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Chemistry > Topic 13 Energetics II > Flashcards

Topic 13 Energetics II Flashcards

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1
Q

Standard enthalpy change of atomisation of an element

A

The enthalpy change measured at a stated temperature and pressure, usually 298K & 100kPa, when one mole of gaseous atoms is formed from an element in its standard state.

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2
Q

Lattice dissociation enthalpy

A

The heat energy required to break up 1 mole of an ionic lattice.

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3
Q

Lattice energy of dissociation

A

Always positive.

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4
Q

Lattice energy of formation

A

Always negative.

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5
Q

Inter-ionic distance

A

Distance between the centres of the two ions.

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6
Q

What factors determine lattice energy?

A
  • The magnitude of the charges on the ions.
  • The sum of the ionic radii.
  • The extent of covalent interactions between the ions.
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7
Q

Equation for the standard lattice energy of NaCl

A

Na+ (g) + Cl- (g) -> NaCl (s) Delta lattice H = -780 kJ mol^-1

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8
Q

Equation for the atomisation of chlorine

A

1/2Cl2 (g) –> Cl (g) Delta at H= +122 kJ mol^-1

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9
Q

Equation for the atomisation of sodium

A

Na(s) –> Na (g) Delta at H= +107 kJ mol^-1

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10
Q

Atomisation energy: +ve or -ve?

A

Always positive!

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11
Q

First electron affinity Eea(1)

A

The energy change when each atom in one mole of atoms in the gaseous state gains an electron to form a -1.

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12
Q

Equation: first electron affinity of oxygen

A

O(g) + e- –> O- (g)

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13
Q

Electron affinity tends to be negative. Except…

A

Nobles gases. Electrons already present in the valence shell cause repulsion. The electron being added would have to occupy a new valence shell.

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14
Q

Second electron affinities tend to be…

A

Positive!

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15
Q

If the sum of the electron affinities and ionisation energies for Ca+O- is more exothermic than that of Ca2+O2-, why is Ca2+O2- formed?

A

The overall lattice energy is much more exothermic for Ca2+O2-.

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16
Q

What assumptions does the theoretical lattice energy calculation (using the principles of electrostatics) make?

A
  • The ions are in contact with one another.
  • The ions are perfectly spherical.
  • The charge on each ion is evenly distributed around the centre, so that each ion can be considered as point charges.
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17
Q

What causes the experimental value to be more negative than the theoretical lattice energy value?

A

Covalent character.

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18
Q

Agreement between the experimental and theoretical lattice energy values…

A

Indicates the ionic model is suitable.

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19
Q

What causes covalency in bonding?

A

Polarisation of the anion by the cation, resulting in distortion of the electron density within the anion, so there is higher electron density near the anion. The cation polarises the anion.

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20
Q

Fajan’s rules: what increases polarisation?

A

A high charge & small size of the cation– high charge density of the cation.
- A high charge and large size of the anion.

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21
Q

Polarising power

A

The ability of the cation to attract electrons from the anion towards itself.

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22
Q

The higher the charge density…

A

…the greater the polarising power.

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23
Q

Despite the lower polarising power of Ag+, there is more covalent character in AgCl than NaCl. Why?

A

Na+ : 1s2 2s2 2p6
Ag+ : [Kr] 3d10
The d10 configuration provides less shielding than a p6 electron configuration. Same applies to Zn.

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24
Q

Enthalpy change of solution

A

The enthalpy change when one mole of an ionic solid dissolves in water to form an infinitely dilute solution.

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25
Q

Enthalpy change of hydration

A

The enthalpy change when one mole of an ion in its gaseous state is completely hydrated by water to form an infinitely dilute solution.

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26
Q

Point of infinite solution

A

When further dilution has no measurable effect on the enthalpy of solution.

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27
Q

Enthalpy change of solution for sodium chloride: equation

A

NaCl (s) –aq–> Na+ (aq) + Cl- (aq)

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28
Q

When is complete hydration said to have occurred?

A

When the solution formed is at infinite dilution.

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29
Q

Enthalpy of hydration for sodium ions: equation

A

Na+ (g) –aq–> Na+ (aq)

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30
Q

Enthalpy of hydration for chloride ions: equation

A

Cl-(g) –aq–> Cl- (aq)

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31
Q

Ion-dipole interaction

A

Attraction between the partial negative of the oxygen atom of the water molecule and the cation.

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32
Q

When transition metals are hydrated…

A

A dative covalent bond is formed between the water molecule and the cation using one of the lone pairs of electrons on the oxygen atom.

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33
Q

When the Cl- ion is hydrated…

A

Ion-dipole interactions form, but some hydrogen bonds also form between the lone pair on the Cl- and the partial positive on the oxygen of H2O.

34
Q

What happens to hydration enthalpy down the group?

A

It gets less negative.

35
Q

Why do hydration enthalpies get less negative down the group?

A

Increasing ionic radius. As ions get larger, the electrostatic force of attraction between them and the water molecules decreases, hence less energy is released upon hydration.

36
Q

Why is hydration enthalpy more negative for +2 than +1 ions?

A

The electrostatic force of attraction between a doubly-charged ion and its water molecules will be greater than with a singly-charged ion and H2O molecules.

37
Q

How does charge density affect hydration enthalpy?

A

The greater the charge density of a cation, the more negative the value of enthalpy of hydration.

38
Q

Enthalpy of solution =

A

Sum of hydration enthalpies - lattice enthalpy.

39
Q

Dimerisation of NO2 in the gas phase

A

2NO2(g) <–> N2O4 (g)

40
Q

At RTP: NH3 (g) + HCl(g) –>

A

NH4Cl (s). A white solid.

41
Q

Spontaneous

A

Process that takes place without continuous intervention.

42
Q

Entropy

A

A property of matter that is associated with the degree of disorder of the particles.

43
Q

2nd Law of Thermodynamics

A

In a spontaneous process, total entropy/entropy of the universe increases.

44
Q

The greater the degree of disorder, the greater the entropy.

A

Gas (highest entropy)> liquid> solid.

45
Q

What is the difference between system entropy & surroundings entropy?

A

The system is the species participating in the reaction. the surroundings is everything else.

46
Q

ΔS total=

A

ΔS system + ΔS surroundings

47
Q

For a reaction to be spontaneous…

A

ΔS total must be +ve.

48
Q

ΔS system =

A

ΣΔ S (products ) - ΣΔ S (reactants)

49
Q

Units of entropy

A

J K^-1 mol^-1

50
Q

Units of enthalpy

A

kJ mol^-1

51
Q

ΔS surroundings

A

-ΔH/T T is in Kelvin.

52
Q

What happens to entropy of the surroundings for an exothermic reaction?

A

It increases as where ΔH is -ve, ΔS surroundings will always be positive.

53
Q

What happens to the entropy of the surroundings for an endothermic reaction?

A

It decreases as when ΔH is +ve, ΔS surroundings will always be -ve.

54
Q

Under what 3 conditions will ΔS total be +ve?

A
  • Both ΔS surroundings & ΔS system are +ve.
  • ΔS surroundings is positive & ΔS system is negative, but the magnitude of ΔS surroundings > the magnitude of ΔS system.
  • ΔS surroundings is -ve and ΔS system is +ve, but the magnitude of ΔS surroundings < the magnitude of ΔS system.
55
Q

Mg(s) + 1/2O2(g) –> MgO (s)
How is this thermodynamically feasible, considering ΔS system is negative as a solid & a gas are reacting to form a solid?

A

The reaction is highly exothermic, so ΔS surroundings is +ve, and the magnitude of ΔS surroundings is greater than of ΔS system, so ΔS total is positive.

56
Q

Why does increasing the number of moles from reactants to products make ΔS system positive?

A

This increases the number of particles which increases the number of ways the particles can be arranged, which increases the entropy of the system.

57
Q

2 processes involved in dissolving an ionic solid

A
  • The lattice structure is broken down.
  • The ions become hydrated.
58
Q

The breaking down of the lattice structure

A

Endothermic. Results in an increased number of particles present, so increases the entropy of the salt.

59
Q

Hydration of the ions

A

Exothermic. Results in water molecules becoming more ordered as they arrange themselves around the ions, so decreases the entropy of the water.

60
Q

When is the ordering of water molecules (thereby reducing the entropy of the water) particularly significant?

A

When dissolving anhydrous solids in water.

61
Q

On what 3 factors does the solubility of the solid (i.e., ΔS total) depend?

A
  • ΔS system.
  • Enthalpy change of solution.
  • The temperature in Kelvin of the water.
62
Q

Why is it assumed that ΔS system does not change with temperature?

A

The entropies of both reactants & products change by similar amounts.

63
Q

ΔG =

A

ΔH - TΔS system

64
Q

What does it mean if ΔG = 0?

A

The reaction is in equilibrium.

65
Q

What does it mean when ΔG is -ve?

A

The reaction is feasible.

66
Q

When is ΔG negative?

A

ΔH > 0 and ΔS system > 0
ΔH < 0, ΔS system < 0 but the magnitude of ΔH > the magnitude of TΔS.
ΔH > 0, ΔS system > 0 but the magnitude of ΔH < the magnitude of TΔS.

67
Q

How can we determine the minimum temperature at which a reaction becomes feasible?

A

ΔG= 0, so ΔH= TΔS system, so T= ΔH/ΔS system

68
Q

How is the equilibrium constant related to Gibbs?

A

ΔG = -RTlnK

K= e^(-ΔG/RT)

69
Q

The thermodynamic equilibrium constant K

A

Has no units.

70
Q

If ΔG is -ve, the exponent is +ve.

A

The equilibrium constant will be greater than 1. The products are favoured.

71
Q

If ΔG is +ve, the exponent is -ve.

A

The equilibrium constant will be less than 1. The reactants are favoured.

72
Q

Demonstrate that the way the equilibrium constant varies with temperature depends on the sign of ΔH.

A

lnK = -ΔH/RT + ΔS system/R

73
Q

ΔG is -ve, but why might the reaction not occur?

A

The reactants are kinetically stable. The activation energy is very high.

74
Q

ΔG is +ve, but why might the reaction still occur?

A

ΔG may become negative under non-standard conditions.

75
Q

Why is HF a weak acid?

A

ΔG is +ve (and it has strong hydrogen bonding). The value of K is small.

76
Q

ΔG = -ve what does this mean for solubility?

A

The salt is soluble.

77
Q

Factors affecting the size of hydration enthalpy:

A

Hydration enthalpy is a measure of energy released when attractions are set up between ions & water molecules. For cations, this could be ion-dipole attractions between the cation & partially negative O of water, or dative covalent bonds. For anions, ion-dipole attractions form but with the partially positive H of the water. Attractions are stronger for smaller, more highly-charged ions.

78
Q

Why are large alkanes soluble in hexane but not water?

A

They cannot hydrogen bond to water.
The London forces in both hexane and the alkane are similar in strength, so the resultant forces in the mixture are similar to those in each liquid.

79
Q

Why are ionic compounds soluble in water but not in hexane?

A

The ions are hydrated in water when dissolved. The enthalpy change of hydration is greater than the energy needed to break apart the lattice.
Insoluble in hexane as any London forces that form between hexane and the ionic compound would be smaller in magnitude than the forces between the ions.

80
Q

Edexcel says lattice energy…

A

…is lattice formation energy.

Chemistry (23 decks)

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