15 HL Energetics / Thermochemistry Flashcards

1
Q

What is ionisation energy?

A

The ionisation energy (ΔHIEꝋ) is the standard enthalpy change that occurs on the removal of 1 mole of electrons from 1 mole of gaseous atoms or positively charged ions

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2
Q

Are ionisation energies exo/endothermic?

A

Ionisation energy is always endothermic as energy is need to overcome the attraction between an electron and the nucleus

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3
Q

What is the first ionisation energy?

A

The first ionisation energy (ΔHIE1ꝋ) is the energy required to remove one mole of electrons from 1 mole of gaseous atoms of an element to form 1 mole of 1+ ions in the gaseous phase

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4
Q

Give an equation for the first ionisation energy for aluminium

A

ΔHIE1ꝋ. Al (g) → Al+ (g) + e– ΔHIE1ꝋ = +577 kJ mol-1

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5
Q

What is the second ionisation energy?

A

The second ionisation energy (ΔHIE2ꝋ) is the energy required to remove 1 mole of electrons from 1 mole of gaseous 1+ ions to form 1 mole of 2+ ions in the gaseous phase

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6
Q

Give the equation for the second ionisation of aluminium

A

ΔHIE2ꝋ Al+ (g) → Al2+ (g) + e– ΔHIE2ꝋ = +1820 kJ mol-1

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7
Q

What is the enthalpy of atomisation?

A

The enthalpy of atomisation (ΔHatꝋ) is the standard enthalpy change that occurs on the formation of 1 mole of separate gaseous atoms an element in its standard state

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8
Q

Is the enthalpy of atomisation always endo/exothermic?

A

The ΔHatꝋ is always endothermic as energy is always required to break any bonds between the atoms in the element or to break the element into its gaseous atoms

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9
Q

What is the sign for the enthalpy of atomisation?

A

Since this is always an endothermic process, the enthalpy change will always have a positive value

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10
Q

Give the enthalpy of atomisation for chlorine.

A

½Cl (g) → Cl (g) ΔHatꝋ = +122 kJ mol -1

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11
Q

Give the enthalpy of atomisation for sodium

A

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

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12
Q

What is electron affinity?

A

The electron affinity (ΔHEAꝋ) of an element is the energy change when 1 mole of electrons is gained by 1 mole of gaseous atoms of an element to form 1 mole of gaseous ions under standard conditions

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13
Q

Give the equation for the first enthalpy of atomisation for chlorine?

A

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

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14
Q

Is the first electron affinity always endo/exothermic?

A

The first electron affinity is always exothermic as energy is released when electrons are attracted to the atoms

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15
Q

Is the SECOND electron affinity always endo/exothermic?

A

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

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16
Q

WHy is the second electron affinity of oxygen endothermic?

A

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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17
Q

What is the lattice enthalpy?

A

The lattice enthalpy (ΔHlatꝋ) is defined as the standard enthalpy change that occurs on the formation of 1 mole of gaseous ions from the solid lattice

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18
Q

Is the lattice enthalpy always endo/exothermic?

A

The ΔHlatꝋ is always endothermic as energy is always required to break any bonds between the ions in the lattice

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19
Q

Therefore what sign does the lattice enthalpy have?

A

Since this is always an endothermic process, the enthalpy change will always have a positive value

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20
Q

Give the equation for the lattice enthalpy of NaCl?

A

NaCl (s) → Na+ (g) + Cl- (g) ΔHlatꝋ = +790 kJ mol -1

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21
Q

What is the standard enthalpy change of solution?

A

The standard enthalpy change of solution (ΔHsolꝋ) is the enthalpy change when 1 mole of an ionic substance dissolves in sufficient water to form an infinitely dilute solution

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22
Q

What is the symbol (aq) used to describe?

A

The symbol (aq) is used to show that the solid is dissolved in sufficient water

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23
Q

Is enthalpy of solution exo/endothermic?

A

BOTH
ΔHsolꝋ can be exothermic (negative) or endothermic (positive)

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24
Q

Give the equation for the enthalpy of solution of LiBr

A

LiBr (s) → LiBr (aq) ΔHsolꝋ = -48.8 kJ mol -1

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25
Q

Give the equation for the enthalpy of solution of KCl

A

KCl (s) → KCI (aq) ΔHsolꝋ = +17.2 kJ mol -1

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26
Q

What is the standard enthalpy change of hydration?

A

The standard enthalpy change of hydration (ΔHhydꝋ) is the enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form an infinitely dilute solution

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27
Q

Give the equation for the enthalpy of hydration of magnesium

A

Mg2+ (g) → Mg2+ (aq) ΔHhydꝋ = -1963 kJ mol -1

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28
Q

Give the equation for the enthalpy of hydration of bromine

A

Br- (g) → Br- (aq) ΔHhydꝋ = -328 kJ mol -1

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29
Q

What do hydration enthalpies measure?

A

Hydration enthalpies are the measure of the energy that is released when there is an attraction formed between the ions and water molecules

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30
Q

Are hydration enthalpies endo/exothermic?

A

Hydration enthalpies are exothermic

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31
Q

When is the term solvation used for enthalpy of hydration?

A

The term solvation is used in place of hydration if water has been replaced by another solvent

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32
Q

What is formed when an ionic solid dissolves in water?

A

When an ionic solid dissolves in water, positive and negative ions are formed

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33
Q

Why do ionic solids dissolve in water?

A

Water is a polar molecule with a δ- oxygen (O) atom and δ+ hydrogen (H) atoms which will form ion-dipole attractions with the ions present in the solution

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34
Q

What parts fo water will be attracted to what parts of the ionic solid?

A

The oxygen atom in water will be attracted to the positive ions and the hydrogen atoms will be attracted to the negative ions

35
Q

What is a born-haber cycle?

A

A Born-Haber cycle is a specific application of Hess’s Law for ionic compounds and enables us to calculate lattice enthalpy, which cannot be found by experiment

36
Q

What is the basic principle used to construct a born-haber cycle?

A

The basic principle of drawing the cycle is to construct a diagram in which energy increases going up the diagram

37
Q

What steps does a born-haber cycle show?

A

The cycle shows all the steps needed to turn atoms into gaseous ions and from gaseous ions into the ionic lattice

38
Q

Where does the alternative route to the ionic lattice begin?

A

The alternative route to the ionic lattice begins from the enthalpy of formation of the elements in their standard states

39
Q
  1. How do you start constructing a Born-Haber Cycle?
A

A good starting point is to draw the elements with their state symbols about a third of the way up the diagram
This is shown as the left hand side of the equation for the process indicated
The location is marked by drawing a horizontal bar or line which represents the starting energy level

40
Q
  1. How do you form gaseous ions?
A

Next, we need to create the gaseous ions
This is a two step process of first creating the gaseous atoms and then turning them into ions
Creating gaseous atoms is a bond breaking process, so arrows must be drawn upwards

41
Q
  1. Give the equation for the formation of gaseous ions from Na and Cl
A

The enthalpy of atomisation of sodium is
Na (s) → Na (g) ΔHatꝋ = +108 kJ mol-1

The enthalpy of atomisation of chlorine is
½Cl2 (g) → Cl (g) ΔHatꝋ = +121 kJ mol-1

42
Q

How can products and energy values be shown on a born-haber cycle?

A

We can show the products of the process on the horizontal lines and the energy value against a vertical arrow connecting the energy levels

43
Q
  1. What happens to the gaseous atoms?
A

Now that the ions are created:
The sodium ion loses an electron, so this energy change is the first ionisation energy for sodium
Na (g) → Na+ (g) + e– ΔHIEꝋ = +500 kJ mol-1

The change is endothermic so the direction continues upwards
The chlorine atom gains an electron, so this is electron affinity
Cl (g) + e– → Cl– (g) ΔHEAꝋ = -364 kJ mol-1

The exothermic change means this is downwards
The change is displaced to the right to make the diagram easier to read

44
Q
  1. What is done with the ions?
A

The two remaining parts of the cycle can now be completed
The enthalpy of formation of sodium chloride is added at the bottom of the diagram
Na(s) + ½Cl2 (g) → NaCl (s) ΔHfꝋ = -411 kJ mol-1

This is an exothermic change for sodium chloride so the arrow points downwards
Enthalpy of formation can be exothermic or endothermic, so you may need to show it above the elements (and displaced to the right) for a endothermic change

45
Q
  1. What is the final change for the born-haber cycle?
A

The final change is lattice enthalpy, which is usually shown as formation. For sodium chloride, the equation is
Na+(g) + Cl–(g) → NaCl (s) ΔHlatꝋ

The cycle is now complete
The cycle is usually used to calculate the lattice enthalpy of an ionic solid, but can be used to find other enthalpy changes if you are given the lattice enthalpy

46
Q

What equation can be used to calculate the enthalpy of formation from a born-haber cycle?

A

ΔHfꝋ = ΔHatꝋ + ΔHatꝋ + IE + EA + ΔHlatꝋ

47
Q

What are the two key factors affecting lattice energy?

A

The two key factors which affect lattice energy, ΔHlatꝋ, are the ionic charge and ionic radii of the ions that make up the crystalline lattice

48
Q

How does the radius of an anion change down a group? Why?

A

The radius of the anion increases as you move down a group
As the distance between the bonded ions increases, the strength of the electrostatic attraction decreases

49
Q

How does lattice enthalpy change down a group?

A

This is reflected by a decrease in the lattice enthalpy
The lattice enthalpy becomes more negative or more exothermic as the ionic radius of the ions increases
This is because the charge on the ions is more spread out over the ion when the ions are larger
The ions are also further apart from each other in the lattice

50
Q

How do electrostatic forces change down a group?

A

The ions are also further apart from each other in the lattice
The attraction between ions is between the centres of the ions involved, so the bigger the ions the bigger the distance between the centre of the ions
Therefore, the electrostatic forces of attraction between the oppositely charged ions in the lattice are weaker

51
Q

How will an increasing ionic charge affect the lattice enthalpy?

A

Increasing the ionic charge will result in an increased attraction between oppositely charged ions
This will increase the energy required to break the lattice apart, and therefore increase the lattice enthalpy (becomes more positive or more endothermic)

52
Q

How is electrostatic attraction affected in terms of increasing ionic charge?

A

The greater the ionic charge, the higher the charge density
This results in stronger electrostatic attraction between the oppositely charged ions in the lattice
As a result, the lattice enthalpy is more endothermic
For example, the lattice energy of calcium oxide (CaO) is more endothermic than the lattice energy of potassium chloride (KCl)

53
Q

What is the equation for calculating the enthalpy of solution?

A

Enthalpy of solution = lattice enthalpy + hydration enthalpy

54
Q

What is the enthalpy of hydration in this equation?

A

The hydration enthalpy is the sum of the hydration enthalpies of each ion
If there is more than one cation or anion, such as in MgCl2, then you must multiply by the appropriate coefficient for that ion

55
Q

How do ionic compounds dissolve?

A

Hydration enthalpies are always negative values (exothermic)
When an ionic solid dissolves in water, positive and negative ions are formed
Water is a polar molecule with a δ- oxygen (O) atom and δ+ hydrogen (H) atoms which will form ion-dipole attractions with the ions present in the solution
The oxygen atom in water will be attracted to the positive ions and the hydrogen atoms will be attracted to the negative ions

56
Q

What is the size of hydration enthalpy determined by? 2

A

The size of the hydration enthalpy is governed by the amount of attraction between the ions and the water molecules

57
Q

How does a smaller ion size affect the attraction between ions and water molecules?

A

The smaller the ion, the stronger the attraction between the ions and the water molecules
As you go down a group, the ionic radius increases so attraction decreases and the the hydration enthalpy will become less exothermic
Overall, a smaller ion gives a more exothermic hydration enthalpy

58
Q

How does a more highly charged ion affect the attraction between ions and water molecules?

A

The more highly charged the ion; the stronger the attraction
The hydration enthalpies of 2+ ions in group 2 are much more exothermic than those of 1+ ions in group 1 as the attraction between the 2+ ions and the water molecules is stronger
Overall, a greater charge on the ion gives a more exothermic hydration enthalpy

59
Q

Calculate the energy change per mole for a reaction where 25 cm3 of 2.00 mol dm-3 hydrochloric acid was neutralised by 25 cm3 of 2.00 mol dm-3 sodium hydroxide. The temperature increased by 13.5 oC.

A

Step 1: Write an equation for the reaction occurring

HCl + NaOH → NaCl + H2O

Step 2: Calculate the energy change for the amount of reactants in the reaction vessel (remember that the mass equals the mass of acid and alkali)

Q = mcΔT
Q = 50 x 4.18 x 13.5
Q = 2821.5 J
Step 3: Calculate the number of moles of HCl (remember that neutralisation has occurred)

Moles of HCl = concentration x volume (dm3)
moles of HCl = 2 x 0.025
moles of HCl = 0.05 moles
Step 4: Calculate ΔH, using -Q as it is an exothermic reaction

ΔH = - Q / moles of HCl
ΔH = - 2821.5 / 0.05
ΔH = -56430 J mol-1
ΔH = -56.4 kJ mol-1

60
Q

What is entropy?

A

The entropy (S) of a given system is the number of possible arrangements of the particles and their energy in a given system
In other words, it is a measure of how disordered or chaotic a system is

61
Q

When will entropy increase?

A

When a system becomes more disordered, its entropy will increase
An increase in entropy means that the system becomes energetically more stable

62
Q

how does entropy change in the equation, CaCO3(s) → CaO(s) + CO2(g)

A

For example, during the thermal decomposition of calcium carbonate (CaCO3) the entropy of the system increases:

  • In this decomposition reaction, a gas molecule (CO2) is formed
  • The CO2 gas molecule is more disordered than the solid reactant (CaCO3), as it is constantly moving around
  • As a result, the system has become more disordered and there is an increase in entropy
63
Q

How does entropy change when a solid melts, e.g. H2O(s) → H2O(l)

A

The water molecules in ice are in fixed positions and can only vibrate about those positions
In the liquid state, the particles are still quite close together but are arranged more randomly, in that they can move around each other
Water molecules in the liquid state are therefore more disordered
Thus, for a given substance, the entropy increases when its solid form melts into a liquid

64
Q

What system will be more favourable in terms of entropy?

A

In both examples, the system with the higher entropy will be energetically favourable (as the energy of the system is more spread out when it is in a disordered state)

65
Q

What is the entropy change when calcium carbonate decomposes?

CaCO3 (s) → CaO (s) + CO2 (g)

Sꝋ298(CaCO3 (s)) = 92.9 J K-1 mol–1
Sꝋ298(CaO (s)) = 39.7 J K-1 mol–1
Sꝋ298(CO2 (g)) = 213.6 J K-1 mol–1

A

Answer:

Step 1: Write out equation to calculate ΔSꝋ298(reaction)

ΔSꝋ298(reaction) = ΣSꝋ298(products) - ΣSꝋ298(reactants)
Step 2: Substitute in formulas and then values for Sꝋ

ΔSꝋ298(reaction) = [Sꝋ298(CaO) + Sꝋ298(CO2)] - Sꝋ298(CaCO3)
ΔSꝋ(reaction) = (39.7 + 213.6) - 92.9
ΔSꝋ(reaction) = +160.4 J K-1 mol–1

66
Q

What determines the feasibility of a reaction?

A

The feasibility of a reaction is determined by two factors, the enthalpy change and the entropy change
he two factors come together in a fundamental thermodynamic concept called the Gibbs free energy (G)

67
Q

State the Gibbs Free Energy equation and state what each symbol means and it sunits

A

ΔGꝋ = ΔHreactionꝋ – TΔSsystemꝋ

The units of ΔGꝋ are in kJ mol–1
The units of ΔHreactionꝋ are in kJ mol–1
The units of T are in K
The units of ΔSsystemꝋ are in J K-1 mol–1(and must therefore be converted to kJ K–1 mol–1 by dividing by 1000)

68
Q

Calculate the free energy change for the following reaction:

2NaHCO3 (s) → Na2CO3 (s) + H2O (l) + CO2 (g)

ΔHꝋ = +135 kJ mol-1
ΔSꝋ = +344 J K-1 mol-1

A

Answer:

Step 1: Convert the entropy value in kilojoules

ΔSꝋ = +344 J K-1 mol-1 ÷ 1000 = +0.344 kJ K-1 mol-1
Step 2: Substitute the terms into the Gibbs Equation

ΔGꝋ = ΔHreactionꝋ – TΔSsystemꝋ
= +135 – (298 x 0.344)
= +32.49 kJ mol-1
The temperature is 298 K since standard values are quoted in the question

69
Q

What does gibbs free enrgy provide an effective way of focusing on?

A

Gibbs free energy provides an effective way of focusing on a reaction system at constant temperature and pressure to determine its spontaneity

70
Q

What sign must gibbs free energy have to be spontaneous?

A

For a reaction to be spontaneous, Gibbs free energy must be have a negative value (ΔGꝋ ≤ 0)

71
Q

Will a reaction be spontaneous when ΔS is positive and ΔH is negative?

A

always

72
Q

Will a reaction be spontaneous when ΔS is negative and ΔH is positive?

A

NEVER

73
Q

Will a reaction be spontaneous when ΔS is negative and ΔH is negative?

A

Spontaneous only at low T

74
Q

Will a reaction be spontaneous when ΔS is positive and ΔH is positive?

A

spontaneous only at high T

75
Q

Explain why a reaction will always be feasible if ΔS is positive and ΔH is negative

A

In exothermic reactions, ΔHreactionꝋ is negative
If the ΔSsystemꝋ is positive:
Both the first and second term will be negative
Resulting in a negative ΔGꝋ so the reaction is feasible
Therefore, regardless of the temperature, an exothermic reaction with a positive
ΔSsystemꝋ will always be feasible

76
Q

Explain why a reaction will only be feasible at low temps if ΔS is negative and ΔH is negative

A

If the ΔSsystemꝋ is negative:
The first term is negative and the second term is positive
At very high temperatures, the –TΔSsystemꝋ will be very large and positive and will overcome ΔHreactionꝋ
Therefore, at high temperatures ΔGꝋ is positive and the reaction is not feasible

77
Q

Explain why a reaction will never be feasible if ΔS is negative and ΔH is positive

A

In endothermic reactions, ΔHreactionꝋ is positive
If the ΔSsystemꝋ is negative:
Both the first and second term will be positive
Resulting in a positive ΔGꝋ so the reaction is not feasible
Therefore, regardless of the temperature, endothermic with a negative ΔSsystemꝋ will never be feasible

78
Q

Explain why a reaction will only be feasible at high temps if ΔS is positive and ΔH is positive

A

If the ΔSsystemꝋ is positive:
The first term is positive and the second term is negative
At low temperatures, the –TΔSsystemꝋ will be small and negative and will not overcome the larger ΔHreactionꝋ
Therefore, at low temperatures ΔGꝋ is positive and the reaction is not feasible
The reaction is more feasible at high temperatures as the second term will become negative enough to overcome the ΔHreactionꝋ resulting in a negative ΔGꝋ

79
Q

What happens to ΔGꝋ at equilibrium?

A

When a reversible reaction reaches equilibrium, the Gibbs free energy is changing as the ratio of reactants to products changes

80
Q

What happens to ΔGꝋ at equilibrium for a non-reversible reaction?

A

For non-reversible reactions:
As the amount of products increases, the reaction moves towards completion
This leads to a decrease in Gibbs free energy

81
Q

What happens to ΔGꝋ at equilibrium for a reversible reaction?

A

For reversible reactions:
As the amount of products increases, the reaction moves towards equilibrium
This causes a decrease in Gibbs free energy

82
Q

What is the relation of K and delta G

A

When the equilibrium constant, K, is determined for a given reaction, its value indicates whether the products or reactants are favoured at equilibrium
ΔG is an indication of whether the forward or backward reaction is favoured

83
Q

Give the equation for the quantitative relation between gibbs and k

A

ΔGꝋ = -RT In K