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Oxford Resources for IB
Chemistry – 2023 Edition
Answers
Reactivity 1.1 – Measuring enthalpy changes
Activity (page 390)
The glass of water is an open system.
The system includes the water and ice in the glass. The surroundings are all matter that exists outside the
glass of iced water. The glass can be considered the boundary between the system and the surroundings.
Heat,
Q
, is a form of energy that is transferred from a warmer body to a cooler body, as a result of the
temperature gradient. The water in the glass is warmer than the ice cubes (H
2
O(s)). Energy will flow from
the water into the ice cubes, warming them, so eventually they will melt. As the surroundings are warmer
than the iced water, energy will flow from the surroundings into the system.
Condensation occurs on the outside of the glass and you observe liquid water. The change of state is from
a gas (water vapour) to a liquid (water). Energy is transferred into the surroundings when water vapour
condenses into liquid water.
Practice question (page 393)
1.
C
Practice questions (page 394)
2.
a.
Q
=
mc
∆
T
= 1 kg × 4.18 kJ kg
–1
K
–1
× 1 K
= 4.18 kJ
b.
Q
=
mc
∆
T
= 1000 kg × 0.385 kJ kg
–1
K
–1
× 1 K
= 385 kJ
3.
a
4.
a
Practice questions (page 396)
5.
Q
=
mc
∆
T
= 30 g × 4.18
J
g
–1
K
–1
× 30
K
= 3762
J
6.
First, convert kJ to J (× 1000)
Next, convert g to kg (÷ 1000)
Q
=
mc
∆
T
T
Q
mc
–1
–1
675 J
0.125 kg × 385 J kg
K
= 14.0
K (3 sf)
1
© Oxford University Press 2023
Skills questions (page 398)
Answers will vary depending on experimental data obtained.
The process shown in Worked example 3 (page 399) should be followed.
ATL questions (page 401)
1./2.
Ignition device: source of ignition to combust the sample
Oxygen supply: ensures complete combustion
Temperature probe: measures temperature
Water bath: absorbs the heat released by combustion in the bomb
Calorimeter bomb: holds sample
Stirrer: even distribution of heat energy throughout the water bath
Jacket: thermal insulator to minimize heat loss to the surroundings.
3.
Answers will vary e.g. good thermal insulation, complete combustion, closed environment minimizes loss
of product, etc.
4.
High specific heat capacity, readily available, liquid over a large temperature range, etc.
End of topic questions (pages 402–403)
1.
Answers will depend on the student, but a possible answer might be: In a chemical reaction, total energy is
conserved. All physical and chemical changes involve energy changes, regardless of their magnitude. When
heat energy is released from a system into the surroundings, the chemical reaction is defined as an exothermic
reaction. When heat energy is absorbed from the surroundings, the chemical reaction is defined as an
endothermic reaction. Endothermic and exothermic reactions can be determined in a school laboratory using a
simple calorimeter. The collection of experimental data in these reactions focuses on a change in temperature
of the reaction solvent, normally water. When the change in the temperature is positive, an exothermic reaction
has taken place. When the change in the temperature is negative, an endothermic reaction has taken place.
2.
D
3.
D
4.
A
5.
B
6.
a.
∆
H
⦵
(rev) = 226
kJ
mol
–1
b.
2NO
2
(g) + H
2
O(l) → HNO
3
(aq) + HNO
2
(aq)
or
4NO
2
(g) + 2H
2
O(l) + O
2
(g) → 4HNO
3
(aq)
Ionic equations would also be acceptable.
7.
a.
3 minutes
b.
The
temperature is still rising while the reaction takes place
c.
Maximum temperature:
54
°C
Assumption
: that temperature would be reached if the reaction were instantaneous
OR
that
temperature would be reached if the reaction occurred without heat loss
Note:
Accept “rate of heat loss is constant”
OR
“rate of temperature decrease is constant”.
d.
Any one of:
excess copper(II) sulfate
AND
mass/amount of zinc is independent variable/being changed.
OR
excess
copper(II) sulfate
AND
with zinc in excess there is no independent variable [as amount of
copper(II) sulfate is fixed]
Excess copper(II) sulfate
AND
having excess zinc will not yield different results in each trial
Excess zinc
AND
results can be used to see if amount of zinc affects temperature rise [so this can be
allowed for]
Excess zinc
AND
reduces variables/keeps the amount reacting constant
2
© Oxford University Press 2023
e.
For
m
:
Density of solution is 1.000 g cm
–3
(same as water)
OR
25.00 cm
3
solution has a mass of 25.00 g
OR
Mass of zinc/reactant is negligible
OR
Mass of contents was 25.00 g
For
c
:
Specific heat capacity of solution is 4.18
J
g
–1
K
–1
(same as water)
OR
Zinc/calorimeter/beaker/thermometer absorbs no heat
f.
lower/less exothermic/less negative
AND
heat loss/some heat not accounted for
OR
lower/less exothermic/less negative
AND
mass of reaction mixture greater than 25.00
g
OR
greater/more exothermic/more negative
AND
specific heat of solution less than water
Note:
Accept “temperature is lower” instead of “heat loss”.
Accept “similar to theoretical value
AND
heat losses have been compensated for”.
Accept “greater/more exothermic/more negative
AND
linear extrapolation overestimates heat loss”.
8.
a.
heat =
mc
∆
T
= 7.8 g × 4.18
J
g
–1
K
–1
× (22.6 – 21.3)
= 42.4
J
b.
–1
4
e
2.
h
4
p
J
f
1
c
.
m
2
u
t
g
nt al y o o
b
s ion
–35.33 J g
= –0.035
kJ
g
–1
Reactivity 1.2 – Energy cycles in reactions
Practice questions (page 407)
1.
bond enthalpy for H–Cl is 431.5 kJ mol
–1
2.
amount of energy required to break bonds of reactants:
(4 × BE C–H) + (2 × BE O–H) = 4 × 414 + 2 × 463
= 2582 kJ mol
–1
amount of energy released during bond formation:
(3 × BE H–H) + (1 × BE C≡O)
= 3 × 436 + 1077
= 2385 kJ mol
–1
∆
H
= 2582 – 2385 = +197 kJ mol
–1
3.
a.
CH
4
(g) +
Cl
2
(g) → CH
3
Cl(
g) + HCl(g)
b.
1898 – 1997 = –99 kJ mol
–1
or 656 – 755 = –99 kJ mol
–1
(if only use bonds that change)
c.
exothermic, because the enthalpy change is negative
d.
products
3
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Practice questions (page 410)
4.
2CO
2
(g) + 3H
2
O(l) → C
2
H
5
OH(l) + 3O
2
(g)
∆
?
= +1367 kJ mol
–1
(reversed equation 1)
2C(s) + 2O
2
(g) → 2CO
2
(g)
∆
?
= 2 × (–394) kJ mol
–1
(doubled equation 2)
3H
2
(g) + 1½O
2
(g) → 3H
2
O(l)
∆
?
= 3 × (–286) kJ mol
–1
(tripled equation 3)
These sum to the equation given in the question.
∆
?
f
⦵
= –279
kJ
mol
–1
5.
Create an enthalpy cycle diagram, with the required equation along the top and 2N
2
, 5O
2
and 6H
2
at the
bottom.
Reverse and double equation 2:
4NH
3
(g) → 2N
2
(g) + 6H
2
(g)
∆
?
= 2 × (+92) kJ mol
–1
Double equation 1:
2N
2
(g) + 2O
2
(g) → 4NO(g)
∆
?
= 2 × (+66) kJ mol
–1
Triple equation 3:
6H
2
(g) + 3O
2
(g) → 6H
2
O(g)
∆
?
= 3 × (–572) kJ mol
–1
4NH
3
(g) + 5O
2
(g) → 4NO(g) + 6H
2
O(l)
∆
?
= –1400 kJ mol
–1
Skills questions (page 411)
1.
a.
2HCl(aq) + K
2
CO
3
(s) → 2KCl(aq) + H
2
O(l) + CO
2
(g)
b.
HCl(aq) + KHCO
3
(s) → KCl(aq) + H
2
O(l) + CO
2
(g)
2.
n
(HCl) = (0.030
cm
3
) × (2
mol
dm
–3
) = 0.06
mol
Mole ratio in first reaction
n
(HCl) :
n
(K
2
CO
3
) = 2
:
1
3
1
2
3 g
138.21 g m
f
s
ol
n
(2 × 0.02) < 0.06.
(
2
×
0.02
)<
0.06
Therefore, HCl is in excess.
Mole ratio in second reaction
n
(HCl) :
n
(KHCO
3
) = 1
:
1
3
1
3 g
H
100.12 g m
f
s
ol
n
0.03 < 0.06. Therefore, HCl is in excess.
3.
Answers will vary depending on experimental data.
4.
2KHCO
3
(s) → K
2
CO
3
(s) + CO
2
(g) + H
2
O(g)
Note that under standard conditions the correct state symbol for water is (l).
5.
6 and 7.
Answers will depend on the experimental data obtained.
8.
Answers will depend on the source of the theoretical value (which can be calculated using enthalpy of
formation data).
9.
The specific heat capacity of the solution is equal to that of water, no thermal energy is lost to the
surroundings, thermal energy was evenly distributed in the solution, mass of water not measured and 1
g =
1
cm
3
is assumed, etc.
4
© Oxford University Press 2023
Activity (page 412)
a.
C
8
H
18
(l) + 12.5O
2
(g) → 8CO
2
(g) + 9H
2
O(l)
∆
H
c
⦵
= –5470 kJ mol
–1
b.
C
6
H
12
O(s) + 8.5O
2
(g) → 6CO
2
(g) + 6H
2
O(l)
∆
H
c
⦵
= –3728 kJ mol
–1
c.
CH
2
O
2
(l) + 0.5O
2
(g) → CO
2
(g) + H
2
O(l)
∆
H
c
⦵
= –255 kJ mol
–1
d.
C
6
H
12
O
6
(s) + 6O
2
(g) → 6CO
2
(g) + 6H
2
O(l)
∆
H
c
⦵
= –2803 kJ mol
–1
e.
C
2
H
5
Cl(g) + 3O
2
(g) → 2CO
2
(g) + 2H
2
O(l) + HCl(g)
∆
H
c
⦵
= –1413 kJ mol
–1
Activity (page 413)
a.
3C(s) + 4H
2
(g) → C
3
H
8
(g)
∆
H
f
⦵
= –105 kJ mol
–1
b.
C(s) +
1.5H
2
(g) +
0.5Cl
2
(g) → CH
3
Cl(g)
∆
H
f
⦵
= –82.0 kJ mol
–1
c.
2C(s) + 3H
2
(g) + 0.5O
2
(g) → C
2
H
6
O(l)
∆
H
f
⦵
= –278 kJ mol
–1
d.
7C(s) + 3H
2
(g) + O
2
(g) → C
6
H
5
COOH(s)
∆
H
f
⦵
= –385 kJ mol
–1
e.
C(s) + 0.5O
2
(g) → CO(g)
∆
H
f
⦵
= –111 kJ mol
–1
f.
C(s) + 2.5H
2
(g) + 0.5N
2
(g) → CH
3
NH
2
(g)
∆
H
f
⦵
= –23 kJ mol
–1
Practice question (page 415)
6.
The standard enthalpy values given in the data booklet are rounded to the nearest whole numbers. If more
precise values for the enthalpies of formation of carbon dioxide and water (–393.5 and –285.8 kJ mol
–1
,
respectively) were used in worked example 3, the final answer would be –173.3 kJ mol
–1
. This case
demonstrates the importance of precision in scientific measurements: imprecise (rounded) source data can
affect both the precision and accuracy of the final result.
Practice questions (page 417)
7.
1:
C(s) + O
2
(g) → CO
2
(g)
∆
H
f
⦵
= –394 kJ mol
–1
2:
H
2
(g) + 0.5
O
2
(g)
→ H
2
O(l)
∆
H
f
⦵
= –286 kJ mol
–1
3:
CH
3
COCH
3
(l) + 4O
2
(g)
→ 3CO
2
(g) + 3H
2
O(l)
∆
H
f
⦵
= –1790 kJ mol
–1
Step 1: triple equation 1
Step 2: triple equation 2
Step 3: reverse equation 3
∆
H
f
⦵
= 3 × (–394) + 3 × (–286) + 1790
= –250 kJ mol
–1
8.
a.
the standard enthalpy of formation
, ∆
H
f
⦵
, is the energy change upon the formation of 1 mol of a
substance from its constituent elements in the standard state. Hydrogen, H
2
, is the element in the standard
state, so its standard enthalpy of formation is zero.
b.
∆
H
f
⦵
= ∑(∆
H
f
⦵
products) – ∑(∆
H
f
⦵
reactants)
= [3 × (0) + (–111)] – [–74 + (–242)]
= 205
kJ
mol
–1
c.
bond enthalpy values are not specific to the compounds
values were experimentally determined/had uncertainties
different sources have slightly different values for bond enthalpies
9.
a.
∆
H
⦵
= (–286) + (–1411) = –1697 kJ mol
–1
b.
step 1 = step 2 – step 3
∆
H
⦵
= (–1697) – (–1561) = –136 kJ mol
–1
5
© Oxford University Press 2023
c.
Accurate:
no approximations were made in the cycle
OR
values are specific to the compounds
OR
Hess’s law is a statement of conservation of energy
OR
method is based on a law
OR
data in table has small uncertainties
Approximate:
values were experimentally determined/had uncertainties
OR
each value has been determined to only three/four significant figures
OR
different sources have slightly different values for enthalpy of combustion
OR
law is valid until disproved
OR
law of conservation of energy is now conservation of mass–energy
OR
small difference between two quite large terms leads to high percentage uncertainty
Practice question (page 421)
10.
∆
H
f
⦵
= (+148) + (+738) + (+1451) + (+249) + (–141) + (+753) + (–3791) = –593
kJ mol
–1
End of topic questions (pages 422–423)
1.
Answers will depend on the student, but a possible answer might be: The law of conservation of energy
states that energy can neither be created or destroyed, it can only be interconverted between different
forms. Hess’s law states that regardless of the route by which a chemical reaction proceeds, the
enthalpy change will always be the same, as long as the initial and final states of the system are the
same. The application of Hess’s Law and the Born-Haber cycle relies on the conservation of energy
law. When we understand the value of the change in enthalpy for the overall chemical, for example the
enthalpy of formation or the lattice enthalpy; we can calculate the value of a missing step in the energy
cycle.
2.
B
3.
D
4.
B
5.
C
6.
B
7.
a.
bonds broken = (N≡N) + 3 × (H–H)
= 945 + (3 × 436) = 2253
bonds made = 2 × 3(N–H)
= 6 × 391 = 2346
∆
H
= 2253 – 2346 = –93
kJ
mol
–1
b.
The reaction is exothermic, as the change in enthalpy is a negative value.
8.
a.
4NH
3
(g) + 3O
2
(g) → 2N
2
(g)+ 6H
2
O(g)
b.
c.
bonds broken = 4 × 3(N–H) + 3(O=O)
= (4 × 3 × 391) + (3 × 498) = 6186
bonds made = 2(N
≡
N) + 6 × 2(O–H)
= (2 × 945) + (6 × 2 × 463) = 7446
6
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∆
H
= 6186 – 7446 = –1260
kJ
mol
–1
7
© Oxford University Press 2023
9.
a.
b
onds broken = 18(C–H) + 7(C–C)+ 12.5(O=O)
= (18 × 414) + (7 × 346 ) + (12.5 × 498) = 16099
bonds made = 8 × 2(C=O) + 9 × 2(O–H)
= (8 × 2 × 804) + (9 × 2 × 463) =
21198
∆
H
= 16099 – 21198 = –5099
kJ
mol
–1
b.
The value of enthalpy of reaction calculated from bond enthalpies is less accurate as the values used
in the calculation are average values and therefore only an approximation.
Bond enthalpy values are only valid for substances in the gaseous state; octane is a liquid.
Reactivity 1.3 – Energy from fuels
Activity (page 425)
Sulfur dioxide is widely used in the industrial production of sulfuric acid, paper manufacturing and the food
industry as a preservative and antioxidant. The environmental impact of the release of sulfur dioxide into the
atmosphere is increased acid deposition and the associated effects such as deforestation, increased acidity of
lakes and rivers affecting the fish industry and the erosion of buildings and structures. Sulfur dioxide can also
be harmful to the human respiratory system.
Practice questions (page 427)
1.
a.
CH
4
(g) + 2O
2
(g) → CO
2
(g) + 2H
2
O(l)
b.
C
3
H
8
(g) + 5O
2
(g) → 3CO
2
(g) + 4H
2
O(l)
c.
C
5
H
12
(l) + 8O
2
(g) → 5CO
2
(g) + 6H
2
O(l)
d.
2C
6
H
14
(l) + 19O
2
(g) → 12CO
2
(g) + 14H
2
O(l)
Practice questions (page 427)
2.
a.
2C
3
H
7
OH(l) + 9O
2
(g) → 6CO
2
(g) + 8H
2
O(l)
b.
2C
5
H
11
OH(l) + 15O
2
(g) → 10CO
2
(g) + 12H
2
O(l)
c.
2C
7
H
15
OH(l) + 21O
2
(g) → 14CO
2
(g) + 16H
2
O(l)
Practice questions (page 428)
3.
a.
C
3
H
7
OH(l) + 3O
2
(g) → 3CO(g) + 4H
2
O(l)
b.
C
5
H
11
OH(l) + 5O
2
(g) → 5CO(g) + 6H
2
O(l)
c.
C
7
H
15
OH(l) + 7O
2
(g) → 7CO(g) + 8H
2
O(l)
Practice questions (page 432)
4.
a.
C
3
H
8
(g) + 5O
2
(g) → 3CO
2
(g) + 4H
2
O(l)
Carbon dioxide is the greenhouse gas.
b.
1 mol of C
3
H
8
is 44.11 g mol
–1
–1
–1
–1
2219 kJ mol
50.31 kJ g
44.11 g mol
50.31 kJ of energy is released when 1 g of propane is combusted.
c.
0.02267 × 3 = 0.0680 mol CO
2
0.0680 × 44.01 = 2.99 g CO
2
d.
2.99 tonnes of carbon dioxide. Increase in the average temperature throughput the World, which is
known as global warming. Increase in extreme weather events. Thawing of glacial ice. Impact on food
production (agriculture and livestock)
Skills questions (page 433)
Responses will vary greatly depending on the database used, and the research question. The guidance given
in Tool 2, Tool 3 and Inquiry cycle chapters should be followed.
8
© Oxford University Press 2023
Practice questions (page 435)
5.
a.
C
6
H
12
O
6
(aq) → 2C
2
H
5
OH(aq) + 2CO
2
(g)
b.
Carbon dioxide, a greenhouse gas, is produced in both fermentation and combustion processes.
However, ethanol-based fuels have a lower carbon footprint as the production of the carbon dioxide is
offset by the absorption of carbon dioxide in the process of photosynthesis.
c.
Possible arguments are listed in table 1 on page 435.
Data-based questions (page 436)
1.
The total biofuel production in terawatt-hours increased slowly for 15 years from 1990. Then production
increased rapidly from 2005 onwards. In 2020, the production declined.
2.
The contribution of the major geographical regions to the production of biofuels has become more evenly
distributed over the past three decades.
3.
Once the major supplier of biofuels when total production was small (1990s), South and Central America’s
market share has decreased from 80% from 20%.
Practice questions (page 440)
6.
a.
A primary electrochemical cell consumes materials and the reaction is non-reversible. Either the
anode, electrolyte or both need to be replaced. Primary cells do not operate well under high current
demand such as electric cars. In a secondary cell, chemical reactions that generate electricity can be
reversed by applying an electric current. Secondary cells can deliver a stronger current flow than a
primary cell.
b.
Lead–acid batteries convert chemical energy into electrical energy. The process is reversible and the
batteries can be recharged. While this type of secondary cell has advantages over traditional primary
cells, there are still problems associated with this type of battery, such as overheating, limited lifespan
and environmental concerns in terms of their disposal.
Hydrogen fuel cells are considered to be a clean energy source. Unlike lead-acid
batteries, hydrogen
cells have the ability to work continuously, generate less hazardous waste on disposal. They produce
higher current per unit mass, have a higher specific energy, and they do not need to be recharged.
c.
The proton exchange membrane (PEM) selectively allows hydrogen ions to diffuse between the
cathode and anode, prevents the passage of other ions, molecules and electrons between these
electrodes. The protons formed at the anode move across the PEM to the cathode, where they
combine with the oxygen gas and electrons. Oxygen gas is reduced to form water as a waste product.
d.
i.
anode half-equation:
CH
3
OH(aq) + H
2
O(l) → CO
2
(g) + 6H
+
(aq) + 6e
−
cathode half-equation:
1.5O
2
(g) + 6H
+
(aq) + 6e
−
→ 3H
2
O(l)
overall reaction:
CH
3
OH(aq) + 1.5O
2
(g) → CO
2
(g) + 2H
2
O(l)
i
i.
Advantage (any one of):
liquid methanol is easier to transport/store than gaseous hydrogen
OR
hydrogen is explosive
longer membrane life (as it operates in aqueous environment)
methanol has greater energy density than hydrogen
Disadvantage (any one of):
lower voltage
lower power per unit mass «of the cell»
lower efficiency
toxic/can be mistaken for ethanol
lower specific energy
Ignore any cost references throughout.
Accept “CO
2
/greenhouse gas produced”
OR
“requires a more highly efficient catalyst”.
Do
not
award marks for converse statements for the advantage and disadvantage.
9
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End of topic questions (page 441)
1.
Answers will depend on the student, but a possible answer might be: The combustion of fossil fuels,
typically coal, oil, natural gas, LPG and other hydrocarbons converts chemical energy into thermal energy. This
thermal energy is then used to generate electrical energy to address our expanding energy needs. The
challenges that face society are the significant environmental impact of increasing levels of carbon dioxide.
Carbon dioxide is a greenhouse gas that traps heat energy inside the Earth’s atmosphere, leading to global
warming. Governments have the challenge of transitioning economies to clean energy alternatives which takes
time, money, expertise and political willpower.
2.
B
3.
B
4.
a.
2CH
4
(g) + 3O
2
(g) → 2CO(g) + 4H
2
O(l)
b.
Carbon monoxide, CO, is a colourless and odourless poisonous gas that attaches irreversibly to the
haemoglobin in the bloodstream. This reduces the oxygen-carrying capacity of the blood. This can lead
to coma and eventually death.
c.
[(4 × 414) + (1.5 × 498)] − [(1 × 1077) + (4 × 463)] = –526 kJ mol
–1
5.
a.
C
2
H
5
OH(l) + 3O
2
(g) → 2CO
2
(g) + 3H
2
O(l)
b.
(showing strong) correlation between (atmospheric) CO
2
concentration/greenhouse gas concentration
and average (global/surface/ocean) temperature
lab evidence that greenhouse gases/CO
2
absorb(s) infrared radiation
computer modelling
ice core data
tree-ring data
ocean sediments / coral reefs / sedimentary rocks data
NOTE: Do
not
accept “global warming” for “average temperature”.
Do
not
accept “traps/reflects heat”
OR
“thermal energy”.
Evidence must be outlined and connected to data.
Accept references to other valid greenhouse gases other than carbon dioxide/CO
2
, such as
methane/CH
4
or nitrous oxide/N
2
O.
c.
Biofuel raw material/sugar/glucose formed by photosynthesis
OR
biofuel raw material/sugar/glucose uses up carbon dioxide during its formation
OR
biofuel from capturing gases due to decaying organic matter formed from photosynthesis
Arguments based on materials coming from a plant source consuming carbon dioxide / carbon are
acceptable.
6CO
2
(g) + 6H
2
O(l) → C
6
H
12
O
6
(aq) + 6O
2
(g)
Accept arguments based on material coming from plant sources consuming carbon dioxide/carbon for
M1.
6.
a.
liquid in cell is less/not corrosive
OR
does not contain lead/toxic chemicals
OR
larger specific energy/charge capacity/current per unit mass
OR
does not have to be charged prior to use / is always ready for use as long as fuel is available
b.
Advantage (any one of):
liquid methanol is easier to transport/store than gaseous hydrogen
hydrogen is explosive
longer membrane life (as it operates in aqueous environment)
methanol has greater energy density than hydrogen
10
© Oxford University Press 2023
Disadvantage (any one of):
lower voltage
lower power per unit mass of the cell
lower efficiency
toxic/can be mistaken for ethanol
lower specific energy
Ignore any cost references throughout.
Accept “CO
2
/greenhouse gas produced”
OR
“requires a more highly efficient catalyst”.
Do
not
award marks for converse statements for the advantage and disadvantage.
7.
a.
(portable) sources of electrical energy/electricity
OR
convert chemical «potential» energy to electrical energy/electricity
b.
primary cells involve irreversible reactions
AND
rechargeable cells involve reversible reactions.
Accept “primary cells have a limited life before going ‘flat’
AND
rechargeable cells can be recharged
when ‘flat’”.
c.
Voltage:
chemical nature of electrodes
OR
electrode reactions
Current:
diffusion rate
OR
internal resistance/resistance of the cell
Accept temperature for either but not both.
Accept concentration for either but not both.
Accept pH for either but not both.
Accept the current depends on the area/separation of the electrodes.
Reactivity 1.4 – Entropy and spontaneity
Practice questions (page 445)
1.
a.
positive, because number of moles of gaseous species increases
b.
negative, because number of moles of gaseous species decreases
c.
positive, because number of moles of gaseous species increases
d.
positive, because number of moles of gaseous species increases
Practice questions (page 447)
2.
B
3.
D
4.
+74.5 J K
−1
mol
−1
Practice questions (page 450)
5.
a.
∆
H
⦵
= [–165 + 2(–297) + 2(–92)] – [–453 + 2(–246)]
∆
H
⦵
= +2 kJ mol
−1
b.
∆
S
⦵
= [209 + 2(248) + 2(187)] – [167 + 2(279)]
∆
S
⦵
= +354 J K
–1
mol
–1
c.
There is a large increase in entropy resulting in
T
∆
S
> ∆
H
at the reaction temperature.
11
© Oxford University Press 2023
Practice questions (page 451)
6.
a.
bonds broken =
4(C–H) + C=C + H-H
= (4 × 414) + 614 + 436 = 2706
bonds made = C–C + 6(C–H)
= 346 + (6 × 414) = 2830
∆
H
= 2706 – 2830 = –124 kJ mol
−1
b.
∆
H
⦵
= Σ(∆
H
f
⦵
products)− Σ(∆
H
f
⦵
reactants)
= [∆
H
f
⦵
(C
2
H
6
)] − [∆
H
f
⦵
(C
2
H
4
) + ∆
H
f
⦵
(H
2
)]
= [−84] − [52 + 0] = −32 kJ mol
−1
c.
The change in entropy value for this reaction should be negative (∆
S
⦵
< 0) as there is a decrease in
the number of moles of gaseous species from 2 mol to 1 mol.
d.
∆
S
⦵
= Σ(∆
S
⦵
products)− Σ(∆
S
⦵
reactants)
= [∆
S
⦵
(C
2
H
6
(g)) + ∆
S
⦵
(H
2
(g))] − [∆
S
⦵
(C
2
H
4
(g))]
= [(+230)) +(131)] − [+220] = +141 kJ mol
−1
e.
∆
G
⦵
= ∆
H
⦵
−
T
∆
S
⦵
= –32 kJ mol
−1
− (298 K × 0.141 kJ K
−1
mol
−1
)
= –74 kJ mol
−1
∆
G
⦵
is always negative (< 0) as ∆
H
⦵
is negative (< 0) and ∆
S
⦵
is positive (> 0).
Skills questions (page 453)
1.
∆
H
for the dissolution process is negative because the process is exothermic. ∆
S
for the dissolution
process is negative because of the regular arrangement of water molecules required to form hydrogen bonds
with the amido groups along the polymer chain.
2.
Answers will vary but should be consistent with the information given below figure 5.
3.
Answers may include: concentration of PNIPAM, volume of either LiCl or LiBr, use of distilled water,
gradual heating to allow for thermal equilibrium between the water bath and sample, etc.
4.
The effect of the presence or absence of an aqueous ionic compound:
a.
Results 5 should be compared to either 1 and 3 or 2 and 4.
b.
In the presence of an aqueous ionic compound, the PNIPAM solution goes cloudy at a lower
temperature. The effect is more pronounced when LiCl is used than LiBr.
The effect of concentration of aqueous ionic compound:
a.
Results 1 and 2 should be compared (or alternatively, 3 and 4).
b.
When the concentration of the ionic compound is higher , the PNIPAM solution goes cloudy at a lower
temperature.
c.
Answers will vary. For example, the ion-dipole interactions between water and the ions in the
electrolyte reduce the solubility of the PNIPAM, causing the temperature at which the solution goes
cloudy to decrease. Students may research and make reference to the Hofmeister effect.
d.
Answers will vary, but students should identify that two points are insufficient for establishing a trend.
e.
Answers will vary. Students could set out to obtain data across a greater range of concentrations,
range of anions, range of cations, etc.
Practice questions (page 456)
7.
a.
∆
H
⦵
= (4 × 33.2) − (2 × 82) = −31.2 kJ mol
−1
b.
∆
S
⦵
= (4 × 240) − [(2 × 220) + (3 × 205)] = −95 J K
−1
mol
−1
c.
∆
G
⦵
= –31.2 – 298 × (–95 / 1000) = –2.89 kJ mol
−1
d.
∆
G
⦵
= –
RT
ln
K
Δ
ln
G
K
–
RT
3
–1
–1
–1
2.89
10
J mol
l
K
n
–
8.31 J K
mol
298
K
ln
K
= 1.17
K
= e
1.17
= 3.22
12
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Skills questions (pages 457–458)
1.
∆
S
is positive due to the formation of a gas.
2.
∆
H
is positive because the process is endothermic. Thermal energy is absorbed to overcome the hydrogen
bonds between water molecules.
3.
At higher temperatures, the spontaneity increases because the magnitude of the
T
∆
S
term increases,
leading to a negative ∆
G
value.
4.
∆
H
= –242 – (–286) = +44 kJ mol
-1
∆
G
= –229 – (–237) = +8 kJ mol
-1
∆
S
= 189 – 70 = +119 J K
-1
mol
-1
5.
∆
G
= 0
Δ
Δ
–
H
T
S
T
= –44 / 0.119 = 370 K (3 sf)
6.
Actual boiling point of water = 373 K
%
3
P
7
c
3
er enta
0
ge err
0
37
373
or
1 0
= 0.804%
8.
Answers will vary but follow the guidance in Tool 3.
9.
Gradient = –∆
S
y
-intercept = ∆
H
x
-intercept = temperature at which the change in spontaneity occurs.
10.
The
x
-intercept must be determined. Use of the equation of the line will give the same answer as Q5
(around 370 K). If obtained visually, values for the
x
-intercept will vary.
11.
The graph is linear because ∆
S
(i.e. the rate of change or gradient) is constant. The relationship is negative
because ∆
S
for this process is positive.
13.
13
© Oxford University Press 2023
14.
∆
H
(
y
-intercept)
∆
S
(gradient)
Spontaneity
positive
negative
non-spontaneous at all temperatures
positive
positive
spontaneous at high temperatures
negative
negative
spontaneous at low temperatures
negative
positive
spontaneous at all temperatures
End of
topic
questions (page 459)
1.
Answers will depend on the student, but a possible answer might be: Chemical systems can
approach equilibrium from either the forward or reverse direction of a balanced chemical equation,
e.g. N
2
(g) +3H
2
(g)
⇌
2NH
3
(g). The direction of change of a chemical reaction is determined by two key
factors, the change in enthalpy and change in entropy (a measure of the disorder of the system).
Exothermic reactions (∆
H
⦵
< 0) are normally spontaneous for a given set of conditions. However, the
entropy change and temperature of the system need to be considered to determine the direction of
chemical change. The combinations for temperature, enthalpy and entropy change will determine if a
reaction is spontaneous or non-spontaneous.
2.
A
3.
B
4.
C
5.
a.
∆
S
⦵
= 270 J K
–1
mol
–1
– 267 J K
–1
mol
–1
– 131 J K
–1
mol
–1
∆
S
⦵
= –128 J K
–1
mol
–1
b.
non spontaneous if ∆
G
⦵
= ∆
H
⦵
–
T
∆
S
⦵
> 0
OR
∆
H
⦵
>
T
∆
S
⦵
.
.
124 4
972 K
0 128
T
6.
a.
∆
G
⦵
= −
RT
ln
K
= −8.31 J K
–1
mol
–1
× 298 K × ln(5.01 × 10
–4
) ÷ 1000 = 18.8 kJ mol
–1
b.
non-spontaneous, because ∆
G
⦵
is positive
7.
a.
∑
S
(RHS) = 313 J K
–1
mol
–1
∑
S
(LHS) = (4 × 198) + 3
= 795 J K
–1
mol
–1
∆
S
⦵
= ∑
S
(RHS) – ∑
S
(LHS) = 313 – 795 = -482 J K
–1
mol
–1
b.
∆
H
⦵
= –633 – 4 × (–110.5) = –191 kJ mol
–1
c.
When ∆
G
⦵
= 0, forward and reverse reaction are equally favourable.
∆
G
⦵
= 0
Δ
H
T
Δ
S
.
191
396 K
123°C
0 482
d.
∆
G
⦵
= ∆
H
⦵
–
T
∆
S
⦵
= (–191) – 298 × (–482 / 1000) = –47.36 kJ mol
–1
∆
G
= ∆
G
⦵
+
RT
ln
Q
= −47364 J mol
−1
+ (8.31 J K
−1
mol
−1
) × 298 K × ln(0.5)
= −49080 J mol
−1
= –49.08 kJ mol
–1
∆
G
is negative, so the reaction is spontaneous at this point in the reaction pathway.
14
© Oxford University Press 2023
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