
The instantaneous rate of consumption of X2O5(g) at t = 7.0 h is 0.061 mol/L ! h .
(e) The observed trend is that the rate of consumption of X2O5(g) decreases over time.
The rate of consumption of a reactant decreases as more and more of the reactant is
converted into products.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-23
61. Given: N2(g) + 2 O2(g) → 2 NO2(g); [NO2(g)]initial = 0.32 mol/L;
[NO2(g)]t = 3 min = 0.80 mol/L
Required: rate of NO2(g) production
Δ[NO 2 (g)] [NO 2 (g)]t =3 min − [NO 2 (g)]t =0 h
Solution:
=
Δt
Δt
0.80 mol/L − 0.32 mol/L
=
3 min − 0 min
Δ[NO 2 (g)]
= 0.2 mol/(L ⋅ min)
Δt
Statement: The overall rate of production of nitrogen dioxide is 0.2 mol/(L ! min) .
62. Given: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g); instantaneous rate of
consumption of NH3(g) is 2.0 × 10−2 mol/(L·s)
Required: (a) instantaneous rate of consumption of oxygen gas
(b) instantaneous rate of formation of water vapour
Analysis: Scale the rates by the inverse of their coefficients in the balanced chemical
equation. Then, substitute the given rate and solve for the required rates.
1 # "[NH 3 (g)] & 1 # "[O 2 (g)] &
Solution: (a) % !
( = 5%!
4$
"t
"t ('
'
$
"[O 2 (g)] 5
= (2.0 ) 10!2 mol/(L * s))
"t
4
"[O 2 (g)]
= 2.5 ) 10!2 mol/(L * s)
"t
1 # "[NH 3 (g)] & 1 # "[H 2O(g)] &
(b) % !
( = 6%
(
4$
"t
"t
'
$
'
"[H 2O(g)] 6
= (2.0 ) 10!2 mol/(L * s))
"t
4
"[H 2O(g)]
= 3.0 ) 10!2 mol/(L * s)
"t
Statement: (a) The instantaneous rate of consumption of oxygen gas is
2.5 ! 10"2 mol/(L # s) .
(b) The instantaneous rate of formation of water vapour is 3.0 ! 10"2 mol/(L # s) .
63. (a) According to the “Methane Gas Volume versus Time” graph in Figure 2,
![CH 4 (g)] [CH 4 (g)]t =7 s " [CH 4 (g)]t =3 s
=
!t
!t
18 mL " 35 mL
=
7.0 s " 3.0 s
![CH 4 (g)]
= "4.3 mL/s
!t
The reaction rate of the decomposition of methane between 3 s and 7 s is 4.3 mL/s.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-24
(b) According to the “Oxygen Gas Volume versus Time” graph in Figure 2,
![O 2 (g)] [O 2 (g)]t =6 s " [O 2 (g)]t = 2 s
=
!t
!t
36 mL " 13 mL
=
6.0 s " 2.0 s
![O 2 (g)]
= 5.8 mL/s
!t
The reaction rate of the production of oxygen gas between 2 s and 6 s is 5.8 mL/s.
(c) Methane is a reactant, because the graph shows that its concentration is decreasing;
oxygen is a product, because the graph shows that its concentration is increasing.
64. The reaction Pb2+(aq) + 2 Cl–(aq) → PbCl2(s) would have a higher reaction rate at
room temperature than the reaction Pb(s) + Cl2(g) → PbCl2(s) because the reactants are
dissolved ions which are free to move about in the solution and collide. A reaction
involving a solid like Pb(s) can only occur at the exposed surface, so the number of
collisions that can occur are limited.
65. The reason why containers of powdered aluminum metal must carry a warning that
the contents are dangerously combustible while many everyday objects made of
aluminum metal are not required to carry the same warning is that the powder is highly
combustible because the small particle size forms a very large surface area. Objects made
of bulk aluminum do not react quickly with oxygen.
66. (a) If 44.2 mL of carbon monoxide gas forms in 30.0 s, the rate of reaction with
respect to carbon monoxide gas production is
![CO(g)] 44.2 mL
=
!t
30.0 s
![CO(g)]
= 1.47 mL/s
!t
(b) (i) The temperature of run 1 was 25 ºC. If the temperature is increased to 30 ºC, the
reaction rate will increase because higher temperature increases the kinetic energy of the
reactant entities so entities move faster and with more energy; the probability of
collisions increases and more entities have enough energy to break their bonds and form
an activation complex.
(ii) In run 1, the solution of formic acid was catalyzed. If the reaction is performed
without the catalyst, the reaction rate will be slower because the activation energy will be
higher, so a smaller fraction of the collisions will be successful at a given temperature
(iii) If the reaction is performed using formic acid that is half as concentrated as in run 1,
the reaction rate will be slower because the probability of collisions between reactant
molecules, and thus of effective collisions, will decrease.
67. Hydrogen peroxide can be stored for months on the shelf but will bubble strongly
when applied to an open cut because there is a compound in blood that acts as a catalyst
for the decomposition of hydrogen peroxide.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-25
68. Answers may vary. Sample answer:
A collision between a hydrogen and a bromine molecule
that is likely to be an effective collision
69. When a magnesium strip is heated in a flame, the reaction is exothermic, and once the
strip is sufficiently heated, the reaction itself provides activation energy to entities, which
speeds up the reaction rate. A strip of magnesium metal can be safely stored at room
temperature for long periods of time but magnesium granules stored at room temperature
will react slowly with the oxygen in the air to form magnesium oxide because the
granules have a large surface area, so all of the magnesium reacts over time.
70. Answers may vary. Sample answer:
Advantages of Biocatalysts over Traditional Catalysts
Biocatalysts
naturally occurring
no toxic by-products
inexpensive
homogeneous
very efficient
Traditional catalysts
must be synthesized
generally toxic metals or salts
often expensive elements
generally heterogeneous
less efficient
71. A highly exothermic chemical reaction will affect the rate of this reaction as follows:
The reaction rate will increase after the reaction starts because the exothermic reaction
provides thermal energy that acts as activation energy.
72. In the reaction of calcium oxide and acetic acid, two ways of increasing the rate at
which calcium oxide dissolves are 1) increasing the concentration of acetic acid to
increase the probability of collisions between reactant molecules, and thus the probability
of effective collisions, and 2) heating the acetic acid to increase the kinetic energy so the
entities move faster and with more energy, increasing the probability of collisions and
causing more entities to have enough energy to form an activation complex.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-26
73. A clock reaction involving the hypochlorite ion, ClO-(aq), is represented by
ClO–(aq) + I–(aq) → Cl–(aq) + IO–(aq)
If the rate is first order with respect to each of the reactants:
(a) (i) when the initial [ClO–(aq)] is doubled, the initial rate will double.
(ii) when the initial [I–(aq)] is halved, the initial rate will halve.
(iii) when the same initial numbers of moles of reactants are placed in a container with
half the volume of water, the initial rate will quadruple because both concentrations
double.
(b) Answers may vary. Sample experiment to study the effect of change in temperature
using the iodine clock reaction in Investigation 6.5.1:
Question: How does temperature affect the iodine clock reaction?
Prediction: The iodine clock reaction rate will increase with temperature.
Experimental design: In the iodine clock reaction, two colourless solutions are combined
and the combined solution initially stays colourless. After a certain amount of time, the
combined solution suddenly changes colour. By changing the reaction temperature, the
effects of temperature on reaction rate will be determined. The time from mixing to the
appearance of the blue-black product will be measured and then graphed.
Equipment and materials:
• chemical safety goggles
• lab apron
• 2 beakers, 250 mL
• 2 graduated cylinders, 10 mL
• 2 large test tubes
• thermometer
• water baths
• kettle
• ice cubes
• timer
• Solution A (contains iodate ions, IO3–(aq)
• Solution B (contains HSO3–(aq) and starch)
Procedure:
1. Put on your safety goggles and lab apron.
2. Prepare a water bath for the solutions. Fill two 250 mL beakers about two-thirds full
with water of a given temperature. For water baths below room temperature, use ice to
chill the water. For warmer baths, use warm water from the tap or a kettle. There
should be enough water in the beakers so that the solutions in the test tubes are well
beneath the water level in the baths.
3. Measure equal volumes of the two solutions, pouring each into its own test tube. Place
the two test tubes in the water baths, allowing them to remain for about 10 min to
allow the solution temperatures to reach the temperature of the water baths.
4. Use the thermometer to monitor water bath temperatures. Try to keep the water bath
temperature within 0.5 °C of the desired temperature. Add more ice or warm water as
necessary. Record the water bath temperature just before mixing the solutions.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-27
5. Once the solutions are at the desired temperatures, prepare to record time and mix the
two solutions. Pour the mixture back and forth between the two test tubes several
times to ensure mixing. Then, place the test tube containing the mixed solutions back
in the water bath. Record the time it takes from initial mixing until the blue-black
colour appears.
6. Follow steps 2 to 5 for at least 4 different temperatures; for example, 5 ºC, 15º C,
25 ºC, and 35 ºC.
7. Plot a graph of temperature versus reaction time.
Safety and disposal precautions: Wear safety glasses and apron. Discard chemicals
following school policy.
74. (a) Given: CO(g) + Cl2(g) → COCl2(g); reaction is first order with respect to
chlorine; k = 1.3 ! 10"2 L/(mol # s)
Required: the order of the reaction with respect to carbon monoxide
Solution: The units of k indicate that the reaction is a second order reaction overall, so
the order of the reaction with respect to carbon monoxide is 1.
(b) If you wanted to increase the reaction rate as much as you could by doubling the
initial concentration of one of the reactants, you could choose either of the reactants,
because the order of reaction is the same with respect to both reactants.
75. Given: zero-order reaction; [A]initial = 1.5 mol/L ; [A]t =120 s = 0.75 mol/L
Required: the rate constant, k
Analysis: In a zero-order reaction, the rate constant equals the rate, since
rate = k[A]0[B]0 = k.
Solution: k = rate
![A]
!t
0.75 mol/L " 1.5 mol/L
=
120 s – 0 s
k = 6.3 # 10"3 mol/(L $ s)
=
Statement: The rate constant, k, is 6.3 ! 10"3 mol/(L # s) .
76. Given: H3O+(aq) + OH–(aq) → 2 H2O(l); k = 1.0 ! 1011 L/(mol " s)
Required: the rate of the reaction when equal volumes of 0.10 mol/L solutions of
hydrochloric acid and sodium hydroxide are mixed.
Analysis: Since hydrochloric acid is a strong acid, the solutions will ionize completely,
and so the concentration of the hydronium ion, H3O+, and of HCl are actually the same.
Similarly, sodium hydroxide is a strong base and will ionize fully, so
[H 3O + (aq)] = [HCl(aq)] = 0.10 mol/L
[OH ! (aq)] = [NaOH(aq)] = 0.10 mol/L
The units of the rate constant are L/(mol ! s) , so the overall reaction order is 2. Hence, the
reaction is first order with respect to both H 3O + (aq) and OH ! (aq) . Thus, the rate law
equation is r = k[H 3O + (aq)][OH ! (aq)] . Substitute the given values in the equation.
Copyright © 2012 Nelson Education Ltd.
Unit 3: Energy Changes and Rates of Reaction
U3-28
Solution: rate = k[H 3O + (aq)][OH ! (aq)]
= [1.0 " 1011 L/(mol # s)](0.10 mol/L)(0.10 mol/L)
rate = 1.0 " 109 mol/(L # s)
Statement: The rate of the reaction is 1.0 ! 109 mol/(L " s) .
77. (a) The reaction C3H8(g) + 5 O2(g) = 3 CO2(g) + 4 H2O(g) is likely to occur in a
series of steps rather than in a single step because a collision involving 6 molecules is
extremely unlikely.
![C3H 3 (g)]
= –4 " 10#2 mol/(L $ s)
(b) Given: C3H8(g) + 5 O2(g) = 3 CO2(g) + 4 H2O(g);
!t
Required: the rate of reaction with respect to oxygen gas and carbon dioxide gas
Analysis: Scale the rates by the inverse of their coefficients in the balanced chemical
equation. Then, substitute the given rate and solve for the required rates.
"[C3H 3 (g)] 1 # "[O 2 (g)] &
= %!
Solution: !
"t
5$
"t ('
"[O 2 (g)]
= 5(–4 ) 10!2 mol/(L * s))
"t
"[O 2 (g)]
= !2 ) 10!1 mol/(L * s)
"t
"[C3H 3 (g)] 1 # "[CO 2 (g)] &
!
= %
(
"t
3$
"t
'
"[CO 2 (g)]
= !3(–4 ) 10!2 mol/(L * s))
"t
"[CO 2 (g)]
= 1 ) 10!1 mol/(L * s)
"t
Statement: The rate of reaction with respect to oxygen gas is 2 ! 10"1 mol/(L # s) , and the
rate of reaction with respect to carbon dioxide gas is 1 ! 10"1 mol/(L # s) .
78. (a) Given: proposed reaction mechanism
2 NO(g) ! N 2O 2 (g)
N 2O 2 (g) + H 2 (g) ! N 2O(g) + H 2O(g)
N 2O(g) + H 2 (g) ! N 2 (g) + H 2O(g)
Required: overall balanced equation
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Unit 3: Energy Changes and Rates of Reaction
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Solution:
2 NO(g) ! N 2O 2 (g)
N 2O 2 (g) + H 2 (g) ! N 2O(g) + H 2O(g)
N 2O(g) + H 2 (g) ! N 2 (g) + H 2O(g)
2 NO(g) + N 2O 2 (g) + H 2 (g) + N 2O(g) + H 2 (g)
! N 2O 2 (g) + N 2O(g) + H 2O(g) + N 2 (g) + H 2O(g)
Overall balanced equation: 2 NO(g) + 2 H 2 (g) ! N 2 (g) + 2 H 2O(g)
(b) The reaction intermediates are N2O2(g) and N2O(g).
(c) If Step 1 is the slow step, the rate law equation for this reaction is r = k[NO(g)]2.
79. (a) Given: 2 NO2Cl(g) → 2 NO2(g) + Cl2(g); experimental data provided in Table 4
Required: the rate law equation
Solution: The rate doubles when [NO2Cl(g)] doubles so the reaction is first order with
respect to NO2Cl(g), and the rate equation is r = k[NO2Cl(g)].
(b) Given: proposed reaction mechanisms; rate = k[NO2Cl(g)]
NO 2Cl ! NO 2 + Cl (slow)
(i)
NO 2Cl + Cl ! NO 2 + Cl2 (fast)
(ii) 2 NO 2Cl ! N 2O 2Cl2
(slow)
N 2O 2Cl2 ! 2 NO 2 + Cl2 (fast)
(iii)
NO 2Cl ! NO 2 + Cl (fast)
NO 2Cl + Cl ! NO 2 + Cl2 (slow)
Required: the incorrect mechanisms, using the rate law
Solution: (i) The first step is the rate-determining step, so the rate law equation
is r = k[NO2Cl(g)], which agrees with the experimentally determined rate law equation.
(ii) The first step is the rate-determining step, so the rate law equation is
r = k[NO2Cl(g)]2, which does not agree with the experimentally determined rate law
equation, so this mechanism is incorrect.
(iii) The second step is the rate-determining step, so the rate law equation is
r = k[NO2Cl(g)][Cl], which does not agree with the experimentally determined rate law
equation, so this mechanism is incorrect.
Statement: Proposed mechanisms (ii) and (iii) are incorrect because they do not agree
with the experimentally determined rate law equation.
(c) No, it is not appropriate to ask which mechanism is correct, because the data only
shows whether the proposed mechanism is consistent with the data. Other mechanisms
could also be consistent.
80. (a) Between run 1 and run 3, the rate quadruples when [NO(g)] doubles and [O2(g)]
remains constant, so the order of reaction with respect to NO(g) is 2.
(b) Between run 1 and run 2, the rate doubles when [O2(g)] doubles and [NO(g)] remains
constant, so the order of reaction with respect to O2(g) is 1.
(c) The total order of the reaction is 1 + 2 = 3.
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Unit 3: Energy Changes and Rates of Reaction
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(d) Given: proposed reaction mechanism; orders of reaction: NO(g) = 2, O2(g) = 1
NO(g) + O 2 (g) ! NO 2 (g) + O(g) (fast)
NO(g) + O(g) ! NO 2 (g)
(slow)
Required: agreement of mechanism with the experimentally determined rate law
Solution: The experimentally determined rate law is r = k[NO(g)]2[O2(g)].
The second step in the proposed mechanism is the rate-determining step, so the rate law
equation is r = k[NO(g)][O(g)], which does not agree with the experimentally determined
rate law equation.
Statement: The mechanism does not agree with the proposed rate law (ii) and (iii) are
incorrect because they do not agree with the experimentally determined rate law
equation.
Evaluation
81. Answers may vary. Sample answers:
(a) Some reasons why the per capita energy use of Canadians is one of the highest of any
country are that most of Canada has very cold winters so we use a lot of energy for heat,
the distance between populated places is large so we use a lot of energy to transport
goods and people from one place to another, and we are a relatively wealthy country, so
people tend to use purchase new clothing and other items (which take fuel to create) and
to use fuel for recreational purposes.
(b) To decrease our energy use, Canadians could keep buildings slightly less warm in
winter and slightly warmer in summer, increase efficiency of heating systems, improve
public transportation systems, add high taxes to fuels to discourage their use, carpool, and
walk or ride bicycles for short errands.
(c) The cold winters and the large distances between population centres in Canada affect
our energy use, but cannot be changed.
(d) I agree that the per capita energy use in Canada will always be one of the highest in
the world, because Canada has a fairly small population and very cold winters. We can
adopt some of the measures mentioned above to conserve some energy, but the cold
winters and great distances will always mean we have to use more energy than others for
basic needs.
82. Answers may vary. Sample answer: (to the Minister of Natural Resources)
Dear Minister,
At current rates of energy use, Canadians are using traditional energy sources faster than
they can be renewed. We desperately need to develop sustainable, large scale
alternatives. I encourage your ministry to promote research into ways of harnessing tidal
energy for use across the country. We are surrounded by three oceans, the force behind
tidal energy is immense, and tidal energy is renewable and predictable. Canada is well
placed to become a leader in this field as we develop the capacity to supply our own
energy needs for the foreseeable future. We need to break our reliance on fossil fuels.
Although this research could be costly in the short term, I believe that Canadian
taxpayers would recognize it as a very worthwhile long-term investment.
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Unit 3: Energy Changes and Rates of Reaction
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83. Answers may vary. Sample answer: No, I would not be concerned if a wind farm was
in my area, because wind farms produce renewable energy that is important to reducing
our global footprint and they have very few risks. We all have to bear some of the
responsibility for creating the energy that we use as a society and I would rather have a
wind farm in my area than an oil well or a nuclear plant, which carry far more
environmental risks.
84. No, fire departments do not need to warn people about storing lumber near potential
sources of sparks the way they do for large masses of paper, because the surface area of
lumber is very small compared to its mass, so a single spark is unlikely to ignite a pile of
lumber.
85. Answers may vary. Sample answers: Industrial processes are likely to increase
reaction rates by using catalysts, increasing temperature, or increasing concentration of
reactants. Increasing concentration is relatively inexpensive but generally works best in
homogeneous reactions. Increasing temperature is effective but higher temperatures can
introduce hazards to material handling and extra fuel costs. Catalysts are safer but some
catalysts are expensive.
86. Answers may vary. Sample answers:
(a) The advantages to society of producing sugars using a biocatalyst such as glucose
isomerase are that the biocatalyst does not use any toxic chemicals and production is less
expensive.
(b) The main disadvantage to society of the use of high fructose corn syrup as a
sweetener is overconsumption of sugar leading to health problems.
(c) I think glucose isomerase has had a positive effect on society because it has reduced
food costs.
87. Answers may vary. Sample answer: I agree with the scientist who proposed
mechanism 1 because it is consistent with the rate law. (Note that mechanism 3 is also
consistent with the rate law.)
88. From the experimental data in Table 6, the reaction is first order with respect to H2(g)
and second order with respect to NO(g). Therefore, step 2 in the proposed mechanism is
the rate-determining step because a fast first step is first order in both reactants,
producing the intermediate for step 2, which is first order with respect to NO(g).
89. (a) It is important to know the rate of decomposition of greenhouses gases in the
atmosphere because greenhouse gases increase the retention of thermal energy until they
decompose.
(b) The lifespan of a molecule is considered in determining the global warming potential
(GWP) of an atmospheric gas because the warming effect is cumulative over time.
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Unit 3: Energy Changes and Rates of Reaction
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Reflect on Your Learning
90. Answers may vary. Sample answer:
Ways to Determine the Enthalpy of a Reaction
Method
Summary of method
calorimetry Reaction run in a
calorimeter and temperature
change is recorded.
bond
Enthalpy change calculated
energy
from energies of reactant
and product chemical bonds
heats of
Calculate enthalpy based on
formation
Hess’s law and the standard
enthalpies of formation.
Strengths
– real data
– fast and simple
calculation
– accurate data
available
– simple
calculation
Weaknesses
– may be difficult to run
the reaction in
calorimeter
– based on average bond
energy, not that of
specific compounds
– heats of formation not
available for every
compound
– result is an estimate
91. Answers may vary depending on individual student experience.
92. Answers may vary depending on individual student experience.
93. Answers may vary. Sample answer: Because our universe is infinite, and because the
human brain has an infinite capacity to wonder about things it does not understand, I do
not believe science will ever reach a point where there are no theoretical constructs. As
we have seen, scientific inquiry extends in many directions, for example: near and far;
large and small; identities, descriptions, processes, and causes. There will always be
something we have not gathered enough direct evidence to prove. Every answer leads to
new questions.
94. Answers may vary. Sample answer: While research that deals directly with societal
issues is crucial, research that is driven solely by the desire to understand is also
important. Much of the research being carried out today with definite societal goals is
based on previous research that was carried out to learn. For example, research into
alternative fuel technologies makes use of centuries of knowledge gained about various
chemicals and about geophysical properties and processes. It could not proceed without
this scaffolding.
Research
95. Answers may vary. Students’ answers should include a method for comparing the
efficiency of the two fuels, evaluating the availability of resources to produce sufficient
ethanol, and the costs and benefits to consumers and to food production.
96. Answers may vary. Sample answer: Coal is used to generate approximately 13 % of
all electrical power in Canada, but its use is not uniform. In some provinces, coal is the
main source of energy, and in other provinces it is not used at all. From the largest
amount of coal used for power generation to the smallest, Canadian provinces rank as
follows: Ontario, Alberta, Saskatchewan, Nova Scotia, New Brunswick, and Manitoba,
with other provinces using insignificant amounts.
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It is important to realize that British Columbia and Alberta produce over 80 % of
Canada’s coal. Nova Scotia is another coal-producing province. Alberta likely uses a lot
of coal to generate power because it produces a lot of coal and does not incur shipping
costs to acquire it. Ontario’s high use of coal to generate power is a result of its great
need for power due to its high population and its historic (but changing) role as the
industrial engine of Canada. British Columbia, Quebec, and Newfoundland and Labrador
have access to large rivers that can be used for hydroelectric power generation.
While Alberta is actively developing new coal-fired power generation plants, provinces
that are not coal producers are looking for ways to reduce their reliance on coal as a result
of concerns about the greenhouse gases emitted when it is burned. Ontario, for example,
is planning to decrease the amount of power generated by burning coal, and is
researching the use of nuclear power instead.
97. Answers may vary. Answers should include the following information:
• While wind turbines in North America and Europe can be huge and grouped in large
“farms”, wind turbines in developing areas of China and India tend to be small and
generate power that will be used locally. This eliminates the need to build infrastructure
to bring electrical power to remote areas.
• Turbines for offshore power generation tend to be larger than those used on land,
because installation is so challenging, the turbines must be as efficient as possible. Since
offshore winds are often of higher speeds, countries with access to ocean or lakes that can
be used in this way tend to construct offshore wind turbines.
• Other than size differences, modern wind turbines have been becoming more and more
similar, using three blades, and turning the blades to catch the wind, for example. These
turbines can be used in both low and high wind conditions, and at both low and high
temperatures.
• In some countries, such as Germany, economic incentives have been provided to
providers of wind power. This has led directly to Germany becoming one of the word’s
leaders in wind-power generation.
98. Answers may vary. Sample answer: The major greenhouse gases are carbon dioxide,
methane, nitrous oxide, and ozone.
Since fossil-fuel combustion causes a large proportion of carbon dioxide emissions,
methods for capturing the carbon dioxide before it enters the atmosphere have focused on
power generating plants. In some cases, a solvent is added to trap carbon dioxide from
the flue emissions and prevent its release. The carbon dioxide trapped is transported to be
stored underground (for example, in sandstone). In other cases, a chemical process can be
used to capture carbon dioxide before it enters the air. This provides a higher yield, but
can only be used where the facility was built specifically to accommodate it. There is
some research ongoing into the development of a hydrogen-powered car that would trap
carbon dioxide emissions and store them in liquid form to prevent their release into the
environment.
Nitrous oxide is also produced during combustion at high temperatures. Many
companies use a selective catalytic reduction process to remove nitrous oxide from the
atmosphere. This involves combining ammonia with the nitrous oxide, then allowing a
catalyst to absorb it, and separating it into nitrogen and oxygen—both harmless
components of the atmosphere.
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Unit 3: Energy Changes and Rates of Reaction
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Significant amounts of methane are liberated from coal seams during mining.
Additional methane enters the atmosphere as a result of raising livestock. Still more
methane may be released as large areas of permafrost thaw with global warming.
Research is being done into regenerative thermal oxidation strategies to reduce the
amount of methane produced during mining. Citizen groups advocate less reliance on
meat, which would reduce farming emissions. All attempts to stop global warming will
reduce emissions from thawing permafrost.
Ozone is formed when nitrous oxides and volatile organic compounds in the
atmosphere interact. Steps to curb nitrous oxide emissions will help reduce ozone levels.
In addition, paints and cleaners have been formulated with lower levels of volatile
organic compounds, to reduce the formation of ozone in the atmosphere.
99. Because biological catalysts are designed to work with molecules of living things,
they may be particularly effective for producing chemicals useful as medicines. In
addition, because they are natural substances, they may not have side effects as severe as
some manufactured catalysts may have.
100. Answers may vary. Answers should include information such as the following:
Adsorption catalysts work by binding reactant molecules temporarily on their surface.
They are used as heterogeneous catalysts; for example, the metal catalyzes the
decomposition of nitrogen oxides in a catalytic converter.
Catalysts that form intermediate compounds react with other chemicals to make an
unstable compound that breaks down to release a new molecule and the catalyst. Enzyme
reactions in cells use intermediate catalysts.
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Unit 3: Energy Changes and Rates of Reaction
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