TOPIC 5: CHEMICAL ENERGETICS

Question 1

[Units]
Enthalpy change, ΔH

Answer 1

kJ/mol or J/mol

Question 2

[Recall]
Exothermic VS Endothermic

Answer 2

ΔH < 0 VS ΔH > 0
Heat released VS Heat absorbed
Surrounding temp increase VS Surrounding temp decrease
Products energetically more stable VS products energetically less stable

Question 3

IMPT THINGS FOR THERMOCHEMICAL EQN / SOLVING QNS

Answer 3

1. STATE SYMBOL
2. MULTIPLY BY MOLE RATIO
3. 3SF
4. SIGN ( + / - )
5. UNITS

Question 4

[Definition]
Standard state

Answer 4

The standard state of a substance at a specified temperature is its pure form at 1 bar ( = 10^5 Pa), and 1 mole/dm^3 for solutions.

*Temperature usually 298K (25°C)

*NOT the stp in Gaseous State

Question 5

[Definition]
Standard enthalpy change of formation

Answer 5

The standard enthalpy change of formation is the ENERGY CHANGE when 1 MOLE of a substance in its standard state is formed from its constituent elements in their standard states, at a specified temperature, usually 298K.

*Standard enthalpy change of formation of ELEMENTS in their standard states is ZERO

Question 6

[Definition]
Standard enthalpy change of combustion

Answer 6

The standard enthalphy change of combustion is the ENERGY RELEASED when 1 MOLE of a substance is burnt in excess oxygen, with all reactants and products in their standard states, at a specified temperature, usually 298K.

*Standard enthalpy change of combustion of O2 (g) / H2O (l) / CO2 (g) is ZERO
O2 cos cannot combust further
H2O and CO2 cos they’re products of combustion

Question 7

[Definition]
Standard enthalpy change of neutralisation

Answer 7

The standard enthalpy change of neutralisation is the ENERGY RELEASED when 1 MOLE of water is formed in the neutralisation reaction between an acid and a base, all in their standard states, at a specified temperature, usually 298K.

Question 8

[Recall]
Why is the enthalpy change of neutralisation between a strong acid and base almost the same for all?

What is the general value of enthalpy change?

Why are reactions with weak acid / base slightly less exothermic than usual?

Answer 8

Complete dissociation, so the reactions between them are effectively the reaction between the aqueous H+ and OH- ions.

-57.0 kJ/mol

Do not dissociate completely, so energy is absorbed to dissociate the undissociated weak acid or base, so less exothermic.

Question 9

[Defintion]
Bond Energy

Answer 9

Bond energy is the AVERAGE ENERGY REQUIRED to break 1 MOLE of a covalent bond between two atoms in the gaseous state.

*2 x ΔHat = BE

Question 10

What is bond dissociation energy?

Answer 10

The energy required to break 1 mole of a PARTICULAR COVALENT BOND in a specific molecule in the gaseous state

*Bond dissociation energy may differ even for the exact same type of bond as the strength is also influenced by its neighbouring atoms

Question 11

[Formulas]
Heat change & Enthalpy change

Answer 11

q = -mcΔT / q = -CΔT

ΔH = q / n = (-mcΔT)/(no. of moles of limiting reagent)

***Values are ALL about the WATER / SOLUTION, NOT the solid / substance
e.g. mass of solution, c of solution etc

Question 12

[Definition] Specific Heat Capacity

Answer 12

The specific heat capacity, c, is definied as the amount of heat energy required to raise the temperature of 1g of the substance by 1K.

Unit of c: J g⁻¹ K⁻¹

Question 13

[Definition] Heat Capacity

Answer 13

Heat capacity, C, is defined as the amount of heat required to raise the temperature of a certain mass of a substance by 1K.

Unit of C: J K⁻¹

Question 14

Assumptions / Sources of errors in calorimetry experiments

Answer 14

Assumed negligible heat loss / heat gain from surrounding air / calorimeter

Density of solution is same as that of water

Incomplete combustion

Wick also burns and releases some heat

Question 15

[Definition]
Hess’ Law

Answer 15

Hess’ Law states that the enthalpy change of a reaction is determined only by the initial and final states and is independent of the reaction pathway taken.

Question 16

3 Ways of Presenting Hess’ Law

Answer 16

Energy Cycle & Energy Level Diagram & Algebraic Method

Question 17

[Recall]
Common Hess’ Law applications for combustion, formation and bond energy

Answer 17

ΔHr (combustion) =
∑mΔHc (reactants) - ∑nΔHc (products)

ΔHr (formation) =
∑mΔHf (products) - ∑nΔHf (reactants)

ΔHr (bond energy) =
∑BE (bonds broken) - ∑BE (bonds formed)

Question 18

Why is experimental value of enthalpy change of bond broken different from that of the theoretical value?

Answer 18

The bond energies given in the data booklet are the average bond energies of that particular covalent bond in different types of molecules. Thus, these values may not apply to the specific molecules in this reaction.

Question 19

[Definition]
Standard enthalpy change of atomisation (element & compound)

Answer 19

The standard enthalpy change of atomisation, for an ELEMENT is the ENERGY REQUIRED to form 1 MOLE of gaseous atoms from the element, all in their standard states, at a specified temperature, usually 298K.

The standard enthalpy change of atomisation, for a COMPOUND is the ENERGY REQUIRED to form gaseous atoms from 1 MOLE of the compound, all in their standard states, at a specified temperature, usually 298K.

*2 x ΔHat = BE

Question 20

[Definition]
Lattice energy

Answer 20

Lattice energy of a compound is the ENERGY RELEASED when 1 MOLE of a solid ionic compound is formed from its consitutent gaseous ions.

Question 21

[Factors]
Factors affecting lattice energy

Answer 21

1. Charge
2. Size

|LE|∝|(q⁺ * q⁻) / (r⁺ + r⁻)|

Question 22

[Definition]
Ionisation Energy (first and second)

Answer 22

The FIRST ionisation energy is the ENERGY REQUIRED to REMOVE 1 MOLE of electrons from 1 MOLE of gaseous atoms to form 1 MOLE of singly charged gaseous cations.

The SECOND ionisation energy is the ENERGY REQUIRED to REMOVE 1 MOLE of electrons from 1 MOLE of singly charged gaseous cations to form 1 MOLE of doubly charged gaseous cations.

*IE ALWAYS positive (aka endothermic)

Question 23

[Definition]
Electron affinity (first and second)

Answer 23

The FIRST electron affinity is the ENERGY CHANGE when 1 MOLE of electrons is added to 1 MOLE of gaseous atoms to form 1 MOLE of singly charged gaseous anions.

The SECOND electron affinity is the ENERGY CHANGE when 1 MOLE of electrons is added to 1 MOLE of singly charged gaseous anions to form 1 MOLE of doubly charged gaseous anions.

*1st EA usually negative (aka exothermic)
- more energy released from attraction than repulsion
*2nd EA usually positive (aka endothermic)
- more energy required to overcome repulsion than attraction

Question 24

[Recall]
Purpose of Born-Haber Cycle

Answer 24

1. Impossble to generate measurable quantities
2. Impossible to harness the ions to react with each other
3. Impossible to measure chg in temp

Question 25

[Defintion]
Standard enthalpy change of hydration

Answer 25

The standard enthalpy change of hydration of an ion is the ENERGY RELEASED when 1 MOLE of free gaseous ions in its standard state is dissolved in water to give a solution of INFINITE DILUTION, at a specified temperature, usually 298K.

Question 26

[Factors]
Factors affecting standard enthalpy change of hydration

Answer 26

Charge density

|ΔHhyd|∝|q/r|

The higher the charge density, the stronger the ion-dipole interactions, the more exothermic it will be

Question 27

[Definition]
Standard enthalpy change of solution

Answer 27

The standard enthalpy change of solution of a substance is the ENERGY CHANGE when 1 MOLE of a substance is completely dissolved in a solvent to give a solution of infinite dilution, at a specified temperature, usually 298K.

*Substance usually ionic compound
*Solvent usually water

Question 28

[Formula]
Standard enthalpy change of solution

Answer 28

ΔHsol = -LE + ΔHhyd (cation) + ΔHhyd (anion)

Question 29

[Definition]
Entropy

Answer 29

Entropy, S, is a measure of the RANDOMNESS or DISORDER in a system, reflected in the number of ways that the energy of a system can be distributed through the motion of its particles.

ΔS = S (final) - S (initial)
ΔS > 0 (more ways to distribute energy)
ΔS < 0 (less ways to distribute energy)

Question 30

[Factors]
Factors affecting entropy

Answer 30

1. Change in temperature
   - Temp increase, Average KE & range of energies increase, more ways energy can be distributed among the particles, entropy increases
2. Change in phase
   - entropy of solid < liquid < gas
   - When melt / vaporize / sublime, particles in final state move more freely and more disordered, increase number of ways the particles and energy can be distributed, entropy increases
3. Change in number of particles
   - more particles more ways the particles and energy can be distributed, entropy increases
4. Mixing of particles (assuming ideal gas)
   - Volume of gas increase, more ways to distribute particles and their energy, entropy increases

Question 31

[Formula]
Gibbs Free Energy

Answer 31

ΔG = ΔH - TΔS

*ΔG < 0 (thermodynamically feasible & spontaneous)
*ΔG = 0 (equilibrium & no net reaction)
- during change in state
*ΔG > 0 (themodynamically not feasible & not spontaneous)

Question 32

[Recall]
Limitations of the use of ΔG to predict spontaneity of reaction

Answer 32

1. All must be in their standard states
2. Temperature must be achievable
3. Kinetics considerations (may be spontaneous but not instaneous - not at observable rate)

Question 33

[Units]
Entropy change, ΔS

Answer 33

Jmol⁻¹K⁻¹ or kJmol⁻¹K⁻¹

Question 34

[Units]
Gibbs free energy, G

Answer 34

kJmol⁻¹ or Jmol⁻¹