Thermodynamics

Question 1

Enthalpy change of formation

Answer 1

The standard enthalpy change of formation of a compound is the energy transferred when 1 mole of the compound is formed from its elements under standard conditions (298K and 100kpa), all reactants and products being in their standard states

Na (s) + ½Cl2 (g)  NaCl (s) [fH = - 411.2 kJ mol-1]

Question 2

Enthalpy of atomization

Answer 2

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

Na (s)  Na(g) [atH = +148 kJ mol-1]

½ O2 (g)  O (g) [atH = +249 kJ mol-1]

Question 3

Enthalpy of sublimation

Answer 3

The enthalpy change for a solid metal turning to gaseous atoms can also be called the Enthalpy of sublimation and will numerically be the same as the enthalpy of atomization

Na (s)  Na(g) [subH = +148 kJ mol-1]

Question 4

Bond dissociation enthalpy (bond energy)

Answer 4

The bond dissociation enthalpy is the standard molar enthalpy change when one mole of a covalent bond is broken into two gaseous atoms (or free radicals)

Question 5

First ionization enthalpy

Answer 5

The first ionisation enthalpy is the enthalpy change required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous ions with a +1 charge

Question 6

Second ionization enthalpy

Answer 6

The second ionisation enthalpy is the enthalpy change to remove 1 mole of electrons from one mole of gaseous 1+ ions to produces one mole of gaseous 2+ ions

Question 7

First electron affinity

Answer 7

The first electron affinity is the enthalpy change that occurs when 1 mole of gaseous atoms gain 1 mole of electrons to form 1 mole of gaseous ions with a –1 charge

O (g) + e-  O- (g) [ea 1H] = -141.1 kJ mol-1 ]

The first electron affinity is exothermic for atoms that normally form negative ions

This is because the ion is more stable than the atom, and there is an attraction between the nucleus and the electro

Question 8

Second electron affinity

Answer 8

The second electron affinity is the enthalpy change when one mole of gaseous 1- ions gains one electron per ion to produce gaseous 2- ions

O – (g) + e-  O2- (g) [ea 2H = +798 kJ mol-1]

The second electron affinity for oxygen is endothermic because it takes energy to overcome the repulsive force between the negative ion and the electron.

Question 9

Enthalpy of lattice formation

Answer 9

The Enthalpy of lattice formation is the standard enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent ions in gaseous form

Na+ (g) + Cl- (g)  NaCl (s) [LattH = -787 kJ mol-1]

Question 10

Enthalpy of lattice dissociation

Answer 10

The Enthalpy of lattice dissociation is the standard enthalpy change when 1 mole of an ionic crystal lattice form is separated into its constituent ions in gaseous form

NaCl (s)  Na+ (g) + Cl- (g) [ LattH = +787 kJ mol-1]

Question 11

Enthalpy of hydration hydH

Answer 11

Enthalpy change when one mole of gaseous ions become aqueous ions

X + (g) + aq  X + (aq) For Li+ hydH = -519 kJ mol-1 or X - (g) + aq  X- (aq) For F- hydH= -506 kJ mol-1

This always gives out energy (exothermic, -ve) because bonds are made between the ions and the water molecules

Question 12

Enthalpy of solution

Answer 12

The enthalpy of solution is the standard enthalpy change when one mole of ionic solid dissolves in a large enough amount of water to ensure that the dissolved ions are well separated and do not interact with one another

NaCl (s) + aq  Na+ (aq) + Cl-(aq)

Question 13

Born Haber cycles

Answer 13

The lattice enthalpy cannot be determined directly. We calculate it indirectly by making use of changes for which data are available and link them together in an enthalpy cycle the Born-Haber cycle

Question 14

DRAW Born-Haber cycle: sodium Chloride

Answer 14

Question 15

DRAW Born-Haber cycle: magnesium Chloride

Answer 15

Question 16

Second electron affinity

Answer 16

second electron affinity for oxygen is endothermic because it take energy to overcome the repulsive force between the negative ion and the electron

Question 17

Trends in Lattice Enthalpies

The strength of enthalpy of lattice formation depends on the following factors:

Answer 17

1. The sizes of the ions

The larger the ions, the less negative the enthalpies of lattice formation (i.e. a weaker lattice). As the ions are larger the charges become further apart and so have a weaker attractive force between them

2. The charges on the ion

The bigger the charge of the ion, the greater the attraction between the ions so the stronger the lattice enthalpy (more negative values).

Question 18

Perfect Ionic Model

Answer 18

Theoretical lattice enthalpies assumes a perfect ionic model where the ions are 100% ionic and spherical and the attractions are purely electrostatic

When the negative ion becomes distorted and more covalent we say it becomes polarised. The metal cation is called polarising if it polarises the negative ion

When 100 % ionic the ions are spherical. The theoretical and the Born Haber lattice enthalpies will be the same

The charge cloud is distorted. The theoretical and the experimental Born Haber lattice enthalpies will differ

Question 19

There is a tendency towards the covalent character in ionic substances when

Answer 19

• the positive ion is small
• the positive ion has multiple charges
• the negative ion is large
• the negative ion has multiple negative charges.

Question 20

Differences between theoretical and Born Haber (experimental) lattice enthalpies

Answer 20

The Born Haber lattice enthalpy is the real experimental value

When a compound shows a covalent character, the theoretical and the born Haber lattice enthalpies differ. The more the covalent character the bigger the difference between the values.

When a compound has some covalent character- it tends towards giant covalent so the lattice is stronger than if it was 100% ionic. Therefore the Born-Haber value would be larger than the theoretical value.

Question 21

Why does Calcium chloride have the formula CaCl2 and not CaCl or CaCl3?

Answer 21

We need to calculate an enthalpy of formation for each case. The one with the most exothermic enthalpy of formation will be the one that forms as it will be the most thermodynamically stable

Question 22

Spontaneous process

Answer 22

A SPONTANEOUS PROCESS (e.g. diffusion) will proceed on its own without any external influence

Question 23

significant increase in the entropy will occur if:

Answer 23

• there is a change of state from solid or liquid to gas
• there is a significant increase in a number of molecules between products and reactants.

Question 24

Calculate entropy

Answer 24

∆S entropy. ˚ = Σ S˚products - ΣS˚reactants

Question 25

Calculate ∆S˚ for the following reaction at 25˚C: 2Fe2O3 (s) + 3C (s)  4Fe (s) + 3CO2 (g)

S [Fe2O3] = 87.4 J K-1 mol-1

S [C] = 5.7 J K-1 mol-1

S [Fe] = 27.3 J K-1 mol-1

S [CO2] = 213.6 J K-1 mol-1

Answer 25

∆S˚ = S˚products - S˚reactants

= (3 x 213.6 + 4 x 27.3) – (2 x 87.4 + 3 x 5.7) = + 558.1 J K-1 mol-1 = + 558 J K-1 mol-1 (3 S.F.)

Question 26

∆G

Answer 26

∆G = ∆H - T∆S

Question 27

For any spontaneous change, ∆G will be

Answer 27

negative

Question 28

If ∆G is negative there is still a possibility, however, that the reaction will not occur or will occur so slowly that effectively it doesn’t happen.

Answer 28

High activation energy

Catalyst

Question 29

Calculating the temperature a reaction will become feasible

Answer 29

T = ∆H / ∆S

Question 30

Hsolution

2 equation

Answer 30

Hsolution =  HL dissociation + sum hydH

H solution = -  HL formation + sum hydH

Question 31

Why hydration enthalpies are exothermic

Answer 31

Hydration enthalpies are exothermic as energy is given out when water molecules bond to the metal ions. The negative ions are attracted to the δ+ hydrogens on the polar water molecules and the positive ions are attracted to the δ - oxygen on the polar water molecules. The higher the charge density the greater the hydration enthalpy (e.g. smaller ions or ions with larger charges) as the ions attract the water molecules more strongly. e.g. Fluoride ions have more negative hydration enthalpies than chloride ions. Magnesium ions have a more negative hydration enthalpy than barium ions

Question 32

What does ΔHSolution tell us?

Answer 32

Generally ΔH solution is not very exo or endothermic so the hydration enthalpy is about the same as lattice enthalpy. In general the substance is more likely to be soluble if the ΔH solution is exothermic. If a substance is insoluble it is often because the lattice enthalpy is much larger than the hydration enthalpy and it is not energetically favourable to break up the lattice, making ΔH solution endothermic.

Question 33

When a solid dissolves

Answer 33

When a solid dissolves into ions the entropy increases as there is more disorder as solid changes to solution and number of particles increases. This positive S can make G negative even if H solution is endothermic, especially at higher temperatures