Physical Chemistry 8: Thermodynamics

1. Standard enthalpy change when
2. one mole of the compound
is formed
3. from its elements under
4. standard conditions and standrard states

Standard conditions: 298K and
100kpa

Eg: Na (s) + 0.5Cl2 (g) → NaCl (s)

1. Standard enthalpy change when
2. one mole of gasesous atom formed from
3. an element in its standard state

M (s) → M (g)

1. Standard enthalpy change when
2. one mole of covalent bond is broken
3. into two gaseous atoms

M2 (g) → 2M (g)

Bond diss = 2 x enthapy of atomisation

Compound can break up too:

CH4 (g) → CH3 (g) + H (g)

1. Standard enthalpy change when
2. one electron is removed from
3. one mole of gasesous atom to form
4. one mole of positively charged gaseous ion

M (g) → M+ (g) + e-

1. Standard enthalpy change when
2. one electron is removed from
3. one mole of positively charged gaseous ion to form
4. one mole of 2+ gaseous ion

M+ (g) → M2+ (g) + e-

1. Standard enthalpy change when
2. one mole of gaseous atom
3. gains one electron to form
3. one mole of negatively charged gaseous ion

M- (g) → M (g) + e-

1. Standard enthalpy change when
2. one mole of negatively charged gaseous ion
3. gains one electron to form
4. one mole of 2- gaseous ion

M2- (g) → M- (g)+ e-

The second electron affinity *endothermic*

WHY? takes energy to overcome the *repulsive* force between the negative ion and the electron

1. Standard enthalpy change when
2. one mole of an ionic crystal lattice forms from its
3. constituent ions in gaseous forms

M+ (g) + P- (g) → MP (s)

1. Standard enthalpy change when
2. one mole of an ionic crystal lattice breaks/separated into its
3. constituent ions in gaseous forms

Always ENDOTHERMIC

MP (s) → M+ (g) + P- (g)

1. Standard enthalpy change when
2. one mole of gaseous ions becomes
3. one mole of aqeous ions

Always EXOTHERMIC ∵ water molecules bond with ions.

-ve ion => to H
+ve ions => O

HIGHER CHARGE DENSITY, HIGHER enthalpy of hydration.

Eg: F has higher enthalpy of hydration than Cl.

M+ (g) + aq → M+ (aq)

M- (g) + aq → M- (aq)

1. Standard enthalpy change when
2. one mole of an ionic solid dissolves
3. in large amounts of water
4. so that ions are well separated and do not interact with each other

MP (s) + aq → M+ (aq) + P- (aq)

Cycle that includes all enthalpy changes to form an ionic compound

1. Lattice form
2. Standard states form
3. Atomisation of metal
4. Atomisation of non metal (or bond dissociation)
5. Ionisation of metal
6. Electron affinity of non metal
7. Back to lattice form by either lattice dissociation or formation

Clockwise arrows = Anti- clockwise arrows

Lattice formation ↓

Lattice dissociation ↑

Most exothermic enthalpy of formation will be most thermodynamically stable.

Theoretical lattice enthalpies assumes a perfect ionic model:

1. ions are 100% ionic
2. spherical
3. attractions are purely electrostatic

BUT not all are purely ionic, some can have Covalent character if:

1. the positive ion is small
2. the positive ion has multiple charges

3. the negative ion is large
4. the negative ion has multiple negative charges

This means that the negative ion is distorted → polarised. +ve ion is ∴ more polarising (since it polarised the -ve ion), If purely ionic then would b perfect spheres.

So lattice is stronger than
if it was 100% ionic → born
haber value would be larger than the theoretical value

If compound shows covalent character:

the theoretical and the
born Haber lattice enthalpies differ so:

the more the covalent character, the bigger the difference between the values

1. size of ion

bigger ion → weaker attractive force → less negative enthalpy

2. charge on ion

smaller charge → weaker attractive force between ions → less negative enthalpy

Proceeds on its
own without any external influence.

Substances with more ways of arranging their atoms and energy (more disordered) have a higher entropy

more disorder → more +ve entropy

1. elements : compounds

2. simpler compounds : complex compounds

3. pure substances : mixtures

1. Solid –> Liquid –> Gas (more disorder, more randomness, less regular arrangement)

2. Ionic solids dissolves in water (hydration) (less order)

3. Moles of gas increases (more gas molecules greater degree of randomness)

4. Temperature increases (more energy more disorder)

=0

no disorder, particles are solid/stationary

∴ cannot be -ve

∆S = Σ Sproducts – ΣSreactants

units: J K-1 mol-1

Gibbs free energy combines enthalpy and entropy and is used to predict feasibility of a reaction.

∆G = ∆H – T∆S

Units are J mol-1

(Sometimes small energy like spark for combustion may be applied but overall no energy is required to be given)

∆G will be negative → feasible

Exception: even if -ve ∆G, reaction may not occur if activation energy is too high

When calcuating ∆G, unit of ∆S is J and ∆H is in kJ, so divide ∆S value by 1000.

∆G = 0

∆G = ∆H – T∆S

∆H is negative (exothermic)

-T∆S is negative

Therefore ∆G is always negative → feasible

As physical phase changes like melting and boiling are equilibria, the ∆G for such changes is zero.

Increasing temp:

1. ∆S more +ve → ∆G will be more -ve → more feasible

2. ∆S is -ve → increasing temp would make it more +ve (since x by negative in equation) → ∆G more +ve → not feasible

3. ∆S is close to 0 → temp will not have a large effect on the
feasibility of the reaction → -T∆S small → ∆G won’t change much

∆G = ∆H – T∆S
y = mx + c

c = ∆H
m = -∆S

x axis = temp in K
y axis = ∆G kJ/mol

feasible when ∆G < 0 = crossing the x axis

graph would not be constant line since entropy changes as states change

Ionic lattice dissolving in water:

1. Ions breaking bonds
2. Forming new bonds with the water

= ∆H hydration + ∆H dissociation

= ∆H hydration – ∆H formation

= ∆H hydration and lattice enthalpy values are quite similar

BUT…

SOLUBLE

∴ salt dissolves at all temperatures because ∆S will be -ve due to more particles

INSOLUBLE

= lattice enthalpy is more stable than hydration enthalpy

∴ salt dissolves at higher temperatures because ∆S will +ve as salt → ions

Relate enthalpies of solution to ∆G to see if feasible.