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Summary Complete Energetics II Revision Notes (A Level Edexcel)

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Comprehensive study guide for Chemistry A Level, made by an Oxford Biochemistry student with all 9s at GCSE and 3 A*s at A Level! Information arranged by spec point. Notes written using past papers, textbooks and more.










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March 20, 2021
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2020/2021
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ENERGETICS II
13A: LATTICE ENERGY
1. be able to define lattice energy as the energy change when one mole
of an ionic solid is formed from its gaseous ions
Lattice energy, ΔHLEϴ – the energy change when one mole of an ionic solid is
formed from its constituent gaseous ions under standard conditions.
 Na+ (g) + Cl- (g)  NaCl (s), ΔHLEϴ = -775 kJ mol-1
 A very large negative value shows that the ionic bonds in the lattice are very
strong.
o More energy is released when the ionic bonds are formed.
o The lattice energy of MgCl2 would be stronger than that of NaCl.
 Born-Haber cycles are used to indirectly determine lattice energy as it cannot
be measured directly.


2. be able to define the terms:
i) enthalpy change of atomisation, ΔatH
Enthalpy change of atomisation, ΔHatϴ – the enthalpy change when one mole of
gaseous atoms is formed from the element in its standard state.
 This is always an endothermic process, as bonds are broken.
 Na (s)  Na (g), ΔatH = + 108 kJ mol-1
 Cl2 (g)  Cl (g), ΔatH = + 121 kJ mol-1


ii) electron affinity
First electron affinity, ΔHEA1ϴ – the enthalpy change when one mole of gaseous
atoms each gain one electron, forming one mole of gaseous, singly charged
anions.
 Cl (g) + e-  Cl- (g), ΔHEA1ϴ = – 364 kJ mol-1
 This is an exothermic process for all non-noble gas elements, as you are
forming an electrostatic force of attraction between the negatively charged
electron and the positively charged nucleus.
Second electron affinity, ΔHEA2ϴ – the enthalpy change when one mole of
gaseous, singly charged anions each gain one electron, forming one mole of
gaseous, doubly charged anions.
 O (g) + e-  O2- (g), ΔHEA1ϴ = + 798 kJ mol-1
 This is an endothermic process, as energy is required to overcome the force
of repulsion between the electron and the negatively charged O - ion.


3. be able to construct Born-Haber cycles and carry out related
calculations

, A Born-Haber cycle is another application of Hess’ Law that can be used to
calculate lattice energy.
 They are often shown as an energy diagram, with exothermic ΔH values
pointing downwards and endothermic ΔH values pointing upwards.
 The steps involved, e.g. for CaCl 2:

Process Equation Enthalpy
change
Atomisation of ΔHatϴ[Ca Ca (s)  Ca (g) Endothermic
cation (s)]
Ionisation of ΔHIE1ϴ[Ca Ca (g)  Ca+ (g) + e- Endothermic
cation (g)]
Ionisation of ΔHIE1ϴ[Ca+ Ca+ (g)  Ca2+ (g) + e- Endothermic
cation (g)]
Atomisation of ΔHatϴ[Cl2 Cl2 (g)  2 Cl (g) Endothermic
anion (g)]
Ionisation of ΔHEA1ϴ[Cl Cl + e- (g)  Cl- (g) Exothermic
anion (g)]
Lattice energy of ΔHLEϴ[CaCl2 Ca2+ (g) + 2 Cl- (g)  Exothermic
CuCl2 (s)] CaCl2 (s)


Calculating enthalpy change values using a Born-Haber cycle:
 Hess’ law tells us that the enthalpy change accompanying a chemical change
is independent of the route by which the chemical change occurs.
o This enables us to calculate certain enthalpy changes using the other
data available.
 N.B. When the stoichiometry is not 1:1, you must multiply the enthalpy
changes accordingly.
o 2 Fe (g) + 1.5 O2 (g)  2 Fe (g) + 3 O (g)
o ΔHatϴ[O2 (g)] = +249.2 kJ mol-1
o You are forming 3 moles of oxygen atoms, so the enthalpy change for
this equation would be 3 x ΔHatϴ[O2 (g)] = 3 x +249.2 kJ mol-1 = +
747.6 kJ mol-1
 We can calculate the lattice energy of NaCl using the data given in the Born-
Haber cycle below:
o ΔHLEϴ[NaCl] = ΔHfϴ[NaCl] – ΔHatϴ[Na] – ΔHIE1ϴ[Na] – ΔHatϴ[Cl2] – ΔHEA1ϴ[Cl]
o ΔHLEϴ[NaCl (s)] = – 411 – 107 – 496 – 122 + 349 = – 787 kJ mol -1

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