thermodynamics

Question 1

define lattice enthalpy of dissociation/formation

Answer 1

lattice enthalpy of dissociation = when 1 mole of a solid ionic lattice is separated into its constituent gaseous ions

lattice enthalpy of dissociation = when 1 mole of a solid ionic lattice is formed from its gaseous ions

lattice enthalpy measures strength of ionic bonding

Question 2

factors affecting size of lattice enthalpy

Answer 2

• ion size
• ion charge
• charge density = ion charge/ion size
  
  • the smallest ions with the biggest size have the highest charge density
  • the higher the charge density, the stronger the electrostatic attractions between oppositely charged ions

Question 3

explain how ion size relates to charge in positive/negative ions

Answer 3

for negative ions:

• size increases as charge increases because more e- than protons =
  
  • weaker electrostatic attraction between outer e- and positive nucleus
  • also increased e- repulsion between inner e- and added e- = bigger ionic radius

for positive ions:

• size decreases as charge increases because less outer e- than protons
  
  • stronger electrostatic attraction between positive nucleus and outer e- = decreased ionic radius

Question 4

compare and explain lattice enthalpies of formation for MgO (-3512 kJmol^-1) and MgCl₂ (-2493 kJmol^-1)

Answer 4

O^2- ions smaller and has bigger charge than Cl^- ion

oxide ions have higher charge density than chloride ions

stronger electrostatic attraction between Mg²⁺ and O^2- tha between Mg²⁺ and Cl^-

more energy released when Mg²⁺ and O^2- form ionic lattice than when Mg²⁺ and Cl^- form ionic lattice

MgO therefore has more exothermic lattice enthalpy of formation

Question 5

enthalpy of atomisation

Answer 5

enthalpy change for the formation of 1 mole of gaseous atoms from its element in its standard state

e.g 1/2 Br₂ (l) → Br (g)

Question 6

bond dissociation enthalpy

Answer 6

enthalpy change when 1 mole of covalent bonds is broken under standard conditions in the gaseous state

e.g F₂(g) →2F (g)

bond dissociation enthalpy for diatomic molecules is twice the enthalpy of atomisation

Question 7

first electron affinity

Answer 7

enthalpy change when 1 mole of gaseous atoms form 1 mole of 1- ions

e.g. F(g) + 1e^- →F^- (g)

Question 8

why is the second electron affinity positive

Answer 8

it is endothermic due to the repulsion between 1- ion and the negative e- being added

Question 9

perfect ionic model

Answer 9

100% ionic bonding with 0% covalent character

both ions are perfect spheres with no distortion

Question 10

fajan’s rules

Answer 10

positve ions that are small and/or highly charged are good at polarising negative ions

negative ions that are large and/or highly charged are easier to be polarised

polarisation of negative ion results in ionic bonding having covalent character

the larger the % diff between theoretical lattice enthalpy (which assumes 100% ionic bonding) and experimental lattice, the the more covalent character the ionic solid has due to more distorted,polarised negatice ions.

Question 11

Born-Haber cycles

calculate the lattice enthalpy of formation of aluminium oxide given that:

Δ_fH^θ (Al₂O₃) = -1669 kJ.mol^-1

Δ_{<span>atm</span>}H^θ (Al) = 314 kJmol^-1

Δ_{<span>atm</span>}H^θ (O) = 248 kJmol^-1

first Δ_IEH^θ (Al) = 577 kJmol^-1

second Δ_IEH^θ (Al) = 1820 kJmol^-1

third Δ_IEH^θ (Al) = 2740 kJmol^-1

first Δ_EAH^θ (O) = -142 kJmol^-1

second Δ_EAH^θ (O) = 844 kJmol^-1

Answer 11

begin with elements in their standard states (these have energy zero)

↑ = positive enthalpy (Endothermic)

↓ = negative enthalpy (Exothermic)

remember all enthalpies must be multiplied by the number of moles in the orginal chemical equation

lattice enthalpy of formation of aluminium oxide given by the equation

2Al³⁺ (g) + 3O^2-(g) → Al₂O₃ (s)

Question 12

enthalpy of solution definition and equation

Answer 12

when 1 mole of a solid ionic lattice dissolves completely in large volume of water so that the dissolved aqueous ions are far enough apart to not interact

e.g NaCl (s) → Na⁺(aq) + Cl^-(aq)

Δ_solH^θ = Δ_LdH^θ +Σ Δ_hydH^θ​(ions) ​

Question 13

factors affecting hydration enthalpy

Answer 13

ion size and ion charge (charge density)

the bigger the charge and the smaller the size = the higher the charge density = ion attracts 𝛿+ H/ 𝛿- O of water molecule more strongly = more exothermic hydration enthalpy

Question 14

calculate enthalpy of solution for CaCl₂ given that:

lattice enthalpy of formation (CaCl₂) = -2237 kJ mol-1

enthalpy of hydration Ca²⁺ = -1650 kJmol-1

enthalpy of hydration Cl^- = -364 kJmol-1

use a born-haber cycle

use Hess’ Law

Answer 14

Question 15

first ionisation energy

Answer 15

enthalpy change when 1 mole of e- is removed from 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions

Question 16

the lattice enthalpy of dissociation of CaO₂ is 3513 kJmol^-1

CaO₂ is insoluble in water explain why

Answer 16

the enthalpy of dissociation is bery endothermic = strong electrostatic attraction between Ca²⁺ and O^2- ions = large amount of energy required to form Ca²⁺ (aq) and O^2-(aq)

Question 17

how does covalent character affect the lattice enthalpy and MP of ionic compounds?

Answer 17

covalent chracter reduces the MP due to weaker IMF

covalent bonds are stronger than ionic bonds so lattice enthalpy of formation will be more exothermic (more negative). lattice enthalpy of dissociation will be more endothermic (bigger)