12 Thermodynamics
12.1 Types of Enthalpy Change
12.2 Born-Haber Cycles
12.3 Enthalpy Changes – Enthalpy of Solution
12.4 Mean Bond Enthalpy
12.5 Entropy
12.1 Enthalpy Change – Ionic Compounds
Learning Objectives:
1. Describe what is meant by the term enthalpy change.
2. Describe the different types of enthalpy changes (formation,
atomisation, ionisation energy, electron affinity, lattice formation,
hydration, solution, bond enthalpy).
3. Calculate the enthalpy changes on forming ionic compounds.
Enthalpy Review
• Enthalpy change is the heat change at constant pressure.
• Standard conditions: 100kPa, 298 K (starting temperature)
• Remember that heat and temperature are not the same.
• Heat is a type of energy and is measured in joules and heat
changes lead to temperature changes, which is measure in
Kelvins.
Types of Enthalpy Changes
• Enthalpy of Formation
• Enthalpy of Atomisation
• First Ionisation Energy/Second Ionisation Energy
• First Electron Affinity/Second Electron Affinity
• Lattice Enthalpy of Formation
• Enthalpy of Lattice Dissociation
• Enthalpy of Hydration
Write down the symbol
• Enthalpy of Solution
and the definition
• Mean Bond Enthalpy
Standard Enthalpy of Formation
change
∆HꝊf
standard conditions
formation
• Enthalpy change when
• one mole of a compound is formed
• from its constituent elements
• under standard conditions
• all reactants and products in their standard
states.
Standard Enthalpy of Atomisation
∆HꝊat
• Enthalpy change when
• one mole of gaseous atoms
• is formed
• from the element
• In it’s standard state
• under standard conditions
First Ionisation Energy
First IE
• Enthalpy change when
• one mole of gaseous atoms
• is converted into
• one mole of gaseous +1 ions
• under standard conditions
Second Ionisation Energy
Second IE
• Enthalpy change when
• one mole of gaseous +1 ions
• is converted into
• one mole of gaseous +2 ions
• under standard conditions
First Electron Affinity
First ∆HꝊea
• Enthalpy change when
• one mole of gaseous atoms
• is converted into
• one mole of gaseous -1 ions
• under standard conditions
Second Electron Affinity
Second ∆HꝊea
• Enthalpy change when
• one mole of gaseous -1 ions
• is converted into
• one mole of gaseous -2 ions
• under standard conditions
Lattice Formation Enthalpy
∆HꝊL
• Enthalpy change when
• one mole of solid ionic compound
• is formed
• from it’s gaseous ions
• under standard conditions
• (always negative, energy released)
Enthalpy of Lattice Dissociation
-∆HꝊL
• Enthalpy change when
• one mole of solid ionic compound
• dissociates into
• it’s gaseous ions
• under standard conditions
• (always positive, energy is absorbed)
Standard Enthalpy of Hydration
∆HꝊhyd
• Enthalpy change when
• one mole of gaseous atoms
• is surrounded by water molecules
• under standard conditions
Standard Enthalpy of Solution
∆HꝊsol
• Enthalpy change when
• one mole of solute
• completely dissolves
• in sufficient solvent to form a solution in which the
molecules are ions do not interact
• under standard conditions
Mean Bond Enthalpy
∆HꝊdiss
• Enthalpy change when
• one mole of gaseous molecules
• breaks a covalent bond
• forming two free radicals
• averaged over a range of compounds
• at standard conditions
For each type…
a) Write an equation to represent the chemical reaction being
described
b) Tell me if the process is likely to be positive or negative
c) Explain why.
Standard Enthalpy of Formation
H2 (g) +
1
O2 (g)
2
∆HꝊf
 H2O (l)
Usually going to be negative.
Molecules usually form because the molecule is more stable
(lower in energy) than the constituent elements.
Remember: Bond making releases energy.
Standard Enthalpy of Atomisation
1
Br2(l) 
2
∆HꝊat
Br (g)
Usually positive.
Molecules form because that is a more stable form, so
gaseous atoms are less stable (higher in energy).
Remember: Bond breaking require energy.
First Ionisation Energy
First IE
Na (g)  Na+(g) + e-
• Positive
• Removing an electron takes energy
Second Ionisation Energy
Second IE
Na+ (g)  Na2+(g) + e-
• Very positive
• Removing electron from positive ion require a lot of
energy.
First Electron Affinity
First ∆HꝊea
O (g) + e-  O- (g)
• Usually Negative
• Energy is gained when electrons are added.
Second Electron Affinity
Second ∆HꝊea
O- (g) + e-  O2- (g)
• Usually Positive
• Because of repulsion, adding the second electron
requires more energy than is gained.
Lattice Formation Enthalpy
∆HꝊL
Na+ (g) + Cl- (g)  NaCl (s)
• Always negative
• Bond making releases energy, more stable in lattice
form.
Enthalpy of Lattice Dissociation
-∆HꝊL
NaCl (s)  Na+ (g) + Cl- (g)
• Always positive
• This is opposite of lattice formation, breaking bonds
requires energy.
Standard Enthalpy of Hydration
∆HꝊhyd
Na+ (g) + aq  Na+ (aq)
Cl- (g) + aq  Cl- (aq)
• Usually negative
• Water molecules stabilise the charges of the ions.
Standard Enthalpy of Solution
∆HꝊsol
NaCl (s) + aq  Na+ (aq) + Cl- (aq)
• Usually slightly positive
• Breaking the bonds of the lattice requires energy,
however, the water molecules stabilise the ions so
overall only small amount of energy absorbed.
Mean Bond Enthalpy
∆HꝊdiss
CH4 (g)  C (g) + 4H (g)
• Always positive
• Bond breaking requires energy.
12.2 Born-Haber Cycles
Learning Objectives:
1. Describe Hess’ Law.
2. Use Born-Haber Cycles to calculate enthalpy changes
Hess’s Law of Thermodynamics
• The enthalpy change for a reaction is the same, no matter what route
is taken.
• For example:
CH4 (g) + O2 (g)  CO2 (g) + H2O (g)
C (s) + H2 (g) + O2 (g)
Born-Haber Cycles
• Born-Haber Cycles are just another method to solve for the unknown
enthalpy change of a chemical reaction by using enthalpy changes
that we DO know.
• It uses a diagram to represent the enthalpy changes on a vertical
scale. Increases in energy are UP ( ) arrows, decreases in energy are
DOWN ( )arrows.
• Molly started out with £0. Then she received £100 for her birthday.
She went out to dinner, this costed £30. Then she bought some new
shoes. At the end of the day to had spent all of her birthday money.
How much did her new shoes cost?
With Birthday Money
∆Mdin = -£30
After Dinner
∆Mbd = +£100
∆Mshu = ? = -£70
Broke
Formation of an Ionic Compound
• Electrons are transferred to atoms to form ions.
• Ions then attract and are arranged into an ionic lattice.
• This is how ionic lattices are formed.
Enthalpy Change in Formation of Ionic
Compounds
1
2
Na (s) + Cl2 (g)  NaCl (s)
∆HꝊf = ?
What are the steps for this complete reaction to occur?
1) Atomisation of Na
2) Atomisation of Chlorine
3) Ionisation of Na
4) Electron affinity of Cl
5) Formation of lattice
Enthalpy Change in Formation of Ionic Compounds
1
2
Na (s) + Cl2 (g)  NaCl (s)
∆HꝊf = ?
What are the steps for this complete reaction to occur?
1) Atomisation of Na
Na (s)  Na (g)
2)
3)
4)
5)
Atomisation of Chlorine
Ionisation of Na
Electron affinity of Cl
Formation of lattice
1
2
Cl2 (g)  Cl (g)
Na (g)  Na+ (g)
Cl (g)  Cl- (g)
Na+ (g) + Cl- (g)  NaCl (s)
Enthalpy Change in Formation of Ionic Compounds
1
2
Na (s) + Cl2 (g)  NaCl (s)
∆HꝊf = ?
What are the steps for this complete reaction to occur?
1) Atomisation of Na
Na (s)  Na (g)
∆HꝊat = +108 kJ/mol
2)
3)
4)
5)
Atomisation of Chlorine
Ionisation of Na
Electron affinity of Cl
Formation of lattice
1
2
Cl2 (g)  Cl (g)
Na (g)  Na+ (g)
Cl (g)  Cl- (g)
Na+ (g) + Cl- (g)  NaCl (s)
∆HꝊat = +122 kJ/mol
first IE = +496 kJ/mol
first EA = -349 kJ/mol
∆HꝊL = -788 kJ/mol
Born-Haber Cycle Formation of NaCl
Na+ (g) + Cl (g)
First EA = -349 kJ/mol
Na+ (g) + Cl- (g)
First IE = +496 kJ/mol
Na (g) + Cl (g)
1
2
1
(s) + 2 Cl2 (g)
∆HꝊat = +122 kJ/mol
Na (g) + Cl2 (g)
∆HꝊat = +108 kJ/mol
Na
∆HꝊL = -788 kJ/mol
∆HꝊf = ?
NaCl (s)
Born-Haber Cycle Formation of NaCl
Na+ (g) + Cl (g)
First EA = -349 kJ/mol
Na+ (g) + Cl- (g)
First IE = +496 kJ/mol
Na (g) + Cl (g)
1
2
1
(s) + 2 Cl2 (g)
∆HꝊat = +122 kJ/mol
Na (g) + Cl2 (g)
∆HꝊat = +108 kJ/mol
Na
∆HꝊL = -788 kJ/mol
∆HꝊf = -411 kJ/mol
NaCl (s)
Example: Lattice Formation Enthalpy of MgCl2
• Write out the overall equation for the formation
of magnesium chloride.
• Write equations for all of the steps in the
formation of magnesium chloride.
• HINT: there are six steps
Draw a Born-Haber Diagram for MgCl2
• HINT: there is a “trick” step, can you catch it? Remember your
definitions
• ∆HꝊat Mg = +148 kJ/mol
• ∆HꝊat Cl = +122 kJ/mol
• First IE Mg= +738 kJ/mol
• Second IE Mg = +1451 kJ/mol
• First EA Cl = -349 kJ/mol
• ∆HꝊf MgCl2 = -641 kJ/mol
Use your Born-Haber
Diagram to Calculate the
Lattice Formation Enthalpy
Mg2+ (g) + 2Cl (g)
Mg2+ (g) + 2Cl- (g)
Second IE = +1451 kJ/mol
Mg+ (g) + 2Cl (g)
First IE = +738 kJ/mol
Mg (g) + 2Cl (g)
2 x ∆HꝊat = +122 kJ/mol
x2
= +244 kJ/mol
∆HꝊat = +148 kJ/mol
∆HꝊL = -2524 kJ/mol
Mg (g) + Cl2 (g)
Mg (s) + Cl2 (g)
∆HꝊf = ?
MgCl2 (s)
2 x First EA = -349 kJ/mol x 2
= -698 kJ/mol
Mg2+ (g) + 2Cl (g)
Mg2+ (g) + 2Cl- (g)
Second IE = +1451 kJ/mol
Mg+ (g) + 2Cl (g)
First IE = +738 kJ/mol
Mg (g) + 2Cl (g)
2 x ∆HꝊat = +122 kJ/mol
x2
= +244 kJ/mol
∆HꝊat = +148 kJ/mol
∆HꝊL = -2524 kJ/mol
Mg (g) + Cl2 (g)
Mg (s) + Cl2 (g)
∆HꝊf = -641 kJ/mol
MgCl2 (s)
2 x First EA = -349 kJ/mol x 2
= -698 kJ/mol
12.3 More Enthalpy Changes
Learning Objectives:
1. Calculate enthalpy change of solution.
2. Describe how lattice enthalpy calculations support models for ionic
bonding.
3. Explain how ions can become polarised.
Enthalpy of Solution
• Ionic solids can dissolve in polar solvents.
• This is called hydration if the solvent is water.
• Hydration is when the water molecules surround ions.
• What are the steps for process of forming a solution?
1. Breaking the ionic lattice (enthalpy of lattice dissociation).
2. Hydrating the positive ions (enthalpy of hydration).
3. Hydrating the negative ions (enthalpy of hydration).
Example: NaCl
Ionic Bonding Models
• For most ionic compounds the theoretical values calculated from
Born-Haber cycles agrees with experimental values.
• This proves that the model for ionic bonding (lattice) is correct.
• However, some ionic compounds have theoretical and experimental
values that DO NOT agree.
• Another model needed to be found to explain these discrepancies.
Polarisation
• ZnSe
• experimental lattice formation enthalpy = -3611 kJ/mol
• theoretical lattice formation enthalpy = -3305 kJ/mol
WHY?
• Zn2+ is very small and has a high + charge
• Se2- is very large and has a high – charge
• Zn2+ moves closely to electron density of Se2- and attracts the e• Since Se2- is large, the e- are far from the nucleus and easily pulled
away
• This distorts the electron cloud surrounding Se2-
Polarisation
• The distortion causes their to be some electron
density shared between the two ions (slightly
covalent nature).
• The Se2- ion is said to be polarised.
• This causes the enthalpy change to be greater
than expected.
When does polarisation happen?
•Cation = small size,
high charge
•Anion = large size,
high charge
12.4 Mean Bond Enthalpy
Learning Objectives:
1. Explain the term mean bond enthalpy.
2. Calculate enthalpy changes using mean bond enthalpy.
3. Explain why this method is not as accurate.
Mean Bond Enthalpy
• The average bond enthalpy term is the average amount of energy
needed to break a specific covalent bond, measured over a wide
variety of different molecules.
• A measure of strength of a covalent bond.
• In comparison, lattice enthalpy is a measure of the strength of an
ionic bond.
Predicting reactions
• Mean bond enthalpies can be used to predict how molecules may react.
• We can predict which bonds may be more likely to break.
Which bond is most likely to break?
C-H
C-C
C-Br
413 kJ/mol
437 kJ/mol
290 kJ/mol
Predicting reactions
• Mean bond enthalpies can also be used to compare reactivities of
different molecules.
• Which haloalkane is more reactive?
C-F
C-Cl
C-Br
C-I
467 kJ/mol
346 kJ/mol
290 kJ/mol
228 kJ/mol
Calculating Approximate Enthalpy Changes
• Hess’s Law can be applied.
• One possible route to products would be to break all bonds in the
reactants and then form all of the bonds for the products.
• The enthalpies for these two processes can then be summed up to
find the total enthalpy change.
• Remember:
bond breaking requires energy (+ value)
bond formation releases energy (- value)
12.5 Why do chemical reactions take place?
Learning Objectives:
1. Explain the concept of entropy.
2. Calculate using enthalpy and entropy whether a reaction will
spontaneously occur.
3. Analyse the effects of temperature on feasibility of a reaction.
Is a reaction feasible or spontaneous?
• Reactions that will take place on their own are called spontaneous.
• If it is possible for a reaction to take place on their own, the reaction
is feasible.
• What determines if a reaction is feasible?
• If ΔH (enthalpy) is negative, the reaction is exothermic
• If ΔS (entropy) is positive, the reaction increases in randomness
Entropy
• Entropy is a mathematical measure of the randomness of a system.
• Change in entropy is represented as ΔS.
• The universe prefers randomness (higher entropy) and is always
moving towards disorder.
• Values for entropy of different substances are determined
mathematically, you will not be expected to calculate these, only how
to use them. (see pg. 179)
Calculating Entropy Changes
• Calculate the difference in entropy from reactants to products to find
the ΔS of a reaction.
ΔS = Sproducts – Sreactants
• If ΔS is positive, entropy is increasing, disorder is increasing. The
products are more disordered than the reactants.
• If ΔS is negative, entropy is decreasing, disorder is decreasing. The
products are less disordered than the reactants.
Gibbs Free Energy
• ΔG represents the Gibbs free energy and combines both enthalpy and
entropy.
• It is used to determine whether or not a reaction is feasible.
ΔG = ΔH – TΔS
• If ΔG is negative (-) the reaction is feasible.
• If ΔG is positive (+) the reaction is NOT feasible.
What happens if ΔG = 0?
• There will be a temperature where ΔG = 0.
• This is the temperature at which the reaction is just feasible.
• In a closed system an equilibrium between products and reactants
occur.
• ΔG = 0 can also be used to calculate ΔS.
• Cases where both forms are equally likely (ie melting point), ΔG = 0.
Thermodynamics does not predict the rate of
a reaction
• Thermodynamics = Kinetics
• Thermodynamics only predicts whether a reaction is feasible.
• It DOES NOT predict how quickly the reaction may take place.
• Kinetics is the branch of chemistry dealing with rate of reaction.