Worked solutions

Chapter 5
Exercises

1 B If the temperature drops, the process 100.0 J

ΔT = = 7.2 °C
must be endothermic. ΔH for endothermic 100.0 g × 0.138 J g−1 °C−1
reactions is always positive. q 100.0 J
ΔT = = = 7.25 K
mc 100.0 g × 0.138 J g−1 K−1
2 B All exothermic reactions give out heat. While = 7.25 °C
there are examples of exothermic reactions
that produces gases, occur quickly and ΔT = Tf – Ti
involve combustion the only statement that Tf = Ti + ΔT = 25.0 °C + 7.25 °C = 32.3 °C
is always true is B: all exothermic reactions
7 A The mass of the burner will decrease as
give out heat.
its contents are burned. The temperature
3 A The equation you need is q = mcΔT. recorded by the thermometer will increase
q as the heat from combustion warms the
Rearranging gives ΔT = . From the
mc water.
rearranged equation we can see that the
8 A An accurate result requires that all the heat
largest temperature increase will occur for
produced by loss of alcohol is used to
the metal with the smallest specific heat
heat the water in the calorimeter. It should
capacity.
be clear that heat loss reduces the rise in
4 D Again use q = mcΔT, recognizing that a temperature of the water (I). Loss of fuel by
temperature change, ΔT, of 2 °C is the same evaporation suggests that more fuel had
as a temperature change of 2 K. been burned (II). If the thermometer touched
q = 5.0 g × 0.385 J g–1 K–1 × 2 K = 3.85 J the bottom of the container then this would
(As there is only 1 significant figure in the give a higher temperature reading and the
value for ΔT (2 K), the answer should also calculated enthalpy of combustion would be
be expressed to 1 significant figure (4 J). higher than predicted (III).
However, this is not one of the possible 9 C Incomplete combustion and heat loss
answers so the best answer is D.) would both result in a lower observed ΔT
5 C Since the mass of each is the same and the and a calculated experimental enthalpy
heat change is the same, the temperature of combustion value lower than expected
change is inversely proportional to the heat compared to the literature value. The effect
q of random errors is accounted for in the
capacity, ΔT = . The substance with the uncertainty calculated for the experimental
mc
lowest heat capacity must therefore have the value (±200 kJ mol–1). If only random errors
highest temperature change and, as they all were present the literature value would
started at room temperature, will have the be within the range determined for the
highest final temperature. Cadmium must experimental value.
have the highest final temperature as it has 10 (a) Note that incorrect units are given for the
the lowest heat capacity. heat capacities in the question. These
q should both have units of J g–1 K–1.
6 q = mcΔT rearranges to give ΔT = .
mc Temperature change: ΔT = 36.50 – 25.85 =
10.65 °C (or K)
1
 Heat change, q = mcΔT exothermic reaction and that ΔHc will therefore
q = m(H2O) × c(H2O) × ΔT(H2O) + be a negative value.
m(Cu) × c(Cu) × ΔT(Cu) –q –4100 J
ΔH⊖c = =
n 1.61 × 10–3 mol
= (200.00 g × 4.18 J g–1 K–1 × 10.65 K) +
= –2.5 × 106 J mol–1 = –2500 kJ mol–1
(120.00 g × 0.385 J g–1 K–1 × 10.65 K)
(If we take the question as referring to
= 8900 J + 492 J white phosphorus molecules, P4, instead
q = 9392 J of phosphorus atoms, the molar mass
Moles of glucose burnt: n(C6H12O6) is 123.88 g mol–1 and the final answer is
m 1.10 g 10000 kJ mol–1.)
= = =
M (C6H12O6) 180.18 g mol−1 The major source of error in such experiments
6.11 × 10–3 mol is heat being transferred to anything other than
In calculating the enthalpy change of the water. Such errors will reduce the value of
combustion ΔHc we have to recognize that the calculated enthalpy. Heat can be lost to the
this is an exothermic reaction and that ΔHc container holding the water (by conduction), to
will therefore be a negative value. the atmosphere surrounding the apparatus (by
convection) or to the surroundings in general (by
Enthalpy of combustion per mole, ΔHc =
radiation).
–9392 J
Other sources of error include incomplete
6.11 x 10–3 mol
= –1.54 × 106 J mol–1 combustion, the measurement of the quantities
of phosphorus and water, impurities in the
= –1540 kJ mol–1
chemicals used, the heat used to ignite the
(b) phosphorus and variation in water’s heat
C6H12O6(s) + 6O2(s) capacity as its temperature changes.
H / kJ mol−

12 q = mcΔT
ΔH = −1540 kJ mol−1 ●● Temperature increase, ΔT = 50.0 °C = 50.0 K
●● Specific heat capacity of water (from Section
2 in the IB data booklet), c = 4.18 J g–1 K–1
●● Volume = 1.00 dm3 = 1000 cm3 and will
6CO2(g) + 6H2O(l)
have a mass of 1000 g (if assume density is
the same as that of water, 1.00 g cm–3)
q = 1000 g × 4.18 J g–1 K–1 × 50.0 K = 209 kJ
Extent of reaction
1000 cm3 of 1.00 mol dm–3 solution contains 1.00
11 q = mcΔT mole of copper sulfate, so the standard enthalpy
●● Temperature increase, ΔT = 6.5 °C = 6.5 K change for this reaction is –209 kJ mol–1.
●● Specific heat capacity of water (from Section As the zinc powder was in excess the molar
2 in the IB data booklet) = 4.18 J g–1 K–1 enthalpy must be based on the moles of CuSO4
●● Mass of water = 150.00 g initially present.
q = 150.0 g × 4.18 J g–1 K–1 × 6.5 K = 4100 J n(CuSO4) = cV = 1.00 mol dm–3 × 1.00 dm3 =
0.0500 g 1.00 mol
0.0500 g is = 1.61 × 10–3 mol of
30.97 g mol–1 In calculating the molar enthalpy change ΔH
phosphorus atoms we have to recognize that this is an exothermic
In calculating the enthalpy change of combustion reaction and that ΔH will therefore be a negative
ΔHc we have to recognize that this is an value.

2
 –q –209 kJ NH4Cl is dissolved) we have to recognize that
ΔH = = = –209 kJ mol–1
n 1.00 mol this is an endothermic reaction and that ΔH will
13 Temperature change ΔT = 32.3 °C – 24.5 °C = therefore have a positive value.
7.8 °C = 7.8 K +q +1650 J

ΔH⊖ = = = +16 500 J mol–1
In the experiment the mass that is being heated n 0.1000 mol
is the combined mass of the NaOH and HCl = +16.5 kJ mol–1
solutions. If we assume that these solutions
(A different answer is obtained if the mass of the
have the same density as water, 1.000 g cm–1,
solution is assumed to be the same as the mass
then 50.00 cm3 of the solutions has a mass
of the water only, i.e. 100.00 g. In this case ΔH =
of 50.00 g. Total mass = 50.00 g + 50.00 g =
+15.7 kJ mol–1.)
100.00 g.
q = mcΔT = 100.00 g × 4.18 J g–1 K–1 × 7.8 K 15 ΔH is change in enthalpy, the heat content of a
system. Enthalpy cannot be measured directly
= 3300 J
but enthalpy changes can be calculated for
Moles of sodium hydroxide, n(NaOH) = cV = chemical reactions and physical processes
50.00 from measured temperature changes using the
0.950 mol dm–3 × dm3 = 0.0475 mol.
1000 equation q = mcΔT, where q is heat change,
In calculating the enthalpy change of
neutralization ΔH we have to recognize that m is the mass of the substance(s) changing
this is an exothermic reaction and that ΔH will temperature, c is the specific heat capacity of
therefore be a negative value. the substance(s) changing temperature and ΔT
is the measured temperature change occurring
–q –3300 J
ΔH = = = –6.9 × 105 J mol–1 = in the substance(s).
n 0.0475 mol
–69 kJ mol–1 16 A Applying Hess’s law the appropriate enthalpy
Assumptions: changes can be calculated:
●● There is no heat loss to the surroundings ΔH (P → S) = ΔH (P → Q) + ΔH (Q → S) =
during the reaction. +50 kJ + –60 kJ = –10 kJ
●● The density of the solution is the same as ΔH (R → Q) = ΔH (R → S) – ΔH (S → Q) =
the density of water, 1.00 g cm–3. +30 kJ – (–60 kJ) = +90 kJ
●● The heat capacity of the solution is the same ΔH (P → R) = ΔH (P → Q) + ΔH (Q → S) –
as the heat capacity of water, 4.18 J g–1 K–1. ΔH (S → R) = +50 kJ + –60 kJ – 30 kJ =
14 heat change, q = mcΔT –40 kJ

In this question the mass that is undergoing a The correct values are I and II. ΔH (P → S) =
temperature change is the mass of the solution –10 kJ and ΔH (R → Q) = +90 kJ.
resulting from 5.035 g of NH4Cl being added to 17 Applying Hess’ law the enthalpy change for
100.00 cm3 of water. the overall equation can be obtained from the
As the density of water is 1.00 g cm–3 the 100.00 first equation and the reverse of the second
cm3 has a mass of 100.00 g. equation:
Mass of solution = 100.00 g + 5.035 g = 105.04 g C(graphite) + O2(g) → ΔH ⊖ = –394 kJ
q = mcΔT = 105.04 g × 4.18 J g–1 K–1 × (21.79 – CO2(g)
25.55) K = –1650 J CO2(g) → CO(g) + ½O2(g) ΔH ⊖ = +283 kJ
m 5.350 g C(graphite) + ½O2(g) → ΔH ⊖ = (–394 + 283)
n(NH4Cl) = = 0.1000 mol
M 53.50 g mol–1 CO(g) kJ = –111 kJ
In calculating the molar enthalpy change ΔH for
this reaction (enthalpy change when 1 mole of
3
18 Applying Hess’ law the enthalpy change for The standard state of Br is Br2(l) not Br2(g),
the overall equation can be obtained from the therefore Br2(g) will not have a standard heat
reverse of the first equation and the second of formation of zero.
equation:
22 D The enthalpy of formation is defined as the
2NO(g) → N2(g) + ΔH ⊖ = –180.5 kJ formation of 1 mole of substance from its
O2(g) constituent elements in their standard states.
N2(g) + 2O2(g) → ΔH ⊖ = +66.4 kJ mol–1 If the elemental substances listed
2NO2(g) correspond to the standard states of
2NO(g) + O2(g) → ΔH ⊖ = (–180.5 + 66.4) the elements then the standard heat of
2NO2(g) kJ = –114.1 kJ formation will be zero at 25.0 °C and 1.00 ×
105 Pa.
19 Applying Hess’ law the enthalpy change for the
The standard state of H is H2(g) not H(g). The
overall equation can be obtained from doubling
standard state of Hg is Hg(l) not Hg(s).
the reverse of the first equation and the second
equation: The standard state of C is C(graphite) not
C(diamond). The standard state of Si is Si(s).
2NO2(g) → N2(g) + ΔH ⊖ = 2 × –33.2 kJmol–1
2O2(g) The only substance listed that does have a
standard heat of formation of zero at 25.0 °C
N2(g) + 2O2(g) → ΔH ⊖ = +9.16 kJ mol–1
and 1.00 × 105 Pa is Si(s).
N2O4(g)
2NO(g) + O2(g) → ΔH ⊖ = (–66.4 + 9.16) kJ 23 D The enthalpy of formation is defined as the
2NO2(g) mol–1 = –57.2 kJ mol–1 formation of 1 mole of substance from its
constituent elements in their standard states.
20 B Applying Hess’ law ΔH3 can be obtained
The enthalpy change for any equation that
from reversing the first equation and halving
starts with compounds, not pure elements in
the second equation:
their standard states, cannot correspond to
CO2(g) → CO(g) + ΔH1 = +283 kJ mol–1 an enthalpy change of formation, therefore A
½O2(g) and C are incorrect.
H2(g) + ½O2(g) → ΔH2 = ½ × –572 kJ If the product of the equation is not a pure
H2O(l) mol–1 substance then the enthalpy change of
CO2(g) + H2(g) → ΔH3 = (+283 + –286) kJ that equation cannot correspond to an
H2O(l) mol–1 = –3 kJ mol–1 enthalpy change of formation. CuSO4(aq)
is a homogeneous mixture and not a pure
21 C The enthalpy of formation is defined as the substance so B is incorrect.
formation of 1 mole of substance from its
Only equation D describes one mole of
constituent elements in their standard states.
a pure substance being formed from its
If the elemental substances listed constituent elements in their standard states
correspond to the standard states of so D is correct.
the elements then the standard heat of
formation will be zero at 25.0 °C and 1.00 × 24 (a) The enthalpy of formation is defined as the
105 Pa. formation of 1 mole of substance from its
constituent elements in their standard states.
Cl2(g), I2(g) and Na(s) are the standard states
The constituent elements of CH3COCH3(l)
of Cl, I and Na so these substances will have
are C, H and O and their standard states are
a standard heat of formation of zero.
C(graphite), H2(g) and O2(g).

4
 3C(graphite) + 3H2(g) + 12O2(g) → 29 B Reaction I is the sublimation of water, which
CH3COCH3(l), ΔH ⊖f = –248 kJ mol–1 is an endothermic process. (It takes energy
(The ΔH ⊖f value can be found in Section 12 to convert a solid to a gas.)
of the IB data booklet.) Reaction II is the deposition of carbon
(b) Under standard conditions of 298 K (25 °C) dioxide, which is an exothermic process. (If
and 1.00 × 105 Pa. If the reaction involves it takes energy to convert a solid to gas then
solutions these have a concentration of 1.00 the opposite process will release energy.)
mol dm–3. Reaction III represents the bond dissociation
of O2, which is an endothermic process. (It
25 Fe3O4(s) + 2C(graphite) → 3Fe(s) + 2CO2(g)
takes energy to break an O=O bond.)
ΔHreaction = ΣΔH ⊖f(products) – ΣΔH ⊖f(reactants)
I and III only are endothermic.
Recognizing that C(graphite) and Fe(s) are
already in their standard states (ΔH ⊖f = 0): 30 A The bond enthalpy is the energy required
to homolytically break 1 mole of bonds
ΔHreaction = 2ΔH ⊖f(CO2(g)) – ΔH ⊖f(Fe3O4(s))
in gaseous molecules under standard
= (2 × –394) – (–1118 ) kJ mol–1 conditions. The bond enthalpy of the H–Cl
= +330 kJ mol–1 bond is therefore represented by A, HCl(g)
H(g) + Cl(g).
26 2NO2(g) → N2O4(g)
B is wrong because it also involves formation
ΔHreaction = ΣΔH ⊖f(products) – ΣΔH ⊖f(reactants)
of H–H and Cl–Cl bonds. C is wrong because
= ΔH ⊖f(N2O4(g)) – 2ΔH ⊖f(2NO2(g)) ions form (the bond has been broken
= +9.2 – (2 × +33.2) kJ mol–1 heterolytically with both electrons going to the
= –57.2 kJ mol–1 chlorine atom). All the products and reactants
must be in the gas phase, so D is wrong.
27 D 2H2O2(l) → 2H2O(l) + O2(g)
ΔHreaction = ΣΔH ⊖f(products) – ΣΔH ⊖f(reactants) 31 C Reaction I is the deposition of carbon
dioxide, which is an exothermic process. (If
Recognizing that O2(g) is already in its
it takes energy to convert a solid to gas then
standard state (ΔH ⊖f = 0):
the opposite process will release energy.)
ΔHreaction = 2ΔH ⊖f(H2O(l)) – 2ΔH ⊖f(H2O2(g))
Reaction II is the sublimation of water, which
= (2 × –286) – (2 × –188) kJ mol–1 is an endothermic process. (It takes energy
= –196 kJ mol–1 to convert a solid to a gas.)

28 2MgO(s) + C(s) → CO2(g) + 2Mg(s) Reaction III represents the bond dissociation
of O2, which is an endothermic process. (It
ΔHreaction = ΣΔH ⊖f(products) – ΣΔH ⊖f(reactants)
takes energy to break an O=O bond.)
Recognizing that C(s) and Mg(s) are already in
II and III only are endothermic.
their standard states (ΔH ⊖f = 0), this assumes
that C(s) is C(graphite), which is the standard for 32 This question needs you to know the structure
carbon. of C2H6 (ethane) as this is the reactant whose
ΔHreaction = ΔH ⊖f(CO2(g)) – 2ΔH ⊖f(MgO(s)) bonds are being broken:
= –394 – (2 × –602) kJ mol–1 H H
= +810 kJ mol–1
Such a highly endothermic reaction is unlikely to H C C H
be feasible. (Instead, magnesium is produced by
electrolysis of its ores.) H H

5
 From this structure we can see that the bonds 35 2H2(g) + O2(g) → 2H2O(g)
broken are six C–H bonds and one C–C bond.
Bonds ΔH / kJ mol–1 Bonds H / kJ mol–1
33 B The bond enthalpy is the energy required broken (endothermic) formed (exothermic)
to homolytically break 1 mole of bonds O=O +498
in gaseous molecules under standard
2H–H 2 × (+ 436) 4 × O–H 4 × (–464)
conditions. The bond enthalpy of the C≡O
bond is therefore represented by B, CO(g) → Total = +1370 = –1856
C(g) + O(g).
ΔH = ΣE(bonds broken) – ΣE(bonds formed) =
A and C are wrong because the products
(+1370 + (–1856)) kJ mol–1 = –486 kJ mol–1
formed must also be in the gaseous state
(Answers will vary slightly depending on the
and C(s) is present. C and D are wrong
values used! The bond enthalpy for the O–H
because the equations also represent the
bond given with the question differs from that in
formation of O=O bonds with O2(g) as a
Section 11 of the IB data booklet.)
product.
36 B The reaction for the hydrogenation of a
34 H2C=CH2 + H2 → H3C–CH3
double bond is
ΔH = ΣE(bonds broken) – ΣE(bonds formed)
–CH=CH– + H2 → –CH2–CH2–
Bonds ΔH / kJ mol –1
Bonds H / kJ mol–1
broken (endothermic) formed (exothermic) Bonds ΔH / kJ mol–1 Bonds ΔH / kJ mol–1
broken (endothermic) formed (exothermic)
C=C +612 C–C –(–347)
C=C +612 C–C 347
4C–H 4 × (+413) 6 × C–H 6 × (–413)
H–H +436 2 × C–H 2 × (–412)
H–H +436
Total = +1048 = –1171
Total = +2700 = –2825
ΔH = ΣE(bonds broken) – ΣE(bonds formed)
ΔH = ΣE(bonds broken) – ΣE(bonds formed) = = (+1048 + (–1171)) kJ mol–1 = –123 kJ
(+2700 + (–2825)) kJ mol–1 = –125 kJ mol–1 mol–1
This is the full method, and is totally correct.
(Note that the bond enthalpies given in the
However, it is simpler to ‘cancel out’ the four
question differ from those in Section 11 of
unchanged C–H bonds:
the IB data booklet.)
Bonds ΔH / kJ mol–1 Bonds H / kJ mol–1
37 C C2H4(g) + F2(g) → CH2FCH2F(g)
broken (endothermic) formed (exothermic)

C=C +612 C–C –(–347 ) Bonds ΔH / kJ mol–1 Bonds ΔH / kJ mol–1

broken (endothermic) formed (exothermic)
2 × C–H 2 × (–413)
C=C +612 C–C –347
H–H +436
4C–H 4 × (+436) 4 × C–H 4 × (–436)
Total = +1048 = –1173
F–F 158 2 × C–F 2 × (–467)
ΔH = ΣE(bonds broken) – ΣE(bonds formed) =
Total = +2514 = –3025
(+1048 + (–1173)) kJ mol–1 = –125 kJ mol–1
(Answers will vary slightly depending on the ΔH = ΣE(bonds broken) – ΣE(bonds formed)
values used! The bond enthalpies given with the = (+2514 + (–3025)) kJ mol–1 = –511 kJ
question differ from those in Section 11 of the IB mol–1
data booklet.) (Note that the bond enthalpies given in
the question on page 234 differ from
6
 those in Section 11 of the IB data booklet, a higher bond order the bonding in O2 is
particularly for the C–F bond!) stronger therefore step I needs more energy
than step III.
38 C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
40 From Table 11 in the IB data booklet the O=O
H H
bond enthalpy is 498 kJ mol–1 (the energy
required to break one mole of O=O bonds).
H C C O H (l) + 3 × O=O(g) →
The energy required to break a single O=O bond
498 kJ mol–1
H H = = 8.27 × 10–22 kJ = 8.27
6.02 × 1023 mol–1
2 × O=C=O(g) + 3 × H–O–H(l))
× 10–19J. A light photon will therefore need a
Bonds ΔH / kJ mol–1 Bonds ΔH / kJ mol–1 minimum energy of 8.27 × 10–19J to break the
broken (endothermic) formed (exothermic) O=O bond in an O2 molecule.
C–C +346 4 × C=O 4 × (–804)
hc
Ephoton =
λ
3 × O=O 3 × (+498) 6 × O–H 6 × (–463) hc 6.63 × 1034 J s × 3 × 108 m s–1
λ = =
O–H +463 Ephoton 8.27 × 10–19 J
= 2.41 × 10 m = 241 nm
–7
C–O +358
Any radiation with a wavelength shorter than
5 × C–H 5 × (+414)
241 nm has sufficient energy to break the O=O
Total = +4731 = –5994 bond in oxygen.

ΔH = ΣE(bonds broken) – ΣE(bonds formed) = 41 The oxygen double bond with a bond order of 2
(+4731 + (–5994)) kJ mol–1 = –1263 kJ mol–1 is stronger than the 1.5 bond in ozone (see page
The calculated value is less exothermic than 194). Thus, less energy is required to dissociate
the enthalpy of combustion given for ethanol in O3 than O2. Longer wavelength radiation of lower
Table 13 in the IB data booklet, which is –1367 energy is needed to dissociate O3.
kJ mol–1. This is because the bond enthalpy 42 A As the electron is attracted to the positively
calculation assumes all species are in the charged nucleus of the Cl atom, the process
gaseous state, but water and ethanol are liquids is exothermic. All the other processes are
in this reaction. The calculated value is less endothermic (i.e. require energy to take
exothermic as it does not account for the energy place).
changes associated with changing ethanol and
water liquids to gases. 43 C Electron affinity is the enthalpy change that
occurs when one mole of gaseous atoms
Another reason is that the tabulated bond
attracts one mole of electrons to form one
enthalpies are average values obtained from a
mole of gaseous anions.
range of compounds and are not specific to the
compounds in this reaction. For potassium the equation that represents
its electron affinity is C: K(g) +e– → K–(g).
39 (a) Step II, as bonds are formed; the other
A is incorrect as it represents the first
steps involve breaking bonds, which is an
ionization of potassium. B does not balance
endothermic process.
(no charges on the left and a total of 2– on
(b) O2 has a double bond so the bond the right.) D is incorrect as the electron is
order is 2. O3 has resonance structures/ being added to a potassium ion and not an
delocalization with bonding intermediate atom.
between double and single bonds; the
bond order is 1.5 (see page 194). As it has
7
44 D The enthalpy of atomization for an element X corresponds to twice the first ionization
is the formation of 1 mole of gaseous atoms energy of potassium, ΔHi⊖(K): 2 × (K(g) →
from the element in its standard state. The K+(g) + e–).
standard state of bromine is Br2(l) so the Y is the sum of the first and second electron
enthalpy of atomization corresponds to affinities of oxygen, ΔHe1⊖(O) + ΔHe2⊖(O)
equation D: 12Br2(l) → Br(g). (the gaseous oxygen atoms are turned into
A is incorrect as bromine is not in its gaseous O2– ions by the successive addition
standard state and forms 2 moles of Br of two electrons: O(g) + e– → O–(g) then O–(g)
atoms. + e– → O2–(g)).
B is incorrect as it forms 2 moles of Br Z is the enthalpy of formation of potassium
atoms. oxide, ΔHf⊖(K2O): 2K(s) + ½O2(g) → K2O(s)
C is incorrect as bromine is not in its (the enthalpy of formation is the enthalpy
standard state. change that occurs when one mole of a
substance is formed from its constituent
45 (a) The lattice enthalpy can be defined as the elements in their standard states).
energy required to turn 1 mole of an ionic
(c) As ΔHlat⊖(K2O) corresponds to the equation
solid into its constituent ions in the gaseous
K2O(s) → 2K+(g) + O2–(g), the desired value
state, therefore the equation corresponding
can be calculated by starting at K2O(S) and
to the lattice enthalpy of potassium oxide is:
summing the enthapies of the individual
K2O(s) → 2K+(g) + O2–(g) processes until reaching the desired
2K(g)  O2(g) products, 2K+(g) + O2–(g).
Y ΔHlat = [–Z + 2(89.2) + W + X + Y] kJ mol–1
2K(g)  2e  O(g)
= –(–361) + (2 × 89.2) + (½ × 498) +
(2 × 419) + (–141 + 753) kJ mol–1
X
= +2238 kJ mol–1
H

2K(g)  O(g) ∆Hlat(K2O)

⊖

46 B Lattice enthalpy is the energy required to

2K(g)
1
 2 O2(g) W
form gaseous ions from an ionic solid: MX(s)
1 2(89.2)
2K(s)  2 O2(g)
→ M+(g) + X–(g). Lattice enthalpy increases in
magnitude with increasing charge of the ions
Z  361
and decreasing ionic radius.
K2O(s)
For the four ionic compounds given, the
(Note that in the diagram for part (b) the greatest lattice enthalpy will occur for MgO
arrow representing the lattice enthalpy has or CaO as these both have 2+ and 2– ions
been drawn for the reverse of this process whereas Na2O has 1+ and 2– ions and KCl
and has been labelled as –ΔHlat⊖(K2O).) has 1+ and 1– ions. As Mg2+ has a smaller
(b) W corresponds to the enthalpy of ionic radius than Ca2+, MgO will have the
atomization of oxygen, ΔHatom⊖(O): 12O2(g) → greatest lattice enthalpy.
O(g). 47 Lattice enthalpy is the energy required to form
It also corresponds to half the bond enthalpy gaseous ions from an ionic solid: MX(s) → M+(g)
of the O=O bond E(O=O) (the bond enthalpy + X–(g). Lattice enthalpy increases in magnitude
is the energy required to break 1 mole of with increasing charge of the ions and
the bonds but that would give two oxygen decreasing ionic radius. The ionic compounds all
atoms hence the need to halve the bond have the same cation so the differences in lattice
enthalpy.) enthalpy are due to the halide ions. As halide
8
 ions become larger as we go down the group NaF contains Na+ ions and F– ions, MgCl2
their increasing size results in a lower charge contains Mg2+ ions and Cl– ions. The higher
density on the anion (the 1– charge is spread charge of the magnesium ion compared to
over a larger volume), therefore the electrostatic the sodium results in a large lattice enthalpy
attraction to the positive Na+ ion will decrease for MgCl2 compared to NaF.
and the lattice enthalpies decrease as we go The greater charge on Mg2+ vs Na+ has a
down the group: ΔHlat⊖(NaF) > ΔHlat⊖(NaCl) > much more significant effect on the lattice
ΔHlat⊖(NaBr) > ΔHlat⊖(NaI). enthalpy than the smaller size of the fluoride
48 Lattice enthalpy is the energy required to ion in NaF compared with the chloride ion in
form gaseous ions from an ionic solid: MX(s) NaCl.
→ M+(g) + X–(g). Lattice enthalpy increases in 51 Lattice enthalpy is the energy required to form
magnitude with increasing charge of the ions gaseous ions from an ionic solid: MX(s) → M+(g)
and decreasing ionic radius. + X–(g). Lattice enthalpy increases in magnitude
As MgO contains Mg2+ and O2– ions it will have a with increasing charge of the ions and
larger lattice enthalpy than NaCl, which contains decreasing ionic radius. As Ag+ has a smaller
Na+ and Cl– ions, as there is significantly greater ionic radius than K+ the lattice enthalpy of AgBr
electrostatic attraction between 2+ and 2– ions is larger than that of KBr as the electrostatic
than between 1+ and 1– ions. attraction between the ions in AgBr is stronger.
The smaller ionic radii of Mg2+ and O2– compared Another reason is that the AgBr bond is not
to Na+ and Cl–, respectively, also contributes to purely ionic and has more covalent character.
an increased electrostatic attraction between the This increases the bond strength above what
ions in MgO and a larger lattice enthalpy, but this would be predicted for a purely ionic bond and
is a lesser effect than that of the greater charge results in a larger lattice enthalpy.
on the ions.
52 K+ and F– have similar ionic radii but the
49 C Lattice enthalpy is the energy required to enthalpy of hydration of the F– ion (–504 kJ
form gaseous ions from an ionic solid: MX(s) mol–1) is significantly more exothermic than the
→ M+(g) + X–(g). Lattice enthalpy increases in enthalpy of hydration of K+ (–340 kJ mol–1). This
magnitude with increasing charge of the ions suggests that there is an additional electrostatic
and decreasing ionic radius. attraction between the F– ion and the polar water
For the four ionic compounds given, the molecules other than ion–dipole interactions.
greatest lattice enthalpy will occur for This extra attraction is hydrogen bonding: due
magnesium bromide (MgBr2) and calcium to the small size of the F– ion, its lone pairs can
bromide (CaBr2) as these both have 2+ and form hydrogen bonds to the hydrogen atoms in
1– ions whereas sodium chloride (NaCl) and the highly polar water molecules.
potassium chloride (KCl) both have 1+ and 1– 53 (a)
ions. As Mg2+ has a smaller ionic radius than K(g)  Cl(g)

Ca2+, MgBr2 will have the greatest lattice

∆Hhyd⊖(K+) +
enthalpy.
∆Hhyd⊖(Cl–)
H ∆Hlat⊖(KCI)
50 A Lattice enthalpy is the energy required to K(aq)  Cl(aq)

form gaseous ions from an ionic solid: MX(s)

∆Hsol⊖(KCI)
→ M+(g) + X–(g). Lattice enthalpy increases in
magnitude with increasing charge of the ions KCI(s)

and decreasing ionic radius. From the energy cycle we can see that:

9
 ΔHsol⊖(KCl) = ΔHlat⊖(KCl) + ΔHhyd⊖(K+) + In reaction D a solid and solution react
ΔHhyd⊖(Cl–) to give products, one of which is a gas.
= (+720 + (–340) + (–359)) kJ mol–1 Gases are more disordered than solids and
solutions, which results in a large increase in
= +21 kJ mol–1
entropy.
(b) (from data booklet) ΔHsol⊖(KCl) = +17.22 kJ
In reaction A the reactants are solutions and
mol–1
21 − 17.22 a solid is formed as one of the products.
% accuracy = × 100% = 22% As solids are more ordered than solutions
17.22
The large inaccuracy is based on the this results in a large decrease in entropy. In
calculated value being found by the reaction B there are two moles of gases for
difference between much larger values. Even both reactants and products so they have
a small variation in either of the large values similar disorder and there will be only a small
will result in a large percentage difference change in entropy. In reaction C there are
in the calculated value compared to the two moles of gaseous reactants but only
experimental value. one mole of gaseous product so this results
in an increase in order and a decrease in
54 B An entropy change close to zero (ΔS ≈ 0)
entropy.
will only occur if the reaction proceeds with
little change in disorder. This will only happen 58 (a) ΔS is negative. Two moles of gaseous
if there is no change of state and/or the products are more ordered and have
reactants and products contain the same less entropy than four moles of gaseous
moles of gases. reactants.
B: H2(g) + Cl2(g)  2HCl(g) is the only (b) ΔS is negative. Three moles of solid and
reaction that does not involve a change four moles of gas change into one mole of
of state or a change in the number of gas solid and four moles of gas. There is a small
molecules. decrease in disorder.
55 C An increase in entropy is associated with an (c) ΔS is positive. A solid reactant is being
increase in disorder. converted into an aqueous solution so there
is a large increase in disorder.
Reaction I decreases entropy because
liquids are more ordered than gases, 59
reaction II increases entropy because two Gas
moles of gas have more entropy than
one mole of gas and reaction III increases
entropy because aqueous ions have more
S

entropy than ions in a lattice.

Liquid
56 A A solid is turning into a gas; there is an
increase in disorder and an increase in Solid
entropy so ΔS is positive. Intermolecular T
forces of attraction are overcome as the
60 N2(g) + 3H2(g) → 2NH3(g)
solid turns into a gas, so ΔH is positive and
Page 252 of the textbook gives these entropy
the change is endothermic. (It takes energy
values:
to convert a solid to a liquid.)
H2(g) = 131 J K–1 mol–1
57 D An increase in entropy is associated with an
N2(g) = 191 J K–1 mol–1
increase in disorder.
NH3(g) = 193 J K–1 mol–1

10
 ΣΔS⊖reaction = ΣS⊖(products) – ΣS⊖(reactants) = H2(g) + ½O2(g) → H2O(s), ΔH = –292 kJ
((2 × +193) – (+191 + 3 × +131)) J K–1 mol–1 = mol–1
–198 J K–1 mol–1 For the reaction H2O(s) → H2O(l)
The value is negative, as expected, since the ΔH⊖ = ΣΔH⊖f(products) – ΣΔH⊖f(reactants)
two moles of gas produced have less entropy
= –286 – (–292) = +6 kJ mol–1
than the four moles of reactant gas.
(b) We need to recall the formula
61 The standard state of methane is CH4(g), so ΔG⊖reaction = ΔH⊖reaction – TΔS⊖reaction
to show its formation from its elements in their
Ice will begin melting at a temperature when
standard states we write:
ΔG = 0. Rearranging the formula we find that
C(graphite) + 2H2(g) → CH4(g) ∆H⊖reaction
T=
Page 252 of the textbook gives these entropy ∆S⊖reaction
values: 6000 J mol–1
= = 273 K = 0 °C (which is
C(graphite) = 5.7 J K–1 mol–1 22.0 J K–1 mol–1
rather what you would expect)
H2(g) = 131 K–1 mol–1
CH4(g) = 186 J K–1 mol–1 (Note that the units for ΔH⊖reaction were
converted from kJ mol–1 to J mol–1 to be
ΣΔS⊖reaction = ΣS⊖(products) – ΣS⊖(reactants)
compatible with the units of ΔS⊖reaction , which
= (+186 – +5.7 + (2 × +131)) J K–1 mol–1 = −82 J are in J K–1 mol–1.
K–1 mol–1
64 A We need to recall the formula
The value is negative, as expected, since one
mole of gas product has less entropy than the ΔG = ΔH – TΔS
two moles of reactant gas. The reaction is spontaneous if ΔG is
negative.
62 C NH4Cl(aq) has a larger entropy than NH4Cl(s)
so the entropy change of the system When T is very low, ΔG ≈ ΔH. For the
(reaction) drives the process. reaction to be ‘spontaneous at low
temperatures’, ΔH must be negative.
A is false – exothermic process are
energetically favourable. B is not true – When T is very high, ΔG ≈ –TΔS. For the
the bonds in NH4Cl(s) are stronger than reaction to be ‘non-spontaneous at higher
the ion–dipole interactions between the temperatures’, ΔS must be negative, making
hydrated ions and water, which is why the –TΔS positive and so making ΔG positive.
process is endothermic. D is wrong because 65 D We need to recall the formula
the decrease in temperature represents
ΔG = ΔH – TΔS
heat loss from the surroundings, which is
associated with a decrease in the entropy of The reaction is spontaneous if ΔG is
the surroundings. negative.
When T is very low, ΔG ≈ ΔH. For the
63 (a) Section 13 of the IB data booklet gives the
reaction to be ‘not spontaneous at low
enthalpy of combustion of hydrogen, which
temperatures’, ΔH must be positive.
is the same as the enthalpy of formation of
When T is very high, ΔG ≈ –TΔS. For the
liquid water:
reaction to be ‘spontaneous at higher
H2(g) + ½O2(g) → H2O(l), ΔH = –286 kJ mol–1
temperatures’, ΔS must be positive, making
The question gives us the enthalpy of –TΔS negative, and so making ΔG negative.
formation of solid water (ice):

11
66 B For a reaction to be spontaneous ΔG spontaneous (ΔG < 0) and where it is non-
has to be negative. At low temperature spontaneous (ΔG > 0).
ΔG(system) ≈ ΔH(system), as TΔS ≈ 0. As ∆H
If ΔG = 0 then T =
ΔH is negative, this reaction will occur at ∆S
low temperatures. At high temperatures 100 kJ mol–1
T=
ΔG(system) ≈ –TΔS(system) as the 100 J K–1 mol–1
temperature is sufficiently high as to make 100 000 J mol–1
T =
the term ΔH(system) negligible. Hence if 100 J K–1 mol–1
= 1000 K
ΔS(system) is negative then –TΔS(system)
will be a positive value and so the reaction is 69 C For a reaction to be spontaneous ΔG
not spontaneous. has to be negative. At high temperatures
67 (a) The reaction requires energy to decompose ΔG(system) ≈ –TΔS(system) as the
the carbonate ion so it is endothermic and temperature is sufficiently high as to make
ΔH is positive. the term ΔH(system) negligible. Hence, if
ΔS(system) is positive then –TΔS(system)
(b) One mole of gas is produced from a solid
will be a negative value, i.e. the reaction is
reactant, so there will be a large increase in
spontaneous.
disorder and ΔS will be positive.
(c) We need to recall the formula 70 CaCO3(s) → CaO(s) + CO2(g)
ΔG = ΔH – TΔS ΔGreaction = ΣΔG⊖reaction (products) – Σ ΔG⊖reaction
(reactants)
The reaction is spontaneous if ΔG is
negative. = –604 + (–394) – (–1129) = +131 kJ mol–1
●● When T is very low, ΔG ≈ ΔH. If ΔH is The positive value of ΔG shows that the reaction
positive, then the reaction will not be is not spontaneous at this temperature.
spontaneous at low temperatures. 71 CaCO3(s) → CaO(s) + CO2(g)
●● When T is very high, ΔG ≈ –TΔS. If ΔS is The worked example gives values of ΔH⊖reaction =
positive (making ΔG negative), then the +178 kJ mol–1 and ΔS⊖reaction = 160.8 J K–1 mol–1.
reaction will be spontaneous at higher
It is necessary to convert the units of ΔS⊖reaction
temperatures.
from J K–1 mol–1 to kJ K–1 mol–1 so that they are
68 D We need to recall the formula consistent with the units of ΔH⊖reaction , which are
ΔG = ΔH – TΔS kJ mol–1. ΔS⊖reaction = 160.8 J K–1 mol–1 = 160.8 ×
10–3 kJ K–1 mol–1.
The reaction is spontaneous if ΔG is
negative. ΔGreaction = ΔH⊖reaction – TΔS⊖reaction
●● When T is very low, ΔG ≈ ΔH. As ΔH = 178 kJ mol–1 – (2000 K × 160.8 × 10–3 kJ K–1
is positive, then the reaction will not be mol–1)
spontaneous at low temperatures. = 178 kJ mol–1 – 321.6 kJ mol–1
●● When T is very high, ΔG ≈ –TΔS. As ΔS = –144 kJ mol–1
is positive (making ΔG negative), then
72 B The definitions of enthalpy change of
the reaction will be spontaneous at
formation and free energy change of
higher temperatures.
formation both refer to the formation of a
The 1000 K figure comes from the substance from its elements in their standard
temperature where ΔG = 0. This is the states. In both cases, if elements are in their
boundary between where the reaction is standard states they need no reaction to

12
 be formed, so the values for ΔH⊖f and ΔG⊖f all temperatures. At low temperature
are zero. S⊖ = 0 describes a situation where ΔG(system) ≈ ΔH(system), as TΔS ≈ 0,
there is perfect order and this only occurs at hence exothermic reactions can occur at
absolute zero (T = –273.15 K) so no element low temperatures. At high temperatures
in its standard state will have S⊖ = 0. ΔG(system) ≈ –TΔS(system) as the
temperature is sufficiently high as to make
73 (a) 2C(graphite) + 3H2(g) + 12O2(g) → C2H5OH(l)
the term ΔH(system) negligible. Hence if
(b) Section 12 of the IB data booklet gives ΔS(system) is positive then –TΔS(system)
the formation entropy for ethanol as will be a negative value, i.e. the reaction
+161 J K–1 mol–1. is spontaneous. It is considered to be
ΔS⊖reaction = ΣS⊖ (products) – ΣS⊖ (reactants) a complete reaction if ΔG(reaction)
= (+161 – (2 × +5.7) + (3 × +65.3) + < –30 kJ mol–1.
(12 × +102.5)) J K–1 mol–1 75 D If ΔG(reaction) = 0 kJ mol–1 then an
= –98 J K mol
–1 –1
equilibrium will be present (see page 262).
(Note that the values for S⊖(H2) and S⊖(O2) At this point the free energy will also be at
given in this question, +65.3 J K–1 mol–1 a minimum and the entropy at a maximum.
and +102.5 J K–1 mol–1, are incorrect. The For a better understanding of this answer
actual values are S⊖(H2) = +131 J K–1 mol–1 see the diagram provided on page 336.
and S⊖(O2) = +205 J K–1 mol–1. Using these
76 B If the reaction has reversed by changing
correct values gives ΔS⊖reaction = –346 J K–1
temperature then ΔG(system) has changed
mol–1.)
sign from positive to negative, or vice versa.
(c) Section 12 of the IB data booklet gives Given that ΔG = ΔH – TΔS, this can only
the formation enthalpy for ethanol as happen if ΔH and ΔS are either both positive
–278 kJ mol–1. or both negative.
ΔG = ΔH⊖reaction – TΔS⊖reaction
ΔH ΔS ΔG = ΔH – TΔS
= –278 kJ mol–1 – (500 K × (–98 × 10–3
kJ K–1 mol–1)) positive negative positive at all T

= –278 kJ mol–1 + 49 kJ mol–1 negative positive negative at all T

= –229 kJ mol –1
positive positive positive at low T
(Note that the answer of ΔG = –105 kJ mol–1 negative at high T
is obtained if ΔS⊖reaction = –346 J K–1 mol–1 is
negative negative negative at low T
used.)
positive at high T
(d) The negative free energy value indicates that
this reaction is spontaneous at 500 K. 77 All three gases will be present as an equilibrium
(e) When T is very low, ΔG ≈ ΔH. If ΔH mixture when ΔGreaction is in the range –30 kJ mol–1
is negative, then the reaction will be to + 30 kJ mol–1.
spontaneous at low temperatures. We can determine the temperatures which
When T is very high, ΔG ≈ –TΔS. If ΔS is correspond to ΔGreaction = –30 kJ mol–1 and
negative then –TΔS will be positive, making ΔGreaction = +30 kJ mol–1 using the equation
ΔGreaction = ΔH⊖reaction – TΔSreaction . Note that this
⊖
ΔG positive, and then the reaction will not be
spontaneous at higher temperatures. requires converting ΔSreaction from units of J K–1
⊖

mol–1 to kJ K–1 mol–1.

74 C For a reaction to be spontaneous at all
temperatures ΔG has to be negative at

13
 First consider when ΔG = –30 kJ mol–1:
ΔGreaction = ΔH⊖reaction – TΔS⊖reaction Practice questions
∆Greaction – ∆H⊖reaction
T = 1 Using the values provided, heat change = mc∆T
∆S⊖reaction
= (10 g × 8.99 × 10–1 J K–1 g–1) × (50 – 20) K.
–30 kJ mol–1 – –123 kJ mol–1
= = 727 K This gives the answer in joules: (10 × 8.99 × 10–1
– (–128 × 10–3 kJ K–1 mol–1)
× 30) J.
Now consider when ΔG = +30 kJ mol–1:
However, the question requires the answer in kJ
ΔGreaction = ΔH⊖reaction – TΔS⊖reaction
so it needs to be divided by 1000: heat change
∆Greaction – ∆H⊖reaction 10 × 8.99 × 10–1 × 30
T = = kJ
∆S⊖reaction 1000
+30 kJ mol–1 – –123 kJ mol–1 Correct answer is D.
= = 1195 K
– (–128 × 10–3 kJ K–1 mol–1)
The three gases will be present at temperatures 2 Negative enthalpy changes occur for exothermic
between 727 K and 1195 K. reactions and processes.
I is the combustion reaction of methanol so is
78 All three gases will be present as an equilibrium
exothermic.
mixture when ΔGreaction is in the range –30 kJ
mol–1 to + 30 kJ mol–1. II is an acid–base neutralization reaction so is
exothermic.
We can determine the temperatures which
correspond to ΔGreaction = –30 kJ mol–1 and III is the condensation of water so is exothermic.
ΔGreaction = +30 kJ mol–1 using the equation Correct answer is D.
ΔGreaction = ΔH⊖reaction – TΔS⊖reaction . Note that this
3 Both beakers have sufficient acid (0.10 mol)
requires converting ΔS⊖reaction from units of J K–1
present to react with all of the Mg so both
mol–1 to kJ K–1 mol–1.
reactions will release the same amount of heat.
First consider when ΔG = –30 kJ mol–1: However, this heat is being absorbed by a
ΔGreaction = ΔH⊖reaction – TΔS⊖reaction smaller mass of solution in beaker A so it will
experience a greater temperature increase as
∆Greaction – ∆H⊖reaction
T = both solutions will have the same heat capacity.
∆S⊖reaction
q
–30 kJ mol–1 – –93 kJ mol–1 q = mcΔT or ΔT =
= = 318 K mc
– (–198 × 10–3 kJ K–1 mol–1)
Now consider when ΔG = +30 kJ mol–1: Correct answer is A.

ΔGreaction = ΔH⊖reaction – TΔS⊖reaction 4 Applying Hess’ Law the enthalpy change for
the overall equation can be obtained from the
∆Greaction – ∆H⊖reaction
T = first equation and the reverse of the second
∆S⊖reaction
equation:
+30 kJ mol–1 – –93 kJ mol–1
= = 621 K Cu2O(s) + ½O2(g) → 2CuO(s) ΔH⊖ = –144 kJ mol–1
– (–198 × 10–3 kJ K–1 mol–1)
The three gases will be present at temperatures Cu(s) + Cu2O(s) → Cu2O(s) ΔH⊖ = –11 kJ mol–1
between 318 K and 621 K. Cu(s) + ½O2(g) → CuO(s) ΔH⊖ = (–144 – 11) kJ mol–1
Correct answer is C.

5 Bond enthalpy is the energy required to break 1

mole of a bond in a gaseous molecule averaged
over similar compounds. The bond breaks
homolytically, with each atom taking one of the

14
 bonded electrons, and no charged species are ΣE(bonds broken) = (3 × E(C–H)) + E(C–O) +
formed. B is the only equation that shows the H– E(O–H) + (1.5 × E(O=O))
Cl bond being broken homolytically to give H(g) = (3 × 414) + 358 + 463 + (1.5 × 498) kJ mol–1
and Cl(g).
= 2810 kJ mol–1
Correct answer is B.
ΣE(bonds formed) = 2 × E(C=O) + 4 ×
6 Applying Hess’ Law the enthalpy change of the E(O–H)
overall equation can be obtained from doubling = (2 × 804) + (4 × 463) kJ mol–1
the first and second equations and combining
= 3460 kJ mol–1
with the third equation:
ΔH = ΣE(bonds broken) – ΣE(bonds formed)
2CH4(g) + 2O2(g) → 2HCHO(l) + 2H2O(l)ΔH⊖ = 2x
= 2810 – 3460 kJ mol–1 = –650 kJ mol–1
2HCHO(l) + O2(g) → 2HCOOH(l)ΔH⊖ = 2y m
(b) (i) n(CH3OH) =
2HCOOH(l) + ½O2(g) → (COOH)2(s) + H2O(l) M(CH3OH)
(80.557 – 80.034) g 0.523 g
ΔH⊖ = z = =
32.05 g mol –1
32.05 g mol–1
2CH4(g) + 3½O2(g) → (COOH)2(s) + 3H2O(l) = 0.0163 mol
ΔH⊖ = 2x + 2y + z
(ii) q = mc∆T = 20.000 g × 4.18 J g–1 K–1 ×
Correct answer is C. (26.4 – 21.5) K = 410 J (2 s.f.) = 0.41kJ
7 Bond enthalpy is the energy required to (iii) In calculating the enthalpy change of
homolytically break 1 mole of a bond in a combustion, ∆Hc, we have to recognize
gaseous molecule averaged over similar this is an exothermic reaction and that
compounds, therefore the equation representing ∆Hc will therefore be a negative value.
the bond enthalpy of the C–Cl bond should only heat change
∆H⊖c =
involve the breaking of one C–Cl bond. Only n(CH3OH)
equation B meets this requirement. –410 J
= = –25 kJ mol–1
Correct answer is B. 0.0163 mol
(c) (i) The enthalpy change calculated from
8 The equation provided involves the burning of 2
bond enthalpies can only be an estimate
moles of CO(g). The enthalpy for the burning of
as the tabulated bond enthalpies are
1.00 mole will be half of the enthalpy associated
average values obtained from similar
with this equation:
compounds and are not specific to the
–564 kJ × 0.5 = –282 kJ compounds in the reaction.
The question asks for the energy released when The tabulated bond enthalpies are also
1.00 mol of CO is burned, not the enthalpy obtained from gaseous compounds but
change, so the answer is 282 kJ. the combustion reaction conducted in
Correct answer is B. the experiment uses liquid methanol.

9 (a) ΔH = ΣE(bonds broken) – ΣE(bonds formed) (ii) Heat released from the reaction is lost to
the surrounding air as well as going into
H
heating the glass test-tube and is not all
H C O H + 1.5 O O going into heating the water.

H Complete combustion may not be

occurring, which would decrease the
O C O + 2 O
H H amount of heat released by the reaction.

15
10 (a) Possible assumptions: heat change –10.1 kJ
(d) ΔH = =
●● All of the heat released by the reaction is n(Cu )
2+
0.0500 mol
transferred to the solution and there is no = –202 kJ mol–1
heat loss to the surroundings. (The enthalpy change is negative as this is
●● The specific heat capacity of the solution an exothermic reaction and heat is released
is equal to the specific heat capacity of by the reaction.)
water. 11 (a) ΔH = ΣE(bonds broken) – ΣE(bonds formed)
●● The density of the solution is 1.00 g cm 3
H H
H H
(i.e. 50.0 cm3 of solution has a mass of
C C + H H H C C H
50.0 g).
H H
●● The temperature changes occur uniformly H H
through the solution.
ΣE(bonds broken) = (4 × E(C–H)) + E(C=C) +
●● The polystyrene cup does not absorb E(H–H)
any of the heat (i.e. it has negligible heat
= (4 × 414) + 614 + 436 kJ mol–1
capacity).
= 2706 kJ mol–1
●● The excess zinc powder does not
absorb any heat (i.e. it has negligible heat ΣE(bonds formed) = (6 × E(C–H)) + E(C–C)
capacity). = (6 × 414) + 346 kJ mol–1
(b) (i) The final temperature, Tfinal, that would = 2830 kJ mol–1
have occurred for an instantaneous ΔH = ΣE(bonds broken) – ΣE(bonds formed)
reaction can be calculated from the = 2706 – 2830 kJ mol–1 = –124 kJ mol–1
equation provided using the time of
(The same value can be obtained by
mixing, t = 100 s.
excluding bonds common to both reactants
Tfinal = –0.050(100) + 78.0 = 73.0 °C and products, and focusing on the bonds
From the data table provided we can unique to both reactants and products:
see that the initial temperature, Tinitial, ΔH = ΣE(bonds broken) – ΣE(bonds formed)
was 24.8 °C. = (E(C=C) + E(H–H)) – (E(C–C) + (2 ×
∆T = Tfinal – Tinitial = 73.0 – 24.8 °C E(C–H))))
= 48.2 °C (= 48.2 K)
(b) ∆Hreaction = ΣH⊖c (reactants) – ΣH⊖c (products)
(ii) That the temperature decreases at
= ΔH⊖c (C2H4(g)) + ΔH⊖c (H2(g)) – ΔH⊖c
a uniform rate (i.e. the linear fit is
(C2H6(g))
appropriate).
= (–1411 – 286) kJ mol–1 – (–1560) kJ mol–1
(iii) Heat change (q) = mc∆T = 50.0 g ×
4.18 J g–1 K–1 × 48.2 K = 1.01 × 104 J = –137 kJ mol–1
= 10.1 kJ (c) The values John used for bond enthalpies
(c) If the solution became colourless then all were average values obtained for various
of the Cu2+ ions have reacted with the zinc compounds whereas Marit used enthalpies
powder. of combustion values that were specific to
50.0 the compounds used in the reaction.
n(Cu2+) = cV = 1.00 mol dm–3 × dm3 =
1000 (d) (i) ΔH = ΣE(bonds broken) – ΣE(bonds
0.0500 mol
formed)
From the balanced equation, n(Zn) = n(Cu2+)
= 0.0500 mol.

16
 H H H H ΣE(bonds broken) = (4 × E(N–H)) + E(N–N) +
H H
C H C
H E(O=O)
C C
H H
C C
H
= (4 × 391) + 158 + 498 kJ mol–1
+ H H
H
C C
H
C C
H = 2220 kJ mol–1
C H
H H
C
H ΣE(bonds formed) = E(N≡N) + (4 × E(O–H))
H H
H H = 945 + (4 × 463) kJ mol–1
ΣE(bonds broken) = (10 × E(C–H)) + = 2797 kJ mol–1
E(C=C) + (5 × E(C–C)) + E(H–H) ΔH = ΣE(bonds broken) – ΣE(bonds formed)
= (10 × 414) + 614 + (5 × 346) + 436 kJ = 2220 – 2797 kJ mol–1 = –577 kJ mol–1
mol–1 13 Consider the Lewis structures of O2 and O3:
= 6920 kJ mol -1
O O O O O O
ΣE(bonds formed) = (12 × E(C–H)) + (6 O O
× E(C–C))
= (12 × 414) + (6 × 346) kJ mol–1
O O
= 7044 kJ mol–1 O
ΔH = ΣE(bonds broken) – ΣE(bonds Bond order in O2 = 2. Because it has two
formed) = (6920 – 7044) kJ mol–1 = resonance structures the bond order in O3 = 1.5.
–124 kJ mol–1
Reaction II will require shorter wavelength as
(The same value can be obtained by the bond in O2 has a higher bond order and is
excluding bonds common to both stronger than the bonds in O3. Ultraviolet light of
reactants and products, and focusing a higher energy (i.e. shorter wavelength) will be
on the bonds unique to both reactants required to break the stronger bond in O2.
and products:
14 Lattice enthalpy is the energy required to form
ΔH = ΣE(bonds broken) – ΣE(bonds gaseous ions from an ionic solid: MX(s) → M+(g)
formed) = (E(C=C) + E(H–H)) – (E(C–C) + X–(g). It increases in magnitude with increasing
+ (2 × E(C–H)))) charge of the ions and decreasing ionic radius.
(ii) Both cyclohexene and cyclohexane For the four ionic compounds given the greatest
are liquids so it is not appropriate to lattice enthalpy will occur for MgO or CaO as
calculate reaction enthalpies for these these both have 2+ and 2– ions whereas NaF
complexes using bond enthalpies as and KF have 1+ and 1– ions. As Mg2+ has a
these are determined for compounds in smaller ionic radius than Ca2+, MgO will have the
the gaseous state. greatest lattice enthalpy.
It would be necessary to determine
Correct answer is A.
the energy required to convert these
liquids to gases, i.e. the enthalpies of 15 A Bond dissociation and enthalpy of
vapourization and incorporate these atomization are both endothermic.
values into the calculations. B Electron affinity is exothermic and ionization
energy is endothermic.
12 ΔH = ΣE(bonds broken) – ΣE(bonds formed) C This reaction is the reverse of endothermic
H H
lattice enthalpy so is exothermic.
N N O
+ O O N N + 2 H D Bond dissociation is endothermic and
H
H H
electron affinity is exothermic.
Correct answer is A.
17
16 Reactions that form gaseous products from IV: ∆H⊖e(Cl) = –349 kJ mol–1
liquid and/or solid reactants have the greatest ∆H⊖lat(MgCl2) = –∆H⊖f(MgCl2) + ∆H⊖atomization(Mg)
increase in entropy (greatest increase in + E(Cl–Cl) + ∆H⊖i1(Mg) + ∆H⊖i2(Mg) + 2∆H⊖e(Cl)
disorder).
= 642 + 148 + 242 + 738 + 1451 + (2 ×
Correct answer is C. –349) kJ mol–1
17 An increase in entropy is associated with an = +2523 kJ
increase in disorder.
(c) Theoretical values are calculated based
A Increasing temperature results in particles
on the premise that the bonding in MgCl2
moving faster and an increase in disorder/
is purely ionic in character. However, the
entropy.
bonding between Mg and Cl also has some
B Gases are more disordered than liquids so
covalent character and this results in the
entropy increases.
experimental lattice enthalpy being higher
C Mixtures are more disordered than pure
than the theoretical lattice enthalpy.
substances so entropy increases.
D A decrease in the number of moles of gases (d) Lattice enthalpy increases in magnitude with
leads to greater order and a decrease in increasing charge of the ions in the salt and
entropy. with decreasing ionic radius.
Correct answer is D. For MgO the ionic attraction is between 2+
and 2– ions so this will result in a higher
18 C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
lattice enthalpy than for MgCl2, where the
∆G⊖reaction = Σ∆G⊖f(products) – Σ∆G⊖f(reactants) attraction is between 2+ and 1– ions.
= ((2 × –394) + (3 × –229)) – (–175 + (3 × 0)) kJ The O2– ion also has a smaller ionic radius
mol–1 than Cl– so this results in a higher lattice
= –1300 kJ mol–1 enthalpy for MgO.
Correct answer is B.

19 2CO(g) + O2(g) → 2CO2(g)

Challenge yourself
∆S⊖reaction = ΣS⊖(products) – ΣS⊖(reactants)
= (2 × 214) – ((2 × 198) + 205) J K–1 mol–1 1 Nitrogen is an element with a positive energy of
= –173 J K mol
–1 –1
combustion:
Correct answer is B. N2(g) + O2(g) → 2NO(g) ΔH⊖c =90 kJ mol–1

20 ∆G⊖reaction = ∆H⊖reaction – T∆S⊖reaction It takes more energy to break the strong N≡N
10.00 triple bond and the O=O bond than is released
= 10.00 kJ mol–1 – (298 K × kJ K–1 mol–1) in the formation of the N≡O triple bond.
1000
= 7.02 kJ mol–1
2 The specific heat capacity of a metal increases
Correct answer is B. with the number of atoms in the sample. For
21 (a) Enthalpy change I is the standard enthalpy of 1-g samples the metal with the lowest molar
atomization of magnesium: ∆H⊖atomization(Mg). mass will have the largest number of atoms
present and the highest specific heat capacity.
Enthalpy change V is the standard enthalpy
Specific heat capacity is approximately inversely
of formation of magnesium chloride:
proportional to the relative atomic mass.
∆H⊖f(MgCl2).
3 After being placed into the calorimeter the heat
(b) From the IB data booklet:
energy lost by the piece of brass is transferred
II: E(Cl–Cl) = +242 kJ mol–1
18
 into the water as well as the aluminium walls of all the bonds being broken and formed belong to
the calorimeter, which are in contact with the molecules in the gaseous state:
water. We therefore need to calculate the heat CH4(g) + 2O2(g) → CO2(g) + 2H2O(g).
transferred to the water and to the aluminium
In reality the combustion reaction forms water in
calorimeter using the appropriate specific heat
the liquid state:
capacities.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Temperature change for water and aluminium
The difference between the two values for
calorimeter (ΔT) = (77.50 – 24.50) °C = 53.00 °C
ΔH⊖c is therefore due to the energy required to
= 53.00 K
vapourize water:
Heat change (water) = mcΔT
H2O(l) →H2O(g)
= 200.00 g × 4.18 J g–1 K–1 × 53.00 K
Applying Hess’s Law we can obtain a value for
= 4.43 × 104 J
the enthalpy change associated with the water
Heat change (aluminium) = mcΔT vapourization using the relevant equations for the
= 80.00 g × 0.900 J g–1 K–1 × 53.00 K two processes. This value can then be used to
= 3.82 × 103 J estimate the strength of the hydrogen bonding in
water.
Total heat energy gained by water and aluminium
= (4.43 × 104 J) + (3.82 × 103 J) = 4.81 × 104 J (1) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = –890 kJ mol–1
Therefore heat energy lost by the brass = 4.81 ×
104 J (2) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
ΔH = –808 kJ mol–1
Heat change (brass) = mcΔT
(2.1) 2H2O(l) → 2H2O(g)
4.81 × 104 J = 21.20 g × 0.400 J g–1 K–1 × ΔT
ΔH = –808 – (–891) = +83 kJ mol–1
ΔT = 5670 K (5670 °C)
(2.2) H2O(l) → H2O(g)
Initial temperature = Tf + ΔT = 77.50 °C + ΔH = +41.5 kJ mol–1
5670 °C = 5748 °C
There are (on average) two hydrogen bonds
The temperature of the flame is 5748 °C. between each molecule so a hydrogen bond is
Assumptions are: approximately 20 kJ mol–1.
●● The brass was the same temperature as the This assumes that all other molecular
flame before it was removed and transferred interactions such as dipole–dipole and London
to the calorimeter. forces are negligible, which is an approximation.
●● No heat is lost by the brass to the air while it
5 Based on the bonding discussions in Chapter 4
is being transferred to the calorimeter.
we can assign a bond order of 1 to each of the
●● The aluminium walls of the calorimeter and C–C single bonds in diamond as each carbon
the water are in thermal equilibrium (have the atom is sp3 hybridized and bonded to four
same temperature). other carbons via single bonds so there are no
●● No heat loss occurs from the calorimeter to possible resonance structures. In graphite the
the surrounding air. C–C bond order is 1.33 as each carbon atom is
sp2 hybridized and has three possible resonance
4 The worked example gives ΔH⊖c = –808 kJ
forms – each with one C=C double bond and
mol–1 and the value for ΔH⊖c provided in Table
two C–C single bonds.
13 of the IB data booklet is –891 kJ mol–1. The
difference in the values is largely to due to the
assumption in the bond enthalpy calculation that

19
 bonds with the water molecules. There is
increased covalent interaction between the
Ag+ ions and the water molecules, which leads
to more exothermic hydration enthalpies as
it will require more energy to overcome these
C C C
attractions.
C C C
C C C C C C
8 Sodium chloride (salt) is an ionic substance that
contains alternating sodium and chlorine ions.
The C–C bonds in graphite are therefore When salt is added to water, the partial charges
intermediate between single and double bonds. on the water molecule are attracted to the Na+
As the C–C bonds in graphite have a higher and Cl– ions. The water molecules work their
bond order (stronger bonds), they are harder to way into the crystal structure and between the
break and graphite is more stable than diamond. individual ions, surrounding them. This disrupts
the ionic attractions between the ions and slowly
6 Free radicals are species that have unpaired
dissolves the salt, but as we have seen the
electrons.
enthalpy change is very small (see page 246
The electron configuration of O is 1s22s22p4 and and Table 19 of the IB data booklet). The reason
the occupancy of the p orbitals can be shown that this is a spontaneous process is that the
using a box diagram: aqueous solution is more disordered and has a
higher entropy, as discussed later in the chapter.

9 When Kc = 1, ΔG⊖reaction = 0.
2p
If ΔG⊖reaction = 0 when Kc = 1, this implies a
As the box diagram shows, oxygen has unpaired logarithmic function: ΔG⊖reaction α lnK.
electrons and will therefore behave as a free When Kc > 1, ΔG⊖reaction < 0, when Kc < 1, ΔG⊖reaction
radical. < 0.
7 With positive ions, there is generally a loose This implies that ΔG⊖reaction and Kc are inversely
electrostatic attraction with the partially related: ΔG⊖reaction α –lnK.
negatively charged oxygen atoms of the water Possible function is therefore ΔG⊖reaction = –A lnKc,
molecules. Positive ions with higher charge where A is a constant with units kJ mol–1.
densities, such as d-block ions, may form The precise relationship discussed in Chapter 7
complex ions with formal coordinate covalent is ΔG⊖reaction = –RT lnKc.

20