Cambridge International A Level Chemistry
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Answers to SAQs
Chapter 19
5
a
1
a standard temperature = 298 K, standard
pressure = 1 × 105 Pa / 100 kPa
b i Mg2+(g) + O2–(g) → MgO(s)
ii K+(g) + Br–(g) → KBr(s)
iii 2Na+(g) + S2–(g) → Na2S(s)
a The bond energy for chlorine is the enthalpy
change Cl2(g) → 2Cl(g). The enthalpy change
of atomisation is 12 Cl2(g) → Cl(g). So the
enthalpy change of atomisation is +244/2 =
+122 kJ mol–1.
b i 12 O2(g) → O(g)
ii Ba(s) → Ba(g)
iii 12 Br2(l) → Br(g)
c 0 kJ mol–1
Because helium exists naturally as single
gaseous atoms, no change is involved in the
process He(g) → He(g).
K+(g) + e– + Br(g)
K+(g) + Br –(g)
∆Hat
1
K(g) + 2 Br2(l)
K(s) +
∆Hea
∆Hi
K(g) + Br(g)
∆Hlatt
∆Hat
1
2 Br2(l)
∆Hf
KBr(s)
2
b i enthalpy change of atomisation of iodine
ii first electron affinity of nitrogen
iii enthalpy change of formation of strontium
chloride
iv lattice energy of cadmium chloride
6
a
Mg2+(g) + O(g) + 2e–
∆ H ea1
1
∆H at Mg2+(g) + 2 O2(g)
+2e–
3
a There must be an input of energy to
overcome the repulsive forces between the
(negative) electron and the negative ion.
b +440 kJ mol–1
c i I(g) + e– → I–(g)
ii S–(g) + e– → S2–(g)
4 a i Cs(g) → Cs+(g) + e–
ii Al2+(g) → Al3+(g) + e–
iii Ca(s) + 12 O2(g) → CaO(s)
iv Fe(s) + 1 12 Cl2(g) → FeCl3(s)
b ΔHʅlatt = ΔHʅf – {ΔHʅat [Na] + ΔHʅi1 [Na]
+ ΔHʅat [ 12 Cl2(g)] + ΔHʅea1 [Cl]}
ΔHʅlatt = (–411) – {(+107) + (+496) + (+122) + (–348)}
= +377 kJ mol–1
ΔHʅlatt = (–411) – (+377) = –788 kJ mol–1
Mg2+(g) + O2–(g)
∆H ea2
+e–
Mg2+(g) + O–(g)
∆H i2
1
∆H latt
Mg+(g) + 2 O2(g)
–
+e
∆H i1
1
Mg(g) + 2 O2(g)
∆H at Mg(s) + 12 O2(g)
∆H f
MgO(s)
b
2Na+(g) + O2–(g)
∆H ea2
∆H at
2Na+(g) + O(g)
+2e–
2Na+(g) + 12 O2(g)
+e–
∆H ea1 2Na+(g) + O–(g)
+2e–
∆H latt
2 × ∆H i1
2Na(g) + 12 O2(g)
2 × ∆H at 2Na(s) +
1
2
O2(g)
∆H f
Na2O(s)
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Cambridge International A Level Chemistry
7
a
Answers to self-assessment questions
is greater for LiF than it is for KBr, accounting
for the greater lattice energy.
2Al3+(g) + 3O2–(g)
2Al3+(g) + 3O(g) + 6e–
2Al3+(g) + 3O–(g) + 3e–
3+
–
10
2Al (g) + 1 12 O2(g) + 6e
2Al2+(g) + 1 12 O2(g) + 4e–
2Al+(g) + 1 12 O2(g) + 2e–
11
2Al(g) + 1 12 O2(g)
2Al(s) + 1 12 O2(g)
Al2O3(s)
a The charge is spread out over a smaller
volume so the charge density is higher.
b Li+ because it has the smallest ionic radius
and therefore has the highest charge density.
c I– because it has the largest ionic radius.
The nitrate ion is an ion with a large ionic
radius so is easily polarised by a small highly
charged cation. Mg2+ has a smaller ionic
radius than Ba2+. Magnesium ions are better
polarisers of nitrate ions than barium ions.
The greater the polarisation, the lower the
thermal stability (the more likely the nitrate
is to decompose).
b ΔHʅlatt = ΔHʅf – {2ΔHʅat [Al] + 2ΔHʅion1+2+3 [Al]
+ 3ΔHʅat [ 12 O2(g)] + 3ΔHʅea1+2 [O]}
ΔHʅlatt = (–1676) – {2 × (+326)
12 a i K2SO4(s) + aq → K2SO4(aq)
+ 2 × (+577 + 1820 + 2740) + 3 × (+249)
or
K2SO4(s) + aq → 2K+(aq) + SO42–(aq)
+ 3 × (–141) + 3 × (+798)}
ii ZnCl2(s) + aq → ZnCl2(aq)
= +13 644 kJ mol–1
or
ΔHʅlatt = (–1676) – (+13 644)
ZnCl2(s) + aq → Zn2+(aq) + 2Cl–(aq)
= –15 320 kJ mol–1
b Sodium chloride and sodium bromide are
soluble in water (because they have values
8 a i BaO
of ΔHʅsol that are negative or slightly positive).
ii MgI2
Silver chloride and silver bromide are insoluble
iii CaO
(because they have large positive values of
b RbCl < LiF < MgO
ΔHʅsol). The data suggest that silver bromide
Lattice energy gets more exothermic the
is less soluble than silver chloride because its
greater the charge on the ions, so MgO > LiF
value of ΔHʅsol is more endothermic. The data
and RbCl.
suggest that sodium chloride is less soluble
Lattice energy gets more exothermic the
than sodium bromide because its value of ΔHʅsol
smaller the ions, so LiF > RbCl.
is slightly positive, whereas the ΔHʅsol of silver
bromide is slightly negative.
9 a For magnesium oxide Q1 × Q2 is 4, for lithium
fluoride Q1 × Q2 is 1. Because ionic radii are
13 a Bond formation is always exothermic.
(almost) unchanged, r2 will be very similar.
Ion–dipole bonds are being formed between
The force between the particles, which is
the gaseous ions and the water molecules.
proportional to Q1 ×2 Q2 , will be four times
b i Na+(g) + aq → Na+(aq)
r
greater for MgO than it is for LiF, accounting
ii Cl–(g) + aq → Cl–(aq)
for the greater lattice energy.
δ+
δ+
c i
H
H
b The ionic charges are the same so Q1 × Q2
δ+ H
δ–
δ+
δ– δ– O H δ+
O
is 1 for both compounds. However, r, the
H O
δ+
Mg2+
H
Br –
separation of the centres of the ions, is much
H
δ+
δ+
O
H
H O δ– δ–
δ+
greater for KBr than it is for LiF. Therefore
δ– O
H δ+
H
H
Q1 × Q2 is much greater for LiF than it is for KBr,
δ+
δ+
2
r
so the attractive force between the particles
Cambridge International AS and A Level Chemistry © Cambridge University Press 2014
Cambridge International A Level Chemistry
δ+
H
– δ– O
Mg2+
δ– δ– O
H
H
δ+ii
δ+
H
H δ+
δ+
δ+ H
δ– O
Br –
O
δ–
H
δ+
H δ+
d Magnesium has a 2+ ion but potassium has
only a 1+ ion. The magnesium ion also has
a smaller radius than the potassium ion.
So the magnesium ion has a greater charge
density than a potassium ion. The greater
the charge density, the greater the attractive
force between the ion and the polar water
molecules and the greater the value of ΔHʅhyd.
14
a
b
c
d
15
a
the enthalpy change of solution of KBr
the enthalpy change of hydration of K+
the lattice enthalpy of KBr
the enthalpy change of hydration of Br–
Mg2+(g) + 2Cl–(g)
b The lattice energy and enthalpy change of
hydration of magnesium sulfate are more
exothermic than those of barium sulfate,
but the difference is more marked for the
enthalpy change of hydration than for lattice
energy. It is the enthalpy change of hydration
of the cations that plays the greatest part in
determining the value of ΔHʅsol.
Because magnesium has a smaller ion than
barium, the enthalpy change of hydration is
more exothermic than for barium.
Overall, the enthalpy change of solution is
less endothermic for magnesium sulfate
than for barium sulfate. This means that
magnesium sulfate is more soluble because
the value of ΔHʅsol is less endothermic (than
for barium sulfate).
∆Hhyd[Mg2+] + 2∆Hhyd [Cl–]
∆Hlatt
MgCl2(s)
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Mg2+(aq) + 2Cl –(aq)
∆Hsol
b ΔHʅlatt + ΔHʅsol = ΔHʅhyd [Mg2+] + 2ΔHʅhyd [Cl–]
so ΔHʅhyd [Mg2+] = ΔHʅatt + ΔHʅsol – 2ΔHʅhyd [Cl–]
ΔHʅhyd [Mg2+] = (–2592) + (–55) – 2 × (–364)
= –2592 + 673
2+
ΔHʅhyd [Mg ] = –1919 kJ mol–1
16
a
Ba2+(g) + SO42–(g)
∆Hhyd [Ba2+] + ∆Hhyd [SO42–]
∆Hlatt [BaSO4]
BaSO4(aq)
∆Hsol [BaSO4]
BaSO4(s)
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