Thermodynamics

1. Thermodynamics

Sunday, 05 April 2026
Thermodynamics
REVISION

2. Types of enthalpy change

Sunday, 05 April 2026
Types of enthalpy change
ΔfH – Enthalpy change of formation
The enthalpy change when 1 mole of a
compound is formed from it’s elements in
their standard states under standard
conditions.
2C(s) + 2H2(g) C2H6(g)
Notice these are
just the opposite
of each other!
ΔdissH – Enthalpy change of dissociation
The enthalpy change when 1 mole of bonds of the
same type of molecule in the gaseous state is broken.
F2(g) 2F(g)
Δea1H – 1st electron affinity
The enthalpy change when 1 mole of gaseous
1- ions are made from 1 mole of gaseous
atoms.
O(g) O-(g)
ΔlatticeH – Lattice enthalpy of formation
The enthalpy change when 1 mole of a solid ionic compound
is formed from it’s gaseous ions under standard conditions.
Ca2+(g) + 2Cl-(g) CaCl2(s)
H – Lattice enthalpy of dissociation
Δlattice
The enthalpy change when 1 mole of a solid
ionic compound is dissociated into it’s
gaseous ions under standard conditions.
CaCl2(s) Ca2+(g) + 2Cl-(g)
Δie1H – Enthalpy change of 1st ionisation
The enthalpy change when 1 mole of
gaseous 1+ ions are made from 1 mole
of gaseous atoms.
Na(g) Na+(g)
Δea2H – 2nd electron affinity
The enthalpy change when 1 mole of
gaseous 2- ions are made from 1 mole of
gaseous 1- ions.
O-(g) O2-(g)
ΔatH – Enthalpy change of
atomisation
The enthalpy change when 1 mole of
gaseous atoms is made from an
element in it’s standard state.
½F2(g) F(g)
Δie2H – Enthalpy change of 2nd
ionisation
The enthalpy change when 1 mole of
gaseous 2+ ions are made from 1
mole of gaseous 1+ ions.
Ca+(g) Ca2+(g)

3. Born-Haber Cycles

Sunday, 05 April 2026
Born-Haber Cycles
Born-Haber cycles are useful to calculate lattice enthalpies. This is because we
can’t calculate this directly from experiments. Here it is structured.
1st electron affinity of
Chlorine Δea1H (exothermic)
1st Ionisation energy of
Lithium Δie1H
(endothermic)
Enthalpy of Atomisation
of Lithium ΔatH
(endothermic)
Enthalpy of
Formation ΔfH
(exothermic)
Enthalpy of Atomisation
of Chlorine ΔatH
(endothermic)
Lattice Enthalpy of Formation of
Lithium Chloride ΔlatticeH (exothermic)
Li+(g) + Cl(g) + eLi(g) + Cl(g)
Li(s) + Cl(g)
Li+(g) + Cl-(g)
Li(s) + ½Cl2(g)
LiCl(s)

4. Born-Haber Cycles

Sunday, 05 April 2026
Born-Haber Cycles
We calculate lattice enthalpies by using the cycle in the same way as Hess’ cycle. Go with
the arrow, keep the sign the same. Go against the arrow, change the sign.
Enthalpy Change
Enthalpy
(kJmol-1)
Enthalpy of formation of LiCl
-409
1st Ionisation of Li
519
Enthalpy of Atomisation of Li
161
Enthalpy of Atomisation of Cl2
242
1st Electron Affinity of Cl
-364
ΔlatticeH = +364-519-161-121 -409
ΔlatticeH = -846kJmol-1
CHECK! Start anywhere on the cycle –
use the arrow rules above. You should
get an answer of ZERO
Li+(g) + Cl(g) + eΔie1H = 519
ΔatH (Li) = 161
ΔatH = (Cl) 242/2 =
121
ΔfH = -409
Li(g) + Cl(g)
Li(s) + Cl(g)
Li(s) + ½Cl2(g)
Δea1H = -364
Li+(g) + Cl-(g)
Start
here
TO CALCULATE
Lattice Enthalpy of
Formation of Lithium
Chloride ΔlatticeH
LiCl(s)
End
here

5. Born-Haber Cycles

Sunday, 05 April 2026
Born-Haber Cycles
We can calculate other parts of the cycle. Remember - go with the arrow, keep the sign the
same. Go against the arrow, change the sign.
Enthalpy Change
Enthalpy
(kJmol-1)
Enthalpy of formation of NaCl
-411
1st Ionisation of Na
496
Lattice Enthalpy of Formation of
NaCl
-787
Enthalpy of Atomisation of Cl2
242
1st Electron Affinity of Cl
-364
ΔatH (Na) = -121 -411+787+364-496
ΔatH (Na) = +123kJmol-1
CHECK! Start anywhere on the cycle –
use the arrow rules above. You should
get an answer of ZERO
Na+(g) + Cl(g) + eΔie1H = 496
TO CALCULATE
ΔatH (Na)
ΔatH = (Cl) 242/2 =
121
ΔfH = -411
Na(g) + Cl(g)
End
here
Na(s) Start
+ Cl(g)
Δea1H = -364
Na+(g) + Cl-(g)
here
Na(s) + ½Cl2(g)
ΔlatticeH = -787
NaCl(s)

6. Born-Haber Cycles

Sunday, 05 April 2026
Born-Haber Cycles
Here is a an extended Born-Haber cycle for ionic compounds made up of 2
double ions – e.g. Mg2+ and O2Notice the 2nd ionisation step. This is
needed to create the Mg2+ ion. This
process is ENDOTHERMIC.
Notice the 2nd electron affinity step.
This is needed to create the O2- ion.
This process is ENDOTHERMIC.
This is because energy is needed to
add an electron to a negative ion.
Repulsive forces between 2 negative
species.
Mg2+(g) + O(g) + 2eMg2+(g) + O2-(g)
Mg+(g) + O(g) + e2+ + O- + eMg
(g)
(g)
Mg + O
(g)
(g)
Mg(s) + O(g)
Mg(s) + ½ O2(g)
MgO(s)

7. Enthalpy Change of Solution

Sunday, 05 April 2026
Enthalpy Change of Solution
ΔsolutionH – the enthalpy change when 1 mole of an ionic substance is dissolved in the
minimum amount of solvent to ensure no further enthalpy change is observed upon
further dilution.
Substance bonds broken to
create free moving ions.
Ionic lattice in solid form
+
- +
+ - +
+
-
+ -- ++
+
- + - -+
+
+ - +
+
δ+
H
δ+
H
δO
Most ionic compounds dissolve in
polar solvents like H2O. The δ+ H is
attracted to negative ions and δoxygen is attracted to positive ions.
The structure starts to break down.
Bonds formed between ions and
water. The ions are hydrated.
The water molecules
surround the ions in a
process called hydration.
For this to happen the new bonds formed must
be the same strength or greater than those
broken.
If not then the substance is very unlikely to
dissolve. Soluble substances tend to have
exothermic enthalpies of solution for this reason.

8. Calculating Enthalpy Change of Solution

Sunday, 05 April 2026
Calculating Enthalpy Change of Solution
Enthalpy change of solution can be calculated from knowing – 1. Lattice
dissociation enthalpy & 2. Enthalpy of hydration
+ -+-++
+- +- +-- -++-+++
- +- +
LiCl(s)
ΔsolutionH
ΔdissH
+846 kJmol-1
Li+(g) + Cl-(g)
ΔsolutionH = +846 -883
ΔsolutionH = -37 kJmol-1
Li+(aq) + Cl-(aq)
ΔhydH
Li+ = -519 kJmol-1
Cl- = -364 kJmol-1
Total = -883 kJmol-1
Enthalpy of hydration (ΔhydH)
– Enthalpy change when 1
mole of aqueous ions is made
from 1 mole of gaseous ions.
We can use a cycle to help work this
out. Remember how it is set out!
Assume that we do the following –
1. Break the solid lattice up into it’s
gaseous ions first (lattice dissociation)
2. Dissolve the gaseous ions in water
(enthalpy of hydration)
Remember – go with the arrow keep
the sign the same. Go against the
arrow you change the sign. (You will
be given data to calculate this)

9. Entropy

Sunday, 05 April 2026
Entropy
Entropy is the measure of disorder in a system
Entropy (S) is the number of ways
energy can be shared out between
particles.
SOLID
LIQUID
The more disorder there is the
higher the level of entropy.
GAS
Solids have the lowest level of
disorder, particles are arranged
neatly in rows. Liquids and then
Gases are more disordered.
The number of particles also affects
entropy change. If a reaction is in the
same state but more moles are produced
then entropy increases. There are more
ways energy can be distributed.
INCREASING DISORDER AND ENTROPY
Example
Dinitrogen tetroxide producing nitrogen
dioxide shows increasing entropy. (more
moles of gas produced)
N2O4(g) 2NO2(g)

10. Entropy

Sunday, 05 April 2026
Entropy
A reaction can be spontaneous (feasible) even if it is enthalpically
unfavourable (i.e. – endothermic)
A reaction will tend towards more
disorder and hence increase entropy.
Increasing entropy is energetically favourable and some reactions that are
enthalpically unfavourable (endothermic) can still spontaneously react if
changes in entropy overcome changes in enthalpy.
Example of a spontaneous endothermic reaction
The reaction between hydrated Barium Hydroxide and
Ammonium Chloride shows how entropy changes overcome
enthalpy changes.
The 2 solids are mixed and a gas, liquid and solid is
produced –
Summary points
1. This reaction is very endothermic and
enthalpically not favourable. (+164 kJmol-1)
2. 3 moles on the left hand side and 13 moles on
the right hand side. Entropically favourable.
Ba(OH)2.8H2O(s) + 2NH4Cl(s) → 2NH3(g) + 10H2O(l) + BaCl2(s)
This reaction can
show temperature
drops of 20°C!
3. Starting with 2 solids but making a gas and
liquid. Increased disorder so entropically
favourable.

11. Entropy

Sunday, 05 April 2026
Entropy
Entropy change (ΔS) can be calculated between reactants and products
ΔS = Sproducts - Sreactants
Example calculation
The reaction between hydrated Barium Hydroxide and Ammonium
Chloride calculate the change in entropy.
Ba(OH)2.8H2O(s) + 2NH4Cl(s) → 2NH3(g) + 10H2O(l) + BaCl2(s)
Products
(2×S⊖[NH3(g)]) + (10×S⊖[H2O(l)]) + S⊖[BaCl2(s)]
(2×192) + (10×70) + 124 = 1208 JK-1mol-1
ΔS = 1208 – 617
ΔS = +591 JK-1mol-1
Units of Entropy (S) = JK-1mol-1
Entropy values are given as standard entropy S⊖
1 mole of substance
100kPa
298K
Reactants
S⊖[Ba(OH)2.8H2O(s)] + (2×S⊖[NH4Cl(s)])
427 + (2×95) = 617 JK-1mol-1
Entropy is positive so
this reaction is
entropically feasible
Compound
S⊖ / JK-1mol-1
Ba(OH)2.8H2O(s)
427
NH4Cl(s)
95
NH3(g)
192
H2O(l)
70
BaCl2(s)
124

12. Gibbs Free Energy

Sunday, 05 April 2026
Gibbs Free Energy
Gibbs Free Energy (ΔG) tells us if a reaction is feasible or not.
ΔG = ΔH - TΔS
Units of change in free
energy (ΔG) = Jmol-1
Enthalpy
Temperature Entropy change
-1
change (Jmol )
(K)
(JK-1mol-1)
The basic rule
A reaction is feasible in theory if ΔG is
NEGATIVE or ZERO.
CAREFUL!
Even if the reaction is calculated to be feasible you may not observe a reaction
occurring. This is due to the activation energy being too high or the rate of
reaction being very slow.
Calculating the free energy (ΔG) for the reaction between barium hydroxide and ammonium chloride at 298K
We know the following from the previous slides –
ΔS = +591 JK-1mol-1
ΔH = +164 kJmol-1 / 164,000 Jmol-1
ΔG is negative so this
reaction is feasible at
this temperature.
ΔG = ΔH – TΔS
ΔG = 164,000 – (298×591)
ΔG = -12,118 Jmol-1

13. Gibbs Free Energy

Sunday, 05 April 2026
REMEMBER - The basic rule
A reaction is feasible in theory if
ΔG is NEGATIVE or ZERO.
Gibbs Free Energy
Temperature may have an effect on reaction feasibility
Let’s look at different scenarios when ΔH
and ΔS is changed and the ΔG value
ΔH value
ΔS value
ΔG value
Feasible?
Negative
Positive
Always Negative
Yes, at any temp
ΔG = ΔH - TΔS
Positive
Negative
Always Positive
Never, at any
temp
Negative
Negative
Negative at
lower temps
Yes, at lower
temps
Positive
Positive
Negative at
higher temps
Yes, at higher
temps
It makes it easier to see this equation in 2
halves –
The ‘ΔH’ part
The ‘TΔS’ part
Look at the first row –
Even if T was at it’s lowest (Zero Kelvin) ΔG would always be
negative. So exothermic reactions with a positive entropy
would be always feasible whatever the temperature.
Look at the second row –
Even if T was at it’s lowest (Zero Kelvin) ΔG would
always by positive. So endothermic reactions with a
negative entropy would never be feasible whatever the
temperature.

14. Gibbs Free Energy

Sunday, 05 April 2026
REMEMBER - The basic rule
A reaction is feasible in theory if
ΔG is NEGATIVE or ZERO.
Gibbs Free Energy
Temperature may have an effect on reaction feasibility
Let’s look at different scenarios when ΔH
and ΔS is changed and the ΔG value
ΔH value
ΔS value
ΔG value
Feasible?
Negative
Positive
Always Negative
Yes, at any temp
ΔG = ΔH - TΔS
Positive
Negative
Always Positive
Never, at any
temp
Negative
Negative
Negative at
lower temps
Yes, at lower
temps
Positive
Positive
Negative at
higher temps
Yes, at higher
temps
Look at the third row –
Freezing water is an exothermic process (ΔH negative)
but we are going to a more ordered arrangement –
liquid to solid (ΔS negative). Freezing only happens
below a certain temperature.
Source –All pictures - Pixabay
Let’s put some numbers to it as an example –
ΔH = -100,000 Jmol-1 ΔS= -100 Jmol-1
@ 500K
@ 1500K
ΔG = -100,000 – (500×-100)
ΔG = -100,000 – (1500×-100)
ΔG = -50,000 Jmol-1
ΔG = +50,000 Jmol-1
FEASIBLE
NOT FEASIBLE

15. Gibbs Free Energy

Sunday, 05 April 2026
REMEMBER - The basic rule
A reaction is feasible in theory if
ΔG is NEGATIVE or ZERO.
Gibbs Free Energy
Temperature may have an effect on reaction feasibility
Let’s look at different scenarios when ΔH
and ΔS is changed and the ΔG value
ΔH value
ΔS value
ΔG value
Feasible?
Negative
Positive
Always Negative
Yes, at any temp
ΔG = ΔH - TΔS
Positive
Negative
Always Positive
Never, at any
temp
Negative
Negative
Negative at
lower temps
Yes, at lower
temps
Positive
Positive
Negative at
higher temps
Yes, at higher
temps
This reaction is
used to make
cinder toffee!
Look at the fourth row –
Decomposition of sodium hydrogen carbonate is an
endothermic process (ΔH positive) but we are going to a
more random arrangement – solid to gas (ΔS positive).
This only happens above a certain temperature.
Source –All pictures – Pixabay and own images
Let’s put some numbers to it as an example –
ΔH = +100,000 Jmol-1 ΔS= +100 Jmol-1
@ 500K
@ 1500K
ΔG = 100,000 – (500×100)
ΔG = 100,000 – (1500×100)
ΔG = +50,000 Jmol-1
ΔG = -50,000 Jmol-1
NOT FEASIBLE
FEASIBLE

16. Gibbs Free Energy

Sunday, 05 April 2026
Gibbs Free Energy
The temperature at which a reaction becomes just feasible can be calculated
A reaction is just feasible when ΔG = O so we can rearrange
the equation to work out the temperature when ΔG = O
Calculate the minimum temperature required for sodium
hydrogen carbonate to decompose.
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) ΔH = 129 kJmol-1
Substance
S⊖ / JK-1mol-1
NaHCO3(s)
102
Na2CO3(s)
136
H2O(g)
189
CO2(g)
214
T = ΔH / ΔS
Notice ΔG has been
Remember
convert ΔH
from kJmol-1
to Jmol-1
Calculate ΔS
ΔS = Sproducts – Sreactants
Sproducts = 136 + 189 + 214 = 539 JK-1mol-1
Sreactants = 2×102 = 204 JK-1mol-1
ΔS = 539 – 204 = 335 JK-1mol-1
omitted. That’s
because it equals ZERO
Assuming ΔG = 0
T = ΔH / ΔS
T = 129,000 / 335
Minimum Temperature
385K