12.4A Energetics & Thermodynamics

Question 1

Bond energy/Bond dissociation energy

Answer 1

— the enthalpy change when 1 mole of covalent bonds between two atoms are broken in gaseous molecules under standard conditions of 298K and 1atm

• Bond energy is always endothermic (E > 0) as energy must be absorbed in order to break bonds
• Bond energy is a measure of covalent bond strength

Question 2

calculate Enthalpy change (ΔH)

Answer 2

∆𝐻_𝑟 = ∑ enthalpy change for bonds broken + ∑ enthalpy change for bonds formed

∆𝐻_𝑟 = ∑ [bond energy of products] − ∑ [bond energy of reactants]

∆𝐻_𝑟 = – Q / n

Question 3

Q: State the meaning of the term mean bond enthalpy for the X—X bond

Answer 3

The enthalpy change to break 1 mol of X—X bonds, averaged over a range of compounds/molecules

Question 4

Enthalpy, H

Answer 4

• The total heat content of a substance
• It cannot be measured directly
• Energy is used (absorbed from the surroundings e.g. melting, boiling) when bonds are broken
• Energy is released (into the surroundings) when bonds are formed

Question 5

Enthalpy change, ∆H

Answer 5

— The amount of heat absorbed or evolved in a chemical reaction

∆𝐻 = (Total enthalpy of products)−(Total enthalpy of reactants)

∆H = - Q / n

Question 6

Calorimetry

Answer 6

— is a technique used to measure enthalpy change (by using calorimeters ex. polystyrene cup, a vacuum flask or metal can)

Question 7

Calorimetry – apparatus & substances required:

Answer 7

• Known amounts of reactants
• Known volume of liquids
• Thermometer (accurate to 0.1°C or 0.2°C)
• Insulated container (with lid if necessary)
• Stopwatch (sometimes)
• Glass rod (sometimes)

Question 8

Calorimetry – underlying principles

Answer 8

Calorimetry relies on the fact that 4.18 J of energy is required to increase the temperature of 1 g of water by 1°C (specific heat capacity, c J g-1 K-1)

amount of heat Q = mc∆T

∆H = - Q / n

Question 9

Examples of reasons for the different values obtained during calorimetry

Answer 9

• heat loss/not all heat transferred to the apparatus/heat absorbed by the apparatus
• incomplete combustion/not completely burned/reaction is not complete
• reactants and/or products may not be in standard states
• MBE do not refer to specific compounds/MBE values vary with different compounds/molecules
• are average/mean values taken from a range of compounds/molecules

Question 10

1. Standard enthalpy change of FORMATION, (∆𝑯_𝒇)^𝜽

Answer 10

— the enthalpy change when 1 mole of a pure compound is formed from its constituent elements in their standard states, under standard conditions of 298K and 1atm

!!! (∆𝑯_𝒇)^𝜽 of elements are assigned value of 0 !!!

H2 (g) + ½ O2 (g) → H2O (l)

Question 11

(∆𝑯_𝒇)^𝜽

Answer 11

(∆𝑯_𝒇)^𝜽 of compounds represents the energy transferred to and from the surroundings when chemical bonds in the elements are broken and new bonds are formed in the compound.
Hence it measures the stability of the compound relative to its constituent elements

Question 12

Q: (∆𝑯_𝒇)^𝜽 of CO2 and CO is -394 kJ mol-1 and -110 kJ mol-1 respectively.
Comment on the relative stability of these oxides with respect to their elements.

Answer 12

• The oxides are more stable relative to the elements as their formation are exothermic reactions, lowering the energy of the system.
• However, CO2 is the preferred product since its formation is more exothermic than that of CO.

Question 13

2. Standard enthalpy change of COMBUSTION, (∆𝑯_𝒄) ^𝜽

Answer 13

— the enthalpy change when 1 mole of a substance is completely burnt in excess oxygen under standard conditions of 298K and 1atm

(∆𝑯_𝒄) ^𝜽 always exothermic (< 0) as heat is always evolved during combustion

C (s) + O2 (g) → CO2 (g)

Question 14

3. Standard enthalpy change of ATOMISATION, (∆𝑯_𝒂𝒕) ^𝜽

Answer 14

— the enthalpy change when 1 mole of isolated gaseous atoms is produced from the element in its standard state under standard conditions of 298K and 1 atm

(∆𝑯_𝒂𝒕) ^𝜽 is always endothermic (> 0) as energy is always absorbed since the bonds in the elements must first be broken before atoms can be formed

C (s) → C (g)

Question 15

4. Standard enthalpy change of NEUTRALISATION, (∆𝑯_𝒏) ^𝜽

Answer 15

— the enthalpy change when 1mole of water is formed when an acid neutralizes a base.
The reaction is carried out in an infinitely dilute solution under standard conditions of 298K and 1 atm

• (∆𝑯_𝒏) ^𝜽 the same for all reactions involving a strong acid and a strong base (ex. HCl, HNO3, H2SO4, NaOH)

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

Question 16

5.1. Enthalpy change of HYDRATION, (∆𝑯_𝒉𝒚𝒅) ^𝜽, of an anhydrous salt

Answer 16

— the enthalpy change when 1 mole of a hydrated salt is formed from 1 mole of anhydrous salt under standard conditions of 298 K and 1 atm

• (∆𝑯_𝒉𝒚𝒅) ^𝜽 is always exothermic (< 0) as energy is always released when bonds are formed between the salt and water molecules

Na2S2O3 (s) + 5 H2O (l) → Na2S2O3 * 5 H2O (s)

Question 17

5.2. Enthalpy change of HYDRATION, (∆𝑯_𝒉𝒚𝒅) ^𝜽, of gaseous ions

Answer 17

— the enthalpy change when 1 mole of a gaseous ions is hydrated

• The ions are dissolved in a large amount of water such that further addition of water produces no further heat changes

Na+ (g) + aq → Na+ (aq)

Question 18

6. Enthalpy change of solution, (∆𝑯_𝒔𝒐𝒍) ^𝜽

Answer 18

— the enthalpy change when 1 mole of a solute is completely dissolved in a solvent to form an infinitely dilute solution under standard conditions of 298 K and 1 atm

Na2SO4 (s) + aq → Na2SO4 (aq)

Question 19

Hess’s Law states that…

Answer 19

— the enthalpy changes in a chemical reaction is independent of the pathway which the reaction takes place, provided that the initial and final states of the system remains the same

!!!ENERGY CYCLE DIAGRAM!!

the concept was later developed into the Law of Conservation of Energy, also the First Law of Thermodynamics – energy can be transformed from one form to another, but cannot be destroyed

Question 20

Bond energy is the enthalpy change when one mole of covalent bonds between two atoms are broken. This does not exist in ionic compounds e.g. NaCl

For ionic compounds, we need to introduce:

Answer 20

1) Enthalpy change of atomisation
2) First ionisation energy
3) First electron affinity
4) Lattice energy

Question 21

1. Enthalpy change of ATOMISATION, (∆𝑯_𝒂𝒕) ^𝜽

Answer 21

— the enthalpy change when 1 mole of isolated gaseous atoms are produced from the element in its standard state under standard conditions of 298K and 1 atm

!!! Always endothermic (> 0) as energy is always absorbed so that bonds within the elements have to be broken to release the atoms

½ Cl2(g) → Cl(g)

Question 22

2. First ionisation energy

Answer 22

— energy required to remove 1 mole of outermost electron from 1 mole of gaseous atoms to produce 1 mole of gaseous ions each with +1 charge

• Always endothermic (> 0) as energy is always absorbed to break bonds between electron and the nucleus

Cs(g) → Cs+(g) + e–

Question 23

3.1. ELECTRON AFFINITY, (∆𝑯_𝒆𝒂) ^𝜽
The FIRST electron affinity, (∆𝑯_𝒆𝒂𝟏) ^𝜽,

Answer 23

— is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous 1– ions under
standard conditions of 298 K and 1 atm

Cl (g) + e– → Cl – (g)

Question 24

3.2. The SECOND electron affinity, (∆𝑯_𝒆𝒂𝟐) ^𝜽,

Answer 24

— is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous 1– ions to form 1 mole of gaseous 2– ions under standard conditions of 298K and 1 atm

Note that:
2nd electron affinities are always endothermic ( (∆𝑯_𝒆𝒂𝟐) ^𝜽 is positive), and so are 3rd electron affinities.

Question 25

Electron affinity example:

Answer 25

• 1st electron affinity of oxygen:
  O(g) + e– →O–(g) (∆𝑯_𝒆𝒂𝟏) ^𝜽 = –141 kJ mol–1
• 2nd electron affinity:
  O–(g) + e– →O2–(g) (∆𝑯_𝒆𝒂𝟐) ^𝜽 = +798 kJ mol–1
• The overall enthalpy change for O(g) + 2e– → O2– (g)
  (∆𝑯_𝒆𝒂𝟏) ^𝜽 + (∆𝑯_𝒆𝒂𝟐) ^𝜽 = (–141) + (+798) = +657 kJ mol–1

Question 26

Q: Suggest why the 2nd and 3rd electron affinities are always endothermic

Answer 26

There must be an input of energy to overcome the repulsive forces between the (negative) electron and the negative ion

Question 27

4. Lattice energy, (∆𝑯_𝑳𝑬) ^𝜽

Answer 27

— Heat change when 1 mole of an ionic solid is formed from its isolated gaseous ions from an infinite distance apart under standard conditions of 298 K and 1 atm

• When ions of opposite charges come together to form a stable solid compound, energy is released due to bond formation. Hence (∆𝑯_𝑳𝑬) ^𝜽 is always exothermic (< 0)
• The stronger the electrostatic attraction between the oppositely charged ions, the more exothermic the lattice energy.
• The more exothermic the lattice energy, the stronger the ionic bonds are, and the more stable the ionic compound.

Na+ (g) +Cl– (g) → NaCl (s)

Question 28

Interpretation of lattice energies

Answer 28

• Lattice energy determined via the Born-Haber cycle is considered as the experimental (thermodynamic) values. Lattice energies can also be calculated using a theoretical model.
  **Good agreement between the experimental and theoretical values indicate the bonding is purely ionic. Great differences between the values is due to polarisation of anion by the cation

Question 29

The Born-Haber Cycle

Answer 29

— is a technique of applying Hess’s Law to the standard enthalpy changes during the formation of an ionic compound

• is a particular type of enthalpy cycle used to calculate lattice energy.
• is an enthalpy level diagram breaking down the formation of an ionic compound into a series of simpler steps

Question 30

The concept of ENTROPY

Answer 30

the universe (and all its happenings) tends to proceed towards a state of greater disorder or randomness
* The natural tendency for greater disorder in Chemistry is that chemical reactions that result in greater disorder will be favoured

Question 31

Entropy, S

Answer 31

— is a measure of the orderliness of a system, [ J mol-1K-1]

• The greater the physical and thermal chaos → the more disorderly the system → the higher the value of S.

SOLID:
Low temperature
Lower thermal chaos
Particles orderly arranged

GAS:
Higher temperature
Higher thermal chaos
Particles disorderly

• The gaseous state has the greatest entropy as its particles have the highest energy, and are moving randomly in all directions. It is highly disordered.
• In larger, more complex molecules, there are more ways for atoms to be arranged in 3-D space, and greater variety of internal motion (e.g. vibrations, bending, stretching). Hence they have greater entropy.

Question 32

Absolute zero & concept of Entropy

Answer 32

• At absolute zero (i.e. 0 Kelvins or -273.15 °C), all matter is in a crystalline solid state. All particles have a perfectly orderly arrangement and absolutely no movement
• In this state, matter has maximum possible order and zero entropy
• However it is physically impossible to achieve absolute zero
• Current record is 100 pK, or 0.000 000 000 1 K

Question 33

Entropy change, ∆S

Answer 33

∆𝑆 = final S of system − initial S of system

• If ∆S > 0, entropy of the system increased.
  System became more disorderly.
• If ∆S < 0, entropy of the system decreased.
  System became less disorderly.

Question 34

When a chemical system is subject to changes, its entropy changes. These changes include:

Answer 34

• Temperature
• Phase (physical state)
• Number of particles, especially for gaseous systems
• Mixing of particles

Question 35

Entropy change, ∆S – effect of temperature

Answer 35

• When temperature increases, both physical chaos and thermal chaos increases
• This results in an increase in entropy ∆S >0

E.g. Heating of ice
As the particles absorb energy, they start to vibrate more, making the arrangement less orderly. The particles also move faster, increasing the disorder of the particles.

The number of ways energy can be distributed in the system increases.

Question 36

Entropy change, ∆S – effect of phase

Answer 36

• Disorder increases as the system changes:
  SOLID → LIQUID → GAS
• The orderly arrangement of particles in a solid is destroyed during phase change
  This results in an increase in entropy ∆S >0

Question 37

Entropy change, ∆S – change in number of particles

Answer 37

• If a change involves increasing the number of particles in the system, the entropy increases and ∆S >0
• The number of ways to arrange a greater number of particles is greater.

E.g. decomposition of laughing gas
N2O (g) → N2 (g) + ½ O2 (g), ∆S > 0

Question 38

Entropy change, ∆S – mixing of particles

Answer 38

• Mixing of particles leads to an increase in entropy since there is higher state of disorder after mixing, therefore ∆S >0
• The number of ways to arrange particles in the system increases upon mixing.
• If the two sets of particles have different energies before mixing, the energy distribution of the system will increase due to collisions upon mixing, increasing thermal chaos.
• High entropy alloys – mixtures of metals

Question 39

Calculating the entropy change

Answer 39

For the system (reactants and products):
Δ S^𝜽 system = Σ S^𝜽_products−Σ S^𝜽_reactants

For the surroundings:
Δ S^𝜽_surroundings = (−ΔН^𝜽_𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛) / Т

The total entropy change is given by:
Δ S^𝜽_total = Δ S^𝜽_system + Δ S^𝜽_surroundings

• Δ S^𝜽_total > 0 — reaction is SPONTANEOUS
• Δ S^𝜽_total = 0 — reaction is EQUILIBRIUM
• Δ S^𝜽_total < 0 — reaction is NON-spontaneous

Question 40

Gibbs Free Energy, ∆G

Answer 40

∆𝐺 = ∆𝐻 − 𝑇∆𝑆

In chemistry, it is used to predict if a chemical reaction will take place spontaneously

• !! ∆G < 0 — reaction takes place SPONTANEOUSLY
• ∆G = 0 — reaction is in EQUILIBRIUM
• !! ∆G > 0 — reaction DOES NOT take place spontaneously

The equation shows the tendency of the system (at constant p and T) to
* Maximise entropy (i.e. increase disorderliness of system)
* Minimise enthalpy (i.e. lower energy of the system)