Edexcel A Level Chemistry:复习笔记5.3.1 Lattice Energy

Lattice Energy

 

Lattice energy

  • As with bond enthalpy, lattice energy (ΔHlattꝋ) can be expressed as a formation or dissociation process
  • As a formation process, it is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions (under standard conditions)
  • The ΔHlattꝋ is therefore exothermic, as when ions are combined to form an ionic solid lattice there is an extremely large release of energy
    • Since this is an exothermic process, the enthalpy change will have a negative value
    • Because of the huge release in energy when the gaseous ions combine, the value will be a very large negative value

     

  • The large negative value of ΔHlattꝋ suggests that the ionic compound is much more stable than its gaseous ions
    • This is due to the strong electrostatic forces of attraction between the oppositely charged ions in the solid lattice
    • Since there are no electrostatic forces of attraction between the ions in the gas phase, the gaseous ions are less stable than the ions in the ionic lattice
    • The more exothermic the value is, the stronger the ionic bonds within the lattice are

     

  • The ΔHlattꝋ of an ionic compound cannot be determined directly by one single experiment
  • Multiple experimental values and an energy cycle are used to find the ΔHlattꝋ of ionic compounds
  • The lattice energy (ΔHlattꝋ) of an ionic compound can be written as an equation
    • For example, sodium chloride is an ionic compound formed from sodium (Na+) and chloride (Cl-) ions
    • Since the lattice energy is the enthalpy change when 1 mole of sodium chloride is formed from gaseous sodium and chloride ions, the equation for this process is:

    Na+(g) + Cl-(g) → NaCl (s)  ΔHlattꝋ = -776 kJ mol -1

Worked Example

Writing equations for lattice energy

Write down the equations which represent the lattice energy of:

(i) Magnesium oxide

(ii) Lithium chloride

Answer

Answer 1:

Mg2+ (g) + O2- (g) → MgO (s)

Answer 2:

Li+ (g) + Cl- (g) → LiCl (s)

Enthalpy change of atomisation

  • The standard enthalpy change of atomisation (ΔHatꝋ) is the enthalpy change when 1 mole of gaseous atoms is formed from its element under standard conditions
  • The ΔHatꝋ is always endothermic as energy is always required to break any bonds between the atoms in the element, to break the element into its gaseous atoms
    • Since this is always an endothermic process, the enthalpy change will always have a positive value

     

  • Equations can be written to show the standard enthalpy change of atomisation (ΔHatꝋ) for elements
  • For example, sodium in its elemental form is a solid
  • The standard enthalpy change of atomisation for sodium is the energy required to form 1 mole of gaseous sodium atoms:

Na (s) → Na (g)           ΔHatꝋ = +108 kJ mol -1

Worked Example

Writing equations for the standard enthalpy change of atomisation

Write down the equations for the standard enthalpy change of atomisation (ΔHatꝋ) for:

(i) Potassium

(ii) Mercury

Answer

Answer 1:

Potassium in its elemental form is a solid, therefore the standard enthalpy change of atomisation is the energy required to form 1 mole of K (g) from K (s)

K (s) → K (g)

Answer 2:

Mercury in its elemental form is a liquid, so the standard enthalpy change of atomisation of mercury is the energy required to form 1 mole of Hg (g) from Hg (l)

Hg (l) → Hg (g)

Electron Affinity

  • The electron affinity (ΔHea) of an element is the energy change when one mole of electrons is gained by one mole of gaseous atoms of an element to form one mole of gaseous ions under standard conditions
  • For example, the first electron affinity of chlorine is:

Cl (g)+ e– → Cl- (g)          ΔHeaꝋ = -364 kJ mol-1

  • The first electron affinity is always exothermic as energy is released when electrons are attracted to the atoms
  • However, the second electron affinity of an element can be endothermic as illustrated by oxygen:

O– (g) + e– → O2- (g)          ΔHeaꝋ = +844 kJ mol-1

  • This is because a large force of repulsion must be overcome between the negatively charged ion and the second electron requiring a large input of energy

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