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A LEVEL - Chemistry (Both) > Topic 13 - Energetics II > Flashcards

Topic 13 - Energetics II Flashcards

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1
Q

In energy taken in or given out when gaseous ions combine to make a solid lattice?

A

Given out

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2
Q

What is the symbol for standard lattice energy?

A

Δ(LE)H°

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3
Q

Define standard lattice energy.

A

The energy change when 1 mole of an ionic solid is formed from its gaseous ions under standard conditions.

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4
Q

What are standard conditions?

A

298K and 100kPa

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5
Q

What is standard lattice enthalpy a measure of?

A

Ionic bond strength

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6
Q

Are lattice enthalpies positive of negative?

A

Negative (in formation of lattice)

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7
Q

Which lattice enthalpy has a greater magnitude, NaCl or MgO?

A

MgO is more negative.

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8
Q

What two factors affect lattice energy?

A
  • Ionic charge

* Ionic size

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9
Q

Why do ionic charge and size affect lattice energy?

A

They affect the strength of the electrostatic attraction, so the bonds formed are stronger, so the lattice energy is higher.

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10
Q

What is the name for the diagram used to determine lattice energies?

A

Born-Haber cycles

See diagram pg 142 of revision guide

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11
Q

On what principle do Born-Haber cycles work?

A

Hess’s Law

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12
Q

State Hess’s law.

A

The total enthalpy change of a reaction is always the same, no matter which route is taken.

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13
Q

What is the enthalpy is atomisation?

A

The enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.

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14
Q

What is the first electron affinity?

A

The enthalpy change when 1 mole of electrons are added to 1 mole of neutral gaseous atoms to form 1 mole of gaseous 1- ions.

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15
Q

Describe a Born-Haber cycle for the formation of NaCl(s).

A

• Start with Na(s) + 1/2Cl₂(g)
PATH 1:
• Arrow down - Enthalpy of formation of NaCl(s)
PATH 2:
• Arrow up - Atomisation enthalpy of chlorine
• Arrow up - Atomisation enthalpy of sodium
• Arrow up - First ionisation energy of sodium
• Arrow down - First electron affinity of chlorine
• Arrow down - Lattice energy of sodium chloride

(See diagram pg 142 of revision guide)

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16
Q

Remember to practise drawing out the Born-Haber cycle for the formation of NaCl(s).

A

See diagram pg 142 of revision guide.

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17
Q

Describe a Born-Haber cycle for the formation of MgCl₂(s).

A

• Start with Mg(s) + Cl₂(g)
PATH 1:
• Arrow down - Enthalpy of formation of MgCl₂(s)
PATH 2:
• Arrow up - 2 x Atomisation enthalpy of chlorine
• Arrow up - Atomisation enthalpy of magnesium
• Arrow up - First ionisation energy of magnesium
• Arrow up - Second ionisation energy of magnesium
• Arrow down - 2 x First electron affinity of chlorine
• Arrow down - Lattice energy of magnesium chloride

(See diagram pg 143 of revision guide)

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18
Q

Remember to practise drawing out the Born-Haber cycle for the formation of MgCl₂(s).

A

See diagram pg 143 of revision guide.

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19
Q

Remember to practise writing out the meaning of each enthalpy type.

A

Do it. Chapters 8 and 13.

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20
Q

What are the two ways to work out a lattice energy?

A
  • Experimental way

* Theoretical way

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21
Q

How can the lattice energies be worked out the experimental way?

A

Using experimental enthalpy values in a Born-Haber cycle.

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22
Q

How can the lattice energies be worked out the theoretical way?

A
  • Assume the ions are spherical and how evenly distributed charge -> i.e. Assume a perfectly ionic model
  • Then you work out how strongly the ions are attracted to each other based on charge and distance between them
  • This gives you a value for the energy change when the ions form the lattice

(NOTE: For the exam, you don’t need to do the second bit)

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23
Q

What is more exothermic, experimental or theoretical lattice energies?

A

Experimental (i.e. the slightly covalent ones)

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24
Q

How can you work out how covalent an ionic lattice is?

A
  • Compare the experimental and theoretical lattice energy
  • If the two are similar -> Highly ionic
  • If the experimental is much more negative -> Slightly covalent
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25
Q

Are theoretical or experimental ionic bonds stronger?

A

Experimental, because they have some covalent character.

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26
Q

Which are more ionic, group 1 or group 2 halides? Why?

A

• Group 1
BECAUSE:
• Group 1 cations have a small charge so they have little polarising power -> So the charge is distributed evenly around the ions
• Group 2 cations have a larger charge, so they have more polarising power and pull electrons from the anion towards themselves -> So the anion is distorted

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27
Q

What is a good representation of the polarising power of a cation?

A

Charge density

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28
Q

What is the equation for charge density?

A

Charge density = Charge / Volume

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29
Q

What sort of cations are most polarising and why?

A
  • Small with a high charge

* Because the positive charge is concentrated in the ion and it can pull electrons towards itself

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30
Q

What sort of anions are most easily polarised and why?

A
  • Large with a high charge

* Because the electrons are further away from the nucleus and there is more repulsion between the electrons

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31
Q

What happens when an ionic bond is polarised enough?

A

A partially covalent bond forms.

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32
Q

What is electronegativity?

A

The ability of an atom to attract the bonding electrons in a covalent bond.

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33
Q

What scale is used to measure electronegativity?

A

Pauling

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34
Q

In terms of electronegativity, what determines the polarity of a bond?

A

The greater the difference in electronegativity, the more polar the bond.

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35
Q

What is the boundary for polar bonds?

A

If the difference in electronegativity is more than 0.4

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36
Q

Predict whether a C-Cl bond will be polar, given that the Pauling electronegativity values of carbon and chlorine are C=2.5 and Cl=3.0.

A
  • 3.0 - 2.5 = 0.5

* 0.5 > 0.4 so the bond will be polar.

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37
Q

What two things happen when a solid ionic lattice dissolves in water? What is the energy change of each?

A

1) Bonds between the ions break -> Endothermic

2) Bonds between the ions and water are made -> Exothermic

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38
Q

In dissolving, when bonds between ions break, what is this essentially the equivalent of?

A

It is the opposite of lattice enthalpy.

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39
Q

In dissolving, what is the name for the broken up ions forming bonds with the water?

A

Enthalpy change of hydration

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40
Q

Remember to practise drawing out the diagram for how dissolving works.

A

See diagram pg 146 of revision guide

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41
Q

What is the symbol for the enthalpy of hydration?

A

Δ(hyd)H

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42
Q

What is the symbol for the enthalpy of solution?

A

Δ(sol)H

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43
Q

Define the enthalpy of hydration.

A

The enthalpy change when 1 mole of gaseous ions dissolves in water.

44
Q

Define the enthalpy of solution.

A

The enthalpy change when 1 mole of solute dissolves in water.

45
Q

Are enthalpies of hydration positive or negative?

A

Negative

46
Q

Give the enthalpy of solution in terms of other enthalpies.

A

Enthalpy of solution = Enthalpies of hydration - Lattice enthalpy

47
Q

When will a substance dissolve and when will it not?

A

It will only dissolve if the energy released by making bonds is roughly the same, or greater, than the energy taken in to break them.

48
Q

What does an exothermic enthalpy of solution suggest?

A

The substance is probably soluble.

49
Q

Describe how you can draw a Hess cycle for dissolving a salt, in order to calculate to enthalpy of solution.

A

1) Put the ionic lattice in the top left. Draw a horizontal arrow to the dissolved ions in the top right. This is the direct route (enthalpy of solution).
2) Draw the gaseous ions below this. Draw an arrow from the gaseous ions to the ionic lattice (lattice energy).
3) Draw an arrow from the gaseous ions to the dissolved ions (enthalpy of hydration of both).

Use Hess’s law to calculate the enthalpy change of solution.

50
Q

Remember to practise drawing out the Hess cycle for dissolving NaCl.

A

Pg 146 of revision guide

51
Q

Remember to practise drawing out the energy level diagram for the dissolving of AgCl.

A

Pg 146 of revision guide

52
Q

How do energy level diagrams differ to energy cycles (Hess cycles)?

A

They are the same, except the substances are arranged vertically in the diagram according to their energies.

53
Q

What two factors affect enthalpy of hydration?

A
  • Ionic charge

* Ionic radius

54
Q

How does ionic charge affect enthalpy of hydration? Why?

A
  • The greater the ionic charge, the more exothermic the enthalpy of hydration
  • The is because the electrostatic attraction between the ion and water is stronger, so more energy is released in making bonds
55
Q

How does ionic radius affect enthalpy of hydration? Why?

A
  • The smaller the ionic radius, the more exothermic the enthalpy of hydration
  • The is because the charge density is higher, so the electrostatic attraction between the ion and water is stronger, so more energy is released in making bonds
56
Q

Compare the enthalpy of hydration of Mg²⁺ and Na⁺.

A

Mg²⁺ is more exothermic

57
Q

What is entropy?

A

A measure of disorder of a system.

58
Q

A high entropy means…

A

High disorder

59
Q

What 3 factors affect entropy?

A
  • Physical state
  • Dissolving
  • Number of particles
60
Q

How does physical state affect entropy?

A
  • Gases have the highest entropy because the particles can move about most freely and randomly
  • Solids have the lowest entropy because their particles are fixed in given positions and can only vibrate about those
61
Q

Explain how entropy changes in this reaction:

2Mg(s) + O₂(g) -> 2MgO(s)

A

It is lowered because a gas is in the reactants, but not in the products.

62
Q

Explain how entropy changes in this reaction:

2CH₃COOH(aq) + (NH₄)₂CO₃(s) -> 2CH₃COONH₄(aq) + H₂O(l) + CO₂(g)

A

It increases because there are no gases in the reactants, but there is a gas in the products.

63
Q

How does dissolving affect entropy?

A

Dissolving a solid increases its entropy because the dissolved particles can move around more freely as they’re no longer held in place.

64
Q

Explain how entropy changes in this reaction:

NH₄NO₃(s) -> NH₄⁺(aq) + NO₃⁻(aq)

A

It increases because the particles are no longer held in place but can move around freely in solution.

65
Q

How does number of particles affect entropy?

A

The more particles you have, the higher the entropy, because there are more ways they and their energies can be arranged.

66
Q

Explain how entropy changes in this reaction:

N₂O₄(g) -> 2NO₂(g)

A

It increases because the number of moles increases, so the number of ways the particles and their energies can be arranged is increased.

67
Q

How does entropy affect stability?

A

A high entropy means high stability.

68
Q

Do particles move to increase or decrease entropy?

A

Increase

69
Q

Why are some reactions feasible even when the enthalpy change in endothermic?

A

The entropy change is positive, so the particles react in order to become more energetically stable.

70
Q

The reaction of sodium hydrogencarbonate with hydrochloride acid is endothermic, but it is feasible. Explain why.

NaHCO₃(s) + H⁺(aq) -> Na⁺(aq) + CO₂(g) + H₂O(l)

A

The entropy increases because:
• A liquid and gas are produced
• More moles are produced
This means that the increase in entropy overcomes the change in enthalpy, making the reaction feasible.

71
Q

What is the symbol for the entropy change of a system?

A

ΔS(system)

72
Q

What is the equation for the entropy change of a system?

A

ΔS(system) = S(products) - S(reactants)

73
Q

What are the units of entropy?

A

J/K/mol

74
Q

Calculate the entropy change for the reaction of ammonia and hydrogen chloride under standard conditions.

NH₃(g) + HCl(g) -> NH₄Cl(s)

S°[NH₃] = 192.3 J/K/mol
S°[HCl] = 186.8 J/K/mol
S°[NH₄Cl] = 94.60 J/K/mol
A

Find the entropy of the products:
• S°(products) = 94.60 J/K/mol
Find the entropy of the reactants:
• S°(reactants) = 192.3 + 186.8 = 379.1 J/K/mol
Finally find the entropy change for the system:
• ΔS(system) = 94.60 - 378.1 = -284.5 J/K/mol

75
Q

Does a positive entropy change guarantee a feasible reaction?

A

It means it is likely to be feasible.

76
Q

Does a negative entropy change guarantee a reaction can’t happen?

A

No - enthalpy, temperature and kinetics also play a part in whether or not the reaction occurs.

77
Q

What is the symbol for the entropy change of the surroundings?

A

ΔS(surroundings)

78
Q

What is the equation for the entropy change of the surroundings?

A

ΔS(surroundings) = -ΔH / T

Where:
• ΔS(surroundings) = Entropy change of surroundings (J/K/mol)
• ΔH = Enthalpy change (J/mol)
• T = Temperature (K)

79
Q

Why does the entropy of the surroundings change in a reaction?

A

Energy is transferred to or from the system.

80
Q

What is the symbol for the total entropy change?

A

ΔS(total)

81
Q

What is the equation for the total entropy change?

A

ΔS(total) = ΔS(system) + ΔS(surroundings)

82
Q

What are the three types of entropy change and what is each?

A
  • ΔS(system) -> Entropy change of the reactants going to the products
  • ΔS(surroundings) -> Entropy change of the surroundings of the system
  • ΔS(total) -> The sum of the system and surrounding entropies
83
Q

When doing calculations of ΔS(surroundings), what is it important to remember?

A

When using ΔH, remember to convert from kJ/mol to J/mol.

84
Q

Calculate the total entropy change for the reaction of ammonia and hydrogen chloride under standard conditions.

NH₃(g) + HCl(g) -> NH₄Cl(s)

ΔH = -315 kJ/mol
S°[NH₃] = 192.3 J/K/mol
S°[HCl] = 186.8 J/K/mol
S°[NH₄Cl] = 94.60 J/K/mol
A

Entropy change of system:
• ΔS(system) = S(products) - S(reactants) = 94.60 - (192.3 + 186.8) = -284.5 J/K/mol
Entropy change of surroundings:
• ΔS(surroundings) = -ΔH / T = -(-315 x 10³) / 298 = +1057 J/K/mol
Total entropy change:
• ΔS(total) = ΔS(system) + ΔS(surroundings) = -284.5 + 1057 = +772.5 J/K/mol

85
Q

Calculate the total entropy change ammonium nitrate crystals being dissolved in water under standard conditions.

NH₄NO₃(s) -> NH₄⁺(aq) + NO₃⁻

ΔH = +25.70 kJ/mol
S°[NH₄NO₃] = 151.1 J/K/mol
S°[NH₄⁺] = 113.4 J/K/mol
S°[NO₃⁻] = 146.4 J/K/mol
A

Entropy change of system:
• ΔS(system) = S(products) - S(reactants) = (146.4 + 113.4) - 151.1 = +108.7 J/K/mol
Entropy change of surroundings:
• ΔS(surroundings) = -ΔH / T = -(25.70 x 10³) / 298 = -86.24 J/K/mol
Total entropy change:
• ΔS(total) = ΔS(system) + ΔS(surroundings) = 108.7 - 86.24 = +22.46 J/K/mol

86
Q

Remember to look at the entropy and enthalpy example on pg 151 of revision guide.

A

Do it.

87
Q

What 3 things determine the feasibility of a reaction?

A
  • Entropy
  • Enthalpy
  • Temperature
88
Q

What does feasible mean?

A

Thermodynamically favourable

NOTE: Check this!

89
Q

Wha is the symbol for Gibbs free energy?

A

ΔG

90
Q

What are the units for Gibbs free energy?

A

J/mol

91
Q

What is the equation for Gibbs free energy, relative to enthalpy?

A

ΔG = ΔH - TΔS(system)

Where:
• ΔG = Gibbs free energy (J/mol)
• ΔH = Enthalpy change (J/mol)
• T = Temperature (K)
• ΔS(system) = Entropy change of system (J/K/mol)
92
Q

Describe how the value of the Gibbs free energy determines whether a reaction is feasible.

A
  • When ΔG is negative -> Feasible
  • When ΔG = 0 -> Just feasible
  • When ΔG is positive -> Not feasible
93
Q

Calculate the free energy change for the following reaction at 298K. Determine whether it is feasible.

MgCO₃ -> MgO + CO₂

ΔH° = +117,000 J/mol
ΔS(system) = +175 J/K/mol
A

ΔG = ΔH - TΔS(system) = 117,000 - (298 x 175) = +64,900 J/mol (to 3 s.f.)

Therefore, the reaction is not feasible at this temperature.

94
Q

Derive the equation for when a reaction becomes feasible.

A
  • When a reaction becomes just feasible, ΔG=0.
  • ΔH - TΔS = 0
  • Therefore, T = ΔH / ΔS(system)
95
Q

At what temperature does this reaction become feasible?

MgCO₃ -> MgO + CO₂

ΔH° = +117,000 J/mol
ΔS(system) = +175 J/K/mol
A

T = ΔH / ΔS(system) = 117,000 / 175 = 669K

96
Q

Describe the values of ΔH and ΔS when the reaction will DEFINITELY be feasible.

A
  • ΔH -> Negative
  • ΔS -> Positive

This would mean that ΔG is definitely negative, so the reaction is feasible.

97
Q

Describe the values of ΔH and ΔS when the reaction will DEFINITELY be not feasible.

A
  • ΔH -> Positive
  • ΔS -> Negative

This would mean that ΔG is definitely positive, so the reaction is not feasible.

98
Q

Describe the value of the equilibrium constant for a theoretically feasible reaction (where ΔG is negative).

A

Greater than 1

99
Q

Describe the value of the equilibrium constant for a theoretically not feasible reaction (where ΔG is positive).

A

Smaller than 1

100
Q

What is the equation for Gibbs free energy, relative to equilibrium constants?

A

ΔG = -RTlnK

Where:
• ΔG = Gibbs free energy (J/mol)
• R = Gas constant = 8.31 J/K/mol
• T = Temperature (K)
• K = Equilibrium constant
101
Q

What are the two equations for Gibbs free energy?

A
  • ΔG = ΔH - TΔS(system)

* ΔG = -RTlnK

102
Q

Ethanoic acid and ethanol were reacted together at 298K and allowed to reach equilibrium. The equilibrium constant was calculated to be 4 at 298K. Calculate the free energy change for the reaction.

CH₃COOH + C₂H₅OH -> CH₃COOC₂H₅ + H₂O

A

ΔG = -RTlnK = -(8.31 x 298) x ln(4) = -3430 J/mol (to 3 s.f.)

103
Q

Hydrogen gas and iodine are mixed together in a sealed flask forming hydrogen iodide. Calculate the equilibrium constant at 763K.

H₂ + I₂ -> 2HI

ΔG = -24287 J/mol

A
  • lnK = ΔG / -RT = -24287 / -(8.31 x 763) = 3.8304
  • K = e^(3.8304) = 46
  • Units for K = No units (this depends on the equation)
104
Q

Does a negative ΔG guarantee a reaction?

A

No, because the reaction could:
• Have a very high activation energy
• Have a very very slow reaction rate so that it is not noticeable

105
Q

Give two reasons why no reaction may be seen with a given reaction with a negative ΔG value.

A

1) High activation energy

2) Slow reaction rate

A LEVEL - Chemistry (Both) (21 decks)

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