Types of Enthalpy:
Physical changes:
• Enthalpy of sublimation
CO2 (s) → CO2 (g) ΔHsub
• Enthalpy of fusion
H2O (s) → H2O (l) ΔHfusion = 6.0 kJ mol-1
also called the enthalpy of melting
• Enthalpy of vapourisation
H2O (l) → H2O (g) ΔHvap = +44.0 kJ mol-1
From Hess’ Law: ΔHsub = ΔHfusion + ΔHvap
E.g.,
Calculate ΔHsub of
H2O (s) → H2O (g)
ΔHsub = ΔHfusion + ΔHvap
= 6.0 + 44.0
= 50.0 kJ mol-1
Chemical changes:
• Enthalpy of combustion
Oxidation of a substance to CO2 and H2O
E.g., C6H12O6 (s) + 6 O2 (g) → CO2 (g) + H2O (l)
ΔH = -2801 kJ mol-1
• Bond dissociation enthalpy
Enthalpy required to break a particular bond in
1 mole of gaseous molecules
AB (g) → A (g) + B (g)
ΔH (A-B)
(A and B may be atoms or groups of atoms)
B
E.g.,
H2O (g) → H (g) + OH (g)
ΔH (HO-H) = 492 kJ mol-1
OH (g) → H (g) + O (g)
ΔH (O-H) = 427 kJ mol-1
Van der Waals Forces are 0.1 to 10 kJ mol-1
A hydrogen bond has an energy of 10 – 40 kJ mol-1
Mean bond dissociation enthalpy:
Average value of the bond dissociation enthalpy
for a series of similar compounds
Also called the bond enthalpy.
Useful for estimating ΔH where data may not be
available.
E.g.,
E (O-H) = 463 kJ mol-1
(From H2O, CH3OH, H3COOH, etc.)
Some bond enthalpies in kJ mol-1
H
C
O
H
436
413
463
C
O
348
360
146
Spontaneous Processes:
A spontaneous process is a physical or chemical
change that occurs by itself
E.g.,
melting of ice above 273.15 K
expansion of gas into a vacuum
2 Na (s) + 2 H2O (g) → 2 NaOH (aq) + H2 (g)
ΔH = –367.5 kJ
Systems prefer to possess the lowest possible energy
For most spontaneous reactions ΔH is negative.
But what about this spontaneous reaction?
Ba(OH)2.8(H2O) (s) + 2 NH4NO3 (s)
→ 2 NH3 (g) + 10 H2O (l) + Ba(NO3)2 (aq)
ΔH = 170.4 kJ
Another quantity besides energy is required to
explain spontaneity.
Free energy change:
ΔG = ΔH – T ΔS
ΔG is the change in the Gibbs function or free
energy of the system
ΔS is the change in the entropy of the system
For the above reaction
Ba(OH)2.8(H2O) (s) + 2 NH4NO3 (s)
→ 2 NH3 (g) + 10 H2O (l) + Ba(NO3)2 (aq)
ΔH = 170.4 kJ and ΔS = 657 J K-1
So at 25°C → T = 298 K
ΔG = +170.4 kJ – (298 K × 0.657 kJ K-1)
= –25.8 kJ
The forward reaction is spontaneous
Entropy:
Entropy is a measure of the randomness or disorder
of a system
Depends on amount, temperature, pressure
Ssolid < Sliquid < Sgas
E.g.,
H2O (s):
H2O (l):
H2O (g):
S = 41 J K-1 mol-1
S = 63 J K-1 mol-1
S = 189 J K-1 mol-1
Entropy of reaction:
The change in entropy for a reaction at a given
temperature and pressure
ΔSr° = ΣnΔS° (products) – ΣnΔS° (reactants)
E.g.,
H2O (s) → H2O (l)
ΔH = +6.01 kJ mol-1
ΔS = 63 – 41 = 22 J K-1 mol-1
Free Energy (G):
The free energy is a thermodynamic quantity to
express reaction spontaneity directly:
G = H – T.S
G depends on the amount, temperature, and
pressure of a substance.
The free energy change:
ΔG = ΔH – T ΔS
ΔG < 0
forward reaction spontaneous
ΔG > 0
reverse reaction spontaneous
ΔG = 0
system at equilibrium
Example:
Consider the melting of ice at 25°C and 0°:
H2O (s) → H2O (l)
ΔH = +6.01 kJ
ΔS = 22 J K-1
25°C: T = 298 K
ΔG = +6.01 – (298 × 0.022) = –0.546 kJ mol-1
0°C: T = 273 K
ΔG = +6.01 – (273 × 0.022) = 0 kJ mol-1
As with enthalpy, there is a free energy of formation
of a substance.
ΔGf° = standard free energy of formation
Knowing the free energies of formation of the
reactants and products, the free energy of a reaction
can be calculated, in an analogous manner to the
enthalpy of reaction.
ΔGr° = ΣnΔGf° (products) – ΣnΔGf° (reactants)