Worked Solutions – IB 12 HL Energetics Review
1 The ionization energy is the amount of energy required to remove one mole of electrons from one
mole of atoms or ions of an element in its gaseous state, forming one mole of positive ions.
The electron affinity is the energy change when one mole of atoms or anions of an element in its
gaseous state gains one mole of electrons, forming one mole of negative ions.
2 a Enthalpy of atomization, ΔHat , of potassium: K(s) → K(g)
b Enthalpy of dissociation (atomization), ΔHD , of bromine gas: Br2(g) → 2Br(g)
c first ionization energy, ΔHI, of potassium: K(g) → K+(g) + e–
d first electron affinity, ΔHE , of bromine: Br(g) + e– → Br–(g)
e Enthalpy of formation, ΔHf, of potassium bromide: K(s) + Br2(l) → KBr(s)
f Lattice enthalpy, ΔHlattice , of potassium bromide: KBr(s) → K+(g) + Br–(g)
3 a Na2O(s)
b Enthalpy of atomization, ΔHat
c first ionization energy, ΔHI
d first and second electron affinities, ΔHE
e Enthalpy of lattice formation, ΔHlf
4
Lattice enthalpy, ΔHlattice (RbCl) = – enthalpy of lattice formation = +692
The lattice enthalpy of rubidium chloride is +692 kJ mol-1
5 a KCl will have a greater lattice enthalpy than KI because the smaller radius of Cl– will cause
the force of attraction between the ions to be greater in KCl.
b MgO will have a greater lattice enthalpy than MgS because the smaller radius of O 2– will
cause the force of attraction between the ions to be greater in MgO.
c MgO will have a greater lattice enthalpy than BaO because the smaller radius of Mg2+ will
cause the force of attraction between the ions to be greater in MgO.
d LiF will have a greater lattice enthalpy than CsF because the smaller radius of Li + will
cause the force of attraction between the ions to be greater in LiF.
e CaCl2 will have a greater lattice enthalpy than KCl because the smaller radius of Ca 2+, together
with its larger charge, will cause the force of attraction between the ions to be greater in CaCl2.
6 The assumption that is made in determining theoretical lattice enthalpy values is that each ion
is a sphere with its charge uniformly distributed over the surface and that the bonding in the
compound is essentially purely ionic.
7 The experimental lattice enthalpy is generally greater than the theoretical lattice enthalpy
because there is some extra bonding between the ions. The smaller ion distorts (polarizes) the
larger ion in the lattice and causes there to be more electrons than expected between the positive
and negative ions. This adds some degree of covalent character to the bond and adds to the value
of the experimental lattice enthalpy in comparison to the theoretical lattice enthalpy.
8
The order of increasing covalent character (increasing percentage difference between the
theoretical and experimental lattice enthalpies) is KF < CaO < LiI < AgBr
9 Percentage difference between experimental and theoretical lattice enthalpies of NaCl
Percentage difference between experimental and theoretical lattice enthalpies of AgCl
The very small difference between the experimental and theoretical lattice enthalpies of NaCl
(0.65%) suggests that the bonding in NaCl is very close to purely ionic, with the Na + and Cl– ions
behaving as spheres with their charges evenly distributed. In comparison, there is a substantial
difference between the experimental and theoretical lattice enthalpies of AgCl (15%). This
suggests that the bonding in AgCl, while remaining ionic, has a significant degree of covalent
character, with the Cl– ion being polarized by the smaller Ag+ ion. A considerable amount of
electron density is between the two ions, rather than being evenly distributed, and the shape of
the Cl– ion is distorted from an ideal spherical shape to a more elliptical one.
10 The order of increasing entropy is:
II 1 mol H2O(s) 0°C, 101.3 kPa
III 1 mol H2O(l) 0°C, 101.3 kPa
V 1 mol H2O(l) 25°C, 101.3 kPa
I 1 mol H2O(l) 100°C, 101.3 kPa
IV 1 mol H2O(g) 100°C, 101.3 kPa
11 a N2(g) + 3H2(g) → 2NH3(g), a negative entropy change
b N2O4(g) → 2NO2(g), a positive entropy change
c HCl(g) + NH3(g) → NH4Cl(s), a negative entropy change
d CaCO3(s) → CaO(s) + CO2(g), a positive entropy change
e C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g), a positive entropy change
12 The order of increasing entropy change is:
II, PCl3(g) + Cl2(g) → PCl5(g) (a decrease: two gaseous particles → one gaseous particle)
IV, C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) (no change: four gaseous particles → four
gaseous particles)
I, COCl2 (g) → CO(g) + Cl2(g) (a small increase: one gaseous particle → two gaseous
particles)
III, 2C6H12(l) + 9O2(g) → 6CO2(g) + 6H2O(g) (a moderate increase: 9 gaseous particles
+ 2 liquid particles → 12 gaseous particles)
13 The entropy of water decreases as it condenses from steam to liquid water.
For the reaction H2O(g) → H2O(l):
ΔS rxn = ΣΔS o (products) – ΣΔS o (reactants)
= [ΔS o (H2O(l))] – [ΔS o (H2O(g))]
= 70 – 189 J K–1 mol–1
= –119 J mol-1 K-1
The entropy change for the reaction H2O(g) → H2O(l) is –119 J K–1 mol–1.
14 a C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g)
ΔS orxn = [4S o (CO2) + 4S o (H2O(g))] – [S o (C4H8) + 6S o (O2)]
= [(4 × 214) + (4 × 189)] – [307 + (6 × 205)] J K–1 mol–1
= 1612 – 1537 J K–1 mol–1
= +75 J K–1 mol–1
The entropy change for the reaction C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g) is +75 J K–1 mol–1.
b 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)
ΔS orxn = [2S o (CO2) + 4S o (H2O(g))] – [S o (CH3OH) + 3S o (O2)]
= [(2 × 214) + (4 × 189)] – [(2 × 127) + (3 × 205)] J K–1 mol–1
= 1184 – 869 J K–1 mol–1
= +315 J K–1 mol–1
The entropy change for the reaction: 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g) is +315 J K–1 mol–
1.
c C2H4(g) + H2(g) → C2H6(g)
ΔS orxn = [S o (C2H6)] – [S o (C2H4) + S o (H2)]
= 230 – 350 J K–1 mol–1
= –120 J K–1 mol–1
The entropy change for the reaction C2H4(g) + H2(g) → C2H6(g) is –120 J K–1 mol–1.
d 2SO2(g) + O2(g) → 2SO3(g)
ΔS orxn = [2S o (SO3)] – [2S o (SO2) + S o (O2)]
= (2 × 256) – [(2 × 249) + 205] J K–1 mol–1
= 512 – 703 J K–1 mol–1
= –191 J K–1 mol–1
The entropy change for the reaction 2SO2(g) + O2(g) → 2SO3(g) is –191 J K–1 mol–1.
15 For the reaction N2(g) + O2(g) → 2NO(g):
the reaction will change from being spontaneous to non-spontaneous when ΔGo = 0.
So, to find the temperature at which this change occurs:
0 = ΔHo – TΔS o
TΔS o = ΔHo
T = ΔHo/ ΔS o
=180.8×103/24.7
= 7320 K
The reaction begins to be spontaneous at temperatures greater than 7320 K.
16 For the reaction H2O(l) → H2O(g):
ΔHorxn = [ΔHof(H2O(g))] – [ΔHof(H2O(l))]
= (–242 kJ mol–1) – (–286 kJ mol–1)
= +44 kJ mol–1
ΔS orxn = [So (H2O(g))] – [So(H2O(l))]
= 189 – 70
= 119 J K–1
The reaction will change from being spontaneous to non-spontaneous when ΔGo = 0.
So, to find the temperature at which this change occurs:
0 = ΔHo – TΔS o
TΔS o = ΔHo
T = ΔHo/ ΔS o
= 44×103/119
= 370 K
The reaction begins to be spontaneous at temperatures greater than 370 K (97°C).
17 a CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
ΔGorxn = ΣΔGof(products) – ΣΔGof(reactants)
= [(1 mol) ΔGof(CO2) + (2 mol)(ΔGof(H2O)] – [(1 mol)(ΔGof(CH4) + (2 mol)(ΔGof(O2)]
= [–394 + (2 × –229)] – [–51 + 0] kJ mol–1
= –852 + 51 kJ mol–1
= –801 kJ mol–1
The free energy change for the reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) is
–801 kJ mol–1.
b C2H4(g) + H2(g) → C2H6(g)
ΔGorxn = ΣΔGof(products) – ΣΔGof(reactants)
= [ΔGof(C2H6)] – [ΔGof(C2H4) + ΔGof(H2)]
= –33 – 68 kJ mol–1
= –101 kJ mol–1
The free energy change for the reaction C2H4(g) + H2(g) → C2H6(g) is –101 kJ mol–1.
c CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
ΔGorxn = ΣΔGof(products) – ΣΔGof(reactants)
= [ΔGof(CCl4) + 4ΔGof(HCl)] – [ΔGof(CH4) + 4ΔGof(Cl2)]
= [–65 + (4 × –95)] – [–51 + 0] kJ mol–1
= –445 + 51 kJ mol–1
= –394 kJ mol–1
The free energy change for the reaction CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g) is –394 kJ mol–1.
18 For the reaction N2(g) + O2(g) → 2NO(g):
ΔGorxn = ΣΔGof(products) – ΣΔGof(reactants)
= [2ΔGof(NO)] – [ΔGof(N2) + ΔGof(O2)]
= 2 × 87 – 0 kJ mol–1
= 174 kJ kJ mol–1
The free energy change for the reaction N2(g) + O2(g) → 2NO(g) is +174 kJ mol–1.
Using ΔGo = ΔHo – TΔS o :
ΔGo = +180.8 × 103 – (298 × 24.7) kJ mol–1
= 180.8 × 103 – 7.36 × 103 kJ mol–1
= 173.4 × 103 kJ mol–1
= 173 kJ mol–1
The free energy change for the reaction N2(g) + O2(g) → 2NO(g) is +173.4 kJ mol–1.
The two values are about the same.
19 Note that while ΔHoc(C3H7OH) = –1167 kJ mol–1, for the balanced equation
2CH3CHO(l) + 5O2(g) → 4CO2(g) + 4H2O(g)
o
–1
ΔH rxn = –2334 kJ mol , since 2 mol of CH3CHO(l) are required to balance the equation.
ΔHorxn (CH3CHO) = ΣnΔHof(products) – ΣmΔHof(reactants)
ΔHorxn (CH3CHO)= [(4 mol)ΔHof(CO2) + (4 mol)ΔHof (H2O)] – [(2 mol)ΔHof( CH3CHO) + (5
mol)ΔHof(O2)]
Rearranging the equation to make (2 mol)ΔHof(C3H7OH) the subject gives:
(2 mol)ΔHof(CH3CHO)= [(4 mol)ΔHof(CO2) + (4mol)ΔHof(H2O)] – [(5 mol)ΔHof(O2)] –
ΔHorxn(CH3CHO)
Substitute the values given for ΔHof and ΔHoc into this equation:
2 × ΔHof(CH3CHO) = [(4 × –394) + (4 × –242)] – [(5 mol) × 0] – (–2334)2 × ΔHof
(CH3CHO) = –2544 + 2334 = –210
ΔHof(CH3CHO(l)) = –105 kJ mol–1
20 For the reaction 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g):
ΔHorxn = [6ΔHof(CO2) + 6ΔHof(H2O)] – [2ΔHof(C3H6) + 9ΔHof(O2)]
= [(6 × –394) + (6 × –242)] – [(2 × +20) + 0] kJ mol–1
= –3816 – 40 kJ mol–1
= –3856 kJ mol–1
The standard enthalpy change for the reaction 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(l) is
– 3856 kJ.
The standard enthalpy change of combustion for propene, C3H6, is − 3856 kJ mol–1/2
= –1928 kJ mol–1.