12 Thermodynamics

12.1 Types of Enthalpy Change 12.2 Born-Haber Cycles12.3 Enthalpy Changes – Enthalpy of Solution12.4 Mean Bond Enthalpy12.5 Entropy

12.1 Enthalpy Change – Ionic CompoundsLearning 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• Enthalpy of Solution• Mean Bond Enthalpy

Write down the symbol and the definition

Standard Enthalpy of Formation

∆HꝊf

•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.

change standard conditions

formation

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 negativec) Explain why.

Standard Enthalpy of Formation

∆HꝊf

H2 (g) + O2 (g) 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

∆HꝊat

Br2(l) 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 IENa (g) Na+

(g) + e-

•Positive•Removing an electron takes energy

Second Ionisation Energy

Second IENa+ (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

After Dinner

Broke

∆Mbd = +£100

∆Mdin = -£30

∆Mshu = ? = -£70

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 CompoundsNa (s) + Cl2 (g) NaCl (s) ∆HꝊ

f = ?

What are the steps for this complete reaction to occur?1) Atomisation of Na2) Atomisation of Chlorine3) Ionisation of Na4) Electron affinity of Cl5) Formation of lattice

Enthalpy Change in Formation of Ionic CompoundsNa (s) + Cl2 (g) NaCl (s) ∆HꝊ

f = ?

What are the steps for this complete reaction to occur?

1) Atomisation of Na Na (s) Na (g)

2) Atomisation of Chlorine Cl2 (g) Cl (g)

3) Ionisation of Na Na (g) Na+ (g)

4) Electron affinity of Cl Cl (g) Cl- (g)

5) Formation of lattice Na+ (g) + Cl- (g) NaCl (s)

Enthalpy Change in Formation of Ionic CompoundsNa (s) + Cl2 (g) NaCl (s) ∆HꝊ

f = ?

What are the steps for this complete reaction to occur?

1) Atomisation of Na Na (s) Na (g) ∆HꝊat = +108 kJ/mol

2) Atomisation of Chlorine Cl2 (g) Cl (g) ∆HꝊat = +122 kJ/mol

3) Ionisation of Na Na (g) Na+ (g) first IE = +496 kJ/mol

4) Electron affinity of Cl Cl (g) Cl- (g) first EA = -349 kJ/mol

5) Formation of lattice Na+ (g) + Cl- (g) NaCl (s) ∆HꝊL = -788 kJ/mol

Born-Haber Cycle Formation of NaCl

Na (s) + Cl2 (g)

Na (g) + Cl2 (g)

Na (g) + Cl (g)

Na+ (g) + Cl (g)

Na+ (g) + Cl- (g)

NaCl (s)

∆HꝊat = +108 kJ/mol

∆HꝊat = +122 kJ/mol

First IE = +496 kJ/molFirst EA = -349 kJ/mol

∆HꝊL = -788 kJ/mol

∆HꝊf = ?

Born-Haber Cycle Formation of NaCl

Na (s) + Cl2 (g)

Na (g) + Cl2 (g)

Na (g) + Cl (g)

Na+ (g) + Cl (g)

Na+ (g) + Cl- (g)

NaCl (s)

∆HꝊat = +108 kJ/mol

∆HꝊat = +122 kJ/mol

First IE = +496 kJ/molFirst EA = -349 kJ/mol

∆HꝊL = -788 kJ/mol

∆HꝊf = -411 kJ/mol

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• ∆HꝊ

f MgCl2 = -641 kJ/mol

Use your Born-Haber Diagram to Calculate the

Lattice Formation Enthalpy

Mg (s) + Cl2 (g)

Mg (g) + Cl2 (g)

Mg (g) + 2Cl (g)

Mg2+ (g) + 2Cl (g)

Mg2+ (g) + 2Cl- (g)

MgCl2 (s)

∆HꝊat = +148 kJ/mol

2 x ∆HꝊat = +122 kJ/mol

x 2= +244 kJ/mol

First IE = +738 kJ/mol

2 x First EA = -349 kJ/mol x 2 = -698 kJ/mol

∆HꝊL = -2524 kJ/mol

∆HꝊf = ?

Second IE = +1451 kJ/mol

Mg+ (g) + 2Cl (g)

Mg (s) + Cl2 (g)

Mg (g) + Cl2 (g)

Mg (g) + 2Cl (g)

Mg2+ (g) + 2Cl (g)

Mg2+ (g) + 2Cl- (g)

MgCl2 (s)

∆HꝊat = +148 kJ/mol

2 x ∆HꝊat = +122 kJ/mol

x 2= +244 kJ/mol

First IE = +738 kJ/mol

2 x First EA = -349 kJ/mol x 2 = -698 kJ/mol

∆HꝊL = -2524 kJ/mol

∆HꝊf = -641 kJ/mol

Second IE = +1451 kJ/mol

Mg+ (g) + 2Cl (g)

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 413 kJ/molC-C 437 kJ/molC-Br 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 467 kJ/molC-Cl 346 kJ/molC-Br 290 kJ/molC-I 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.