Energetics

Ms. Peace

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Lesson 1

Endothermic and Exothermic

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Lesson 1: Endothermic and Exothermic

▶ Objectives:

▶ Reflect on prior knowledge of the energy changes in chemical reactions

▶ Understand the difference between endothermic and exothermic reactions

▶ Draw enthalpy level diagrams showing the relative stabilities of products and reactants

▶ Complete an experiment investigating endothermic and exothermic reactions

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Heat and Temperature

▶ Heat is a form of energy that flows from something at a higher temperature to something at a lower temperature

▶ Temperature is a measure of average kinetic energy of particles

▶ Internal energy is the name given to the total amount of energy in a substance. Total energy is conserved in a chemical reaction

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Exothermic vs endothermic

▶ Exothermic reactions release energy▶ Usually they get hotter (without an external energy supply)

▶ However they can give out other forms of energy such as light or electricity instead

▶ Generally they are self-sustaining, stopping only when the reactants run out

▶ Endothermic reactions absorb energy▶ Usually they get cooler, absorbing heat from the surroundings

▶ However they can absorb other forms of energy such as light or electricity instead

▶ Generally they require an external energy source to keep them going and will stop either when the energy supply is halted or they run out of reactants

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Enthalpy, H (kJ/mol)▶ This is a measure of the energy locked

up inside chemicals

▶ We can only measure changes in enthalpy, ∆H, not enthalpy itself

▶ Substances with lower enthalpy are more stable than those with higher enthalpy

▶ Enthalpy level diagrams show the changes in enthalpy over the course of a reaction.

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H H

Enthalpy level diagrams

▶ ∆H is negative (H goes down)▶ Activation energy small▶ Products more stable▶ Break weaker bonds, make stronger bonds▶ Chemical energy turned into heat energy

▶ ∆H is positive (H goes up)▶ Activation energy large▶ Products less stable▶ Break stronger bonds, make weaker bonds▶ Heat energy turned into chemical energy

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Simplified diagrams▶ These are the level needed by the IB

▶ Less useful as they tell you nothing about activation energy and so on

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Enthalpy, ∆H▶ ΔH values are usually expressed under standard conditions,

given by ΔH°,including standard states ▶ Standard state refers to the normal, most pure stable state

of a substance measured at 100 kPa. Temperature is not a part of the definition of standard state, but 298 K is commonly given as the temperature of interest.

▶ Students can assume the density and specific heat capacities of aqueous solutions are equal to those of water, but should be aware of this limitation.

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Enthalpy Changes in a Solution

▶ Enthalpy changes of neutralization (∆Hn) is the enthalpy change when one mole of water molecules are formed when an acid (H+) reacts with an alkali (OH-) under standard condition▶ H+ (aq) + OH- (aq) H2O (l)

▶ Enthalpy change of solution (∆Hsol) is the enthalpy change when one mole of solute is dissolved in excess solvent to form a solution of ‘infinite dilution under standard conditions▶ NH4NO3 (s) NH4

+ (aq) + NO3- ∆H=25.7kJ/mol

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Lesson 2

Measuring Enthalpy Changes

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▶ Which is true for a chemical reaction in which the products have a higher enthalpy than the reactants?

Reaction ∆HA. endothermic positiveB. endothermic negativeC. exothermic positiveD. exothermic negative

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Lesson 2: Endothermic and Exothermic

▶ Objectives:

▶ Understand the technique of calorimetry, including the assumptions underpinning it

▶ Calculate enthalpy changes from experimental data

▶ Complete a calorimetry experiment

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Specific heat capacity, C▶ The specific heat capacity of a substance is the amount

of energy required to raise the temperature of one gram by one Kelvin.▶ q=mcΔT

▶ Specific heat capacity is different for different substances:

Substance Specific Heat CapacityJ K-1 g-1

Water 4.18Ethanol 2.44

Air 1.00Iron 0.450

Copper 0.385

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Calorimetry▶ Calorimetry is used to measure the amount

of heat released/absorbed in a reaction.

▶ The reaction is used to heat some water, and the temperature change measured

▶ If we know the mass of water used, the specific heat capacity of the water and the temperature change, we can calculate the heat change.

▶ If we are being accurate, we should also take into account the heat capacity of the calorimeter itself as this also heats up.

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International Mindedness▶ The SI unit of temperature is the Kelvin (K), but the

Celsius scale (°C), which has the same incremental scaling, is commonly used in most countries. The exception is the USA which continues to use the Fahrenheit scale (°F) for all non-scientific communication.

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Theory of Knowledge▶ What criteria do we use in judging discrepancies

between experimental and theoretical values? Which ways of knowing do we use when assessing experimental limitations and theoretical assumptions?

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Utilization

▶ Measuring enthalpy changes is used in determining energy content of important substances in food and fuels.

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Lesson 3

Hess’ Law

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Lesson 3: Hess’ Law

▶ Objectives:

▶ Meet Hess’ Law

▶ Use Hess’ Law to design and conduct an experiment

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Hess’ Law▶ The enthalpy change for a reaction that is carried out

in a series of steps is equal to the sum of the enthalpy changes for the individual steps.

▶ Determination of the enthalpy change of a reaction that is the sum of multiple reactions with known enthalpy changes

▶ Enthalpy of formation data can be found in the data booklet in section 12.

▶ ΔH reaction =Σ(ΔHf products) - Σ(ΔHf reactants)

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Hess’ Law▶ The enthalpy change of a reaction is independent of the pathway

of that reaction▶ i.e. All that matters is the start and finish points

▶ Note: add when going ‘with’ an arrow, subtract when going against an arrow.

A + B C + D

E + F

∆H1

∆H2 ∆H3

∆H1 = ∆H2 + ∆H3

A + B C + D

E + F

∆H1

∆H2 ∆H4

∆H1 = ∆H2 - ∆H3 + ∆H4

G + H∆H3

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Why is this useful

▶ It is not always possible to directly measure the enthalpy change we want.▶ It may be an endothermic reaction that needs a constant

heat supply▶ It may be that the reaction doesn’t ‘stop’ where you need it

to▶ It may be that the reaction is simply too slow

▶ Hess cycles allow us to measure enthalpy changes indirectly

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Example 1: C(s) + O2(g) → CO2(g) ∆Hο = -394 kJ mol–1,

Mn(s) + O2(g) → MnO2(s) ∆Hο = -520 kJ mol–1

Determine ∆H, in kJ, for: MnO2(s) + C(s) → Mn(s) + CO2(g)

C(s) + O2(g) → CO2(g) ∆Hο = -394 kJ mol–1

+ MnO2(s)→Mn(s) + O2(g) ∆Hο = 520 kJ mol–1

520 kJ mol–1+ (-394kJ mol–1) = 126 kJ mol–1

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Example 1Using the equations below:

C(s) + O2(g) → CO2(g) ∆Hο = -394 kJ mol–1, Mn(s) + O2(g) → MnO2(s) ∆Hο = -520 kJ mol–1

Determine ∆H, in kJ, for: MnO2(s) + C(s) → Mn(s) + CO2(g)

MnO2(s) + C(s) Mn(s) + CO2(g)

Mn(s) + O2(g) + C(s)

∆Hr

∆H1 = -520

∆Hr = -∆H1 + ∆H= -(-520)+(-394)= +126 kJ mol-1

∆H2 = -394

Tip: We need to look at this a bit like a chemical jigsaw puzzle.

Drawing coloured boxes around the equations for which you have ∆H values, can help you see how the puzzle will fit together

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Try This:Given: 2C(s) + O2(g) 2CO(g) ∆Hο = -222 kJ mol–1 ∆H1

C(s) + O2(g) CO2(g) ∆Hο = -394 kJ mol–1 ∆H2 Calculate the enthalpy change for the reaction

2CO(g) + O2 2CO2(g)

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The Answer!

2CO(g) 2C(g)+ O2(g) ∆Hο = +222 kJ mol–1 + 2C(s) + 2O2(g) 2CO2(g) ∆Hο = 2(-394)kJ mol–1

2CO(g) + O2 2CO2(g) 222 kJ mol–1 + (-784 kJ mol–1) = -566kJ mol–1

2CO(g)+ O2(g) 2CO2(g)

2C(s)+ O2(g)+ O2(g)

∆Hο= 222 kJ mol–1 + (-784 kJ mol–1) = -566kJ mol–1

∆Hο

Direct Route

-222kJ mol–1

2(-394)kJ mol–1Remember the reactants are products are “flopped” so change your sign

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▶ Hess’ Law:▶ The enthalpy change is independent of the path of a reaction

▶ Hess Cycles:▶ When building them, look for the common features of the

reactions

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International Mindedness▶ Recycling of materials is often an effective means of

reducing the environmental impact of production, but varies in its efficiency in energy terms in different countries.

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Theory of Knowledge▶ Hess’ Law is an example of the application of the

Conservation of Energy. What are the challenges and limitations of applying general principles to specific instances?

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Utilization▶ Hess’s Law has significance in the study of nutrition,

drugs, and Gibbs free energy where direct synthesis from constituent elements is not possible.

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Lesson 4

Bond Enthalpies

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Lesson 4: Bond Enthalpies

▶ Objectives:

▶ Understand bond enthalpies and what they mean

▶ Use bond enthalpies to calculate enthalpy changes

▶ Use bond enthalpies to help you predict the outcome of an experiment

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Average Bond Enthalpies▶ Bond enthalpies are defined in terms of breaking bonds.

▶ Bond-forming releases energy and bond-breaking requires energy

▶ Average bond enthalpy is the energy needed to break one mol of a bond in a gaseous molecule averaged over similar compounds

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The bond breaking reaction:

▶ For O=O: O2(g) → 2O(g) ΔH = 498 kJ mol-1

▶ For H-F: HF(g) → H(g) + F(g) ΔH = 568 kJ mol-1

▶ For O-H: ½H2O(g) → H2(g) + ½O2(g) ΔH = 464 kJ mol-1

▶ NOT: H2O(l) → 2H(g) + O(g)▶ Note: in this case, since this is breaking 2 O-H bonds

▶ From gaseous covalent bonds to gaseous atoms

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Average Bond Enthalpy▶ The average energy required to break one mole of a chemical bond to

produce two moles of gaseous atoms at 298K.

▶ Check Table 11 of the Data Booklet

▶ Bond enthalpies are always positive

These are average values, they do not take into account:▶ Variations in strengths of the same bond in different compounds

▶ For example: the strength of a C-H bond will vary slightly depending on whether it is in methane, ethene, glucose and so on

▶ Energy changes due to state changes (from intermolecular forces)▶ Energy changes due to dissolution in water

▶ This means enthalpy changes calculated from bond enthalpies will generally be less accurate than those calculated by other means

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H H

Bond Enthalpy and Enthalpy Change▶ As the enthalpy increases, bonds are being broken, as it

decreases, bonds are being made.

▶ ∆H is negative▶ Weaker bonds broken▶ Stronger bonds made▶ Bond making

▶ ∆H is positive▶ Stronger bonds broken▶ Weaker bonds made▶ Bond breaking

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Bond Enthalpies in Calculations – Method 1▶ Determine the enthalpy change for the following reaction

H2(g) + Cl2(g) → 2 HCl(g)

2 H(g) + 2 Cl(g)

∆H = 679-864 = -185 kJ mol-1

Note: the experimentally measured value is also -185 kJ mol-1

1 x H-H + 1 x Cl-Cl = 436 + 243 = 679

2 x H-Cl = 2 x 432 = 864

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Bond Enthalpies in Calculations – Method 2▶ Determine the enthalpy change for the following reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

∆H = BONDS BROKEN - BONDS MADE = (4 x C-H + 2 x O=O) - (2 x C=O + 4 x O-H)

= (4 x 413 + 2 x 498) - (2 x 746 + 4 x 464)= 2648 – 3348= -700 kJ mol-1

Note: the experimentally measured value is -890 kJ mol-1

▶ Why the big difference?

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Key Points

▶ Bond enthalpies tell you the amount of energy required to break a bond▶ They are always positive (endothermic)

▶ Bond enthalpies are always positive

▶ Enthalpies of reaction calculated from bond enthalpies are less accurate due to their being averages.

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Lesson 6-7HL Only

Enthalpies of Combustion and Formation

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▶ Define the term average bond enthalpy.

▶ Use the information from Table 10 of the Data Booklet to calculate the enthalpy change for the complete combustion of but-1-ene, according to the following equation.

▶ C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g)

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Lessons 6-7: Enthalpies of Combustion and Formation

▶ Objectives:

▶ Understand enthalpies of combustion and formation, and use them to calculate enthalpy changes.

▶ Plan and deliver a lesson on the above.

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Welcome to my world!▶ Each half of the class will spend today planning and tomorrow

delivering a 25-30 min lesson on the objectives below. It must include:▶ A presentation – with worked examples▶ An activity (with at least three calculation problems)▶ A review (answers etc)▶ Notes to give out, including a short homework task!

▶ Tip: it makes sense to spend 20 minutes at the start reading up on your topic (text-book best), and then discussing between you so that you all understand.

HALF-AEnthalpy of Formation

HALF-BEnthalpy of Combustion

Definitions (of ∆Hor and ∆Ho

f)∆Ho

f of elements in their standard statesUse of ∆Ho

f in calculating ∆Hor

Definitions (of ∆Hor and ∆Ho

c)Use of ∆Ho

c in calculating ∆Hor

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Key Points

▶ ∆Hor is the enthalpy change of a reaction for molar

quantities under standard conditions

▶ ∆Hor can be calculated from enthalpies of

combustion/formation using Hess Cycles▶ With formation, the arrows point up▶ With combustion, the arrows point down

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Lesson 8

Born-Haber Cycles

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▶ The standard enthalpy change of three combustion reactions is given below in kJ.

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) ∆Ho = –31202H2(g) + O2(g) → 2H2O(l) ∆Ho = –572C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) ΔHo = –1411

▶ Based on the above information, calculate the standard change in enthalpy, ∆Ho, for the following reaction.

C2H6(g) → C2H4(g) + H2(g)

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Lessons 8: Lattice Enthalpy

▶ Objectives:

▶ Understand the term lattice enthalpy

▶ Use Born-Haber cycles to calculate lattice enthalpy

▶ Identify and explain trends in lattice enthalpy

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Bond Enthalpies

▶ Look at the covalent bond enthalpies on Table 10 in the Data Booklet.

▶ Why do you think there are no ionic bond enthalpies?

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Haber Cycle

▶ Born-Haber Cycles are energy cycles for the formation of certain ionic compounds

▶ A Born-Haber cycle can be used to calculate quantities that are difficult to measure directly such as lattice energies

▶ The formation of an ionic compound as a sequence of steps whose energies can be determined.

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Energy Cycles

▶ Enthalpy of atomization is the enthalpy change that occurs when one mole of gaseous atoms is formed from the element in the standard state under standard conditions

▶ Electron affinity is the enthalpy change that occurs when an electron is added to an isolated atom in the gaseous state:

▶ Lattice enthalpy is the enthalpy change that occurs from the conversion of an ionic compound into its gaseous ions

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Lattice Enthalpy, ΔHlat▶ This is the equivalent of ‘bond strength’ for ionic compounds

▶ It is the enthalpy change when one mole of an ionic compound is converted to gaseous ions.▶ This is an endothermic process, requiring energy to be put in.

MX(s) →M+(g) + X-(g)

Compound Lattice EnthalpykJ mol-1

LiF 1049LiBr 820KF 829

CaF2 2651

Note: Most places define lattice enthalpy the opposite way round, i.e:

M+(g) + X-(g) → MX(s)

The values would be the same magnitude, but with a negative sign to show they are exothermic.

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Lattice EnthalpyTHERE ARE TWO DEFINITIONS OF LATTICE ENTHALPY

1. Lattice Formation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice is formed from its isolated gaseous ions.’

Example Na+(g) + Cl¯(g) Na+ Cl¯(s)

2. Lattice Dissociation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice dissociates into isolated gaseous ions.’

Example Na+ Cl¯(s) Na+(g) + Cl¯(g)MAKE SURE YOU CHECK WHICH IS BEING USED

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Lattice Formation Enthalpy1. Lattice Formation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice is formed from its isolated gaseous ions.’Values highly EXOTHERMIC strong electrostatic attraction between oppositely charged ions a lot of energy is released as the bond is formed relative values are governed by the charge density of the ions.Example Na+(g)+Cl¯(g) Na+ Cl¯(s)

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Lattice Dissociation Enthalpy2. Lattice Dissociation Enthalpy ‘The enthalpy change when ONE MOLE of an ionic lattice dissociates into isolated gaseous ions.’Values highly ENDOTHERMIC strong electrostatic attraction between oppositely charged ions a lot of energy must be put in to overcome the attraction relative values are governed by the charge density of the ions.Example Na+ Cl¯(s) Na+(g) + Cl¯(g)

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Calculating Lattice EnthalpyYou CANNOT MEASURE LATTICE ENTHALPY DIRECTLY,It is CALCULATED USING A BORN-HABER CYCLE

greater charge densities of ions = greater attraction = larger lattice enthalpy

EffectsMelting point: the higher the lattice enthalpy, the higher the melting point of an ionic compound

Solubility: solubility of ionic compounds is affected by the relative values of Lattice and Hydration Enthalpies

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Lattice Enthalpy Values

Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions.

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Lattice Enthalpy Values

The sodium ion has the same charge as a potassium ion but is smaller. It has a higher charge density so will have a more effective attraction for the chloride ion. More energy will be released when they come together.

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Born Haber Cycles

▶ A Born-Haber cycle is an enthalpy level diagram breaking down the formation of an ionic compound into a series of simpler steps.

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Born-Haber Cycle for Sodium Chloride kJ mol-1

Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——>NaCl(s) – 411

Enthalpy of sublimation of sodium Na(s) ——> Na(g) + 108

Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) + 121

1st Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯ + 500

Electron Affinity of chlorine Cl(g) + e¯ ——> Cl¯(g) – 364

Lattice Enthalpy of NaCl Na+(g) + Cl¯(g) ——> NaCl(s) ?

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Born-Haber Cycle-NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s)

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Born-Haber Cycle-NaCl

Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of sublimation of sodium Na(s) ——> Na(g)

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Born-Haber Cycle-NaCl

Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of sublimation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g)

Born-Haber Cycle-NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of sublimation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) 1st Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯

Born-Haber Cycle-NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of sublimation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) 1st Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯ Electron Affinity of chlorine Cl(g) + e¯ ——> Cl¯(g)

Born-Haber Cycle-NaCl Enthalpy of formation of NaCl Na(s) + ½Cl2(g) ——> NaCl(s) Enthalpy of sublimation of sodium Na(s) ——> Na(g) Enthalpy of atomisation of chlorine ½Cl2(g) ——> Cl(g) 1st Ionisation Energy of sodium Na(g) ——> Na+(g) + e¯ Electron Affinity of chlorine Cl(g) + e¯ ——>Cl¯(g) Lattice Enthalpy of NaCl Na+(g) + Cl¯(g) ——> NaCl(s)

Born-Haber Cycle-MgCl21. Enthalpy of formation of MgCl2 Mg(s) + Cl2(g) ——> MgCl2(s)2. Enthalpy of sublimation of Mg Mg(s) ——> Mg(g)3. Enthalpy of atomisation of Cl ½Cl2(g) ——> Cl(g) x24. 1st Ionisation Energy of Mg Mg(g) ——> Mg+(g) + e¯5. 2nd Ionisation Energy of Mg Mg+(g) ——> Mg2+(g) + e¯6 Electron Affinity of Cl Cl(g) + e¯ ——>Cl¯(g) x27. Lattice Enthalpy of MgCl2 Mg2+(g) + 2Cl¯(g) ——> MgCl2(s)

Born-Haber Cycle-NaClApply Hess’ Law

= - + + + +

The minus shows you are going in the opposite direction to the definition= - (411) + (108) + (121) + (500) + (364)= - 682 kJ mol-1

OR…

Ignore the signs and just use the values;

If you go up you add, if you come down you subtract the value

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Born-Haber Cycles▶ Lattice enthalpies are

difficult to measure directly, so we use Born-Haber cycles▶ These are just a

specialised type of Hess cycle

▶ To calculate ΔHlat :▶ Start at the bottom and

work round clockwise, adding and subtracting according to the arrows.

solid ionic compound

elements in their standard states

metal and atomised non-metal

atomised metal and atomised non-metal

ionised metal and atomised non-metal

ionised metal and ionised non-metal

Enthalpy of atomisation (metal)

Ionisation energy/s (Metal)

Enthalpy of atomisation (non-metal)

Enthalpy of formation

ΔHlat

= ?

Electron affinity/s (Non-metal)

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International MindednessThe importance of being able to obtain measurements of

something which cannot be measured directly is significant everywhere. Borehole temperatures, snow cover depth, glacier

recession, rates of evaporation and precipitation cycles are among some indirect indicators of global warming. Why is it

important for countries to collaborate to combat global problems like global warming

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Comparing Lattice Enthalpies

▶ Study Table 13 in the data booklet.

▶ What is the relationship between lattice enthalpy and the charge on an ion?▶ Explain why this occurs▶ Use suitable pairs of lattice enthalpies to illustrate your answer

▶ What is the relationship between lattice enthalpy and the size of an ion:▶ Explain why this occurs▶ Use suitable pairs of lattice enthalpies to illustrate your answer

▶ Compare the experimental values in Table 13 with the theoretical values in table 14.▶ What do you think accounts for the difference? Research online to see if you

were correct.

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Key Points

▶ Lattice enthalpy is the equivalent of bond enthalpy in ionic compounds

▶ Born-Haber cycles are a specialist type of Hess cycle▶ Each individual step in the reaction is included

▶ Lattice enthalpy:▶ Increases with ionic charge▶ Decreases with ionic size

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Lesson 9

Spontaneity and Entropy

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Refresh▶ Which equation represents the electron affinity of calcium?

A. Ca(g) →Ca+(g) + e–

B. Ca(g) →Ca–(g) + e–

C. Ca(g) + e– → Ca–(g)D. Ca+(g) + e– → Ca(g)

▶ Which reaction has the most negative ∆Hο value?

A. LiF(s) → Li+(g) + F–(g)B. Li+(g) + F–(g) → LiF(s)C. NaCl(s) → Na+(g) + Cl–(g)D. Na+(g) + Cl–(g) → NaCl(s)

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Lessons 9: Entropy and Gibbs Free Energy

▶ Objectives:

▶ Understand the term ‘spontaneous’ in relation to a chemical reaction

▶ Understand the term entropy, and predict the sign of changes in entropy

▶ Understand the term Gibb’s Free Energy, and it’s connection to the spontaneity of a reaction

▶ Conduct calculations involving entropy and Gibb’s free energy

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Spontaneity▶ A spontaneous reaction is one that keeps

itself going▶ i.e. It doesn’t require an external energy input

▶ Some reactions are spontaneous at all temperatures:▶ Although if you lower the temperature enough,

the rate might be too low to matter

▶ Some reactions only become spontaneous above a certain temperature.

▶ Some reactions are only spontaneous below a certain temperature.

▶ Burning wood is a spontaneous reaction

▶ The thermal decomposition of copper carbonate is a non-spontaneous reaction

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Gibb’s Free Energy: Measuring Spontaneity

▶ Spontaneity is measured by Gibb’s Free Energy:▶ -ve means spontaneous▶ +ve means non-spontaneous

∆G = ∆H - T∆S

▶ Where:▶ ∆G is the change in Gibb’s free energy▶ ∆H is the enthalpy change▶ T is the temperature in K▶ ∆S is the entropy change

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Entropy, S▶ Entropy (S) refers to the distribution of available energy among the

particles. The more ways the energy can be distributed the higher the entropy. ▶ The disorder of a system

▶ Units: J K-1 mol-1 (Joules per Kelvin per mole)

LOW ENTROPY HIGH ENTROPY

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Entropy of PhasesEntropy of gas>liquid>solid under same conditions.

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Entropy Changes, ∆S▶ The entropy of a closed system* can only remain fixed or

increase.* A closed system is one where the total energy content is fixed, nothing can enter or leave.

•Lower Entropy•Pure substances•Slow moving particles•Particles close together•Fewer particles•Solid

•Higher Entropy•Mixtures of substances•Fast moving particles•Particles spread out•More particles•Gas

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Factors Affecting Entropy

▶ An increase in the number of moles of gas ▶ ΔS: positive; Entropy: increases

▶ A decrease in the number of moles of gas ▶ ΔS: negative; Entropy: decreases

▶ An increase in temperature▶ ΔS: positive; Entropy: increases

▶ An increase in pressure▶ ΔS: positive; Entropy: increases

▶ The entropy of a mixture is higher than the entropy of a pure substance

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Calculating EntropyEntropy change = total entropy of products - total entropy of reactants

ΔS = ΣSproducts - ΣSreactants

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Calculating Entropy Change, ∆So

▶ For example: ▶ Calculate the standard entropy change when pentane

(So=261 JK-1mol-1) is ‘cracked’ to form propane (So=270 JK-1mol-1) and ethene (So=220 JK-1mol-1) :

▶ C5H12(l) → C2H4(g) + C3H8(g)▶ ∆So = (270 + 220) – 261 = 229 J K-1 mol-1

▶ The positive nature of this result should be expected, as you have increased the number of molecules, and changed state from liquid to gas

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Calculate ∆So for:For each one, explain why ∆So has the sign it does∆So are written underneath the formula if not in data booklet.

▶ C6H14(l) → C2H4(g) + C4H10(g)

▶ CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) 205 214 189

▶ 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) 214 189 209 205

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Key Points▶ Entropy is a measure of ‘disorder’

▶ Entropy tends to increase

▶ Entropy increases as:▶ Substances mix▶ Particles move faster▶ Particles spread out▶ The number of particles increases▶ State changes from solid → liquid → gas

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Lesson 10

Entropy and Gibb’s Free Energy

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Lessons 10: Calculating Gibb’s Free Energy

▶ Objectives:

▶ Calculate Gibb’s Free Energy from ∆G = ∆H - T∆S

▶ Calculate Gibb’s Free Energy from Hess Cycles

▶ Predict and explain the effect of temperature changes on Gibb’s Free Energy

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Gibbs Free Energy

ΔG is a convenient way to take into account both the direct entropy change resulting from the transformation of the chemicals, and the indirect entropy change of the surroundings as a result of the gain/loss of heat energy

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Free Energy

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Free Energy

Example: Find the standard free energy of combustion of methane given the standard free energies of formation of methane, carbon dioxide, water and oxygen.

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∆G = ∆H - T∆S▶ Is the reaction of ethene with hydrogen (to form ethane),

spontaneous at room temperature (298K)?▶ C2H4(g) +H2(g) → C2H6(g)

▶ ∆Ho = -137 kJ mol-1

▶ ∆So = -121 J K-1 mol-1

▶ Calculate Gibbs Free Energy▶ ∆G = ∆H - T∆S▶ ∆G = -137 – (298 x -121)/1000 Divide by 1000 to convert to kJ▶ ∆G = -149 – (-36) = -101 kJ mol-1

▶ Evaluate the answer▶ The answer is negative, which means the reaction is spontaneous at

room temperature

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The Effect of Changing the Temperature

∆G = ∆H - T∆S

▶ What happens if we increase the temperature?▶ Since ∆H and both ∆S are both negative, ∆G can be either

positive or negative depending on the temperature▶ A high temperature will make the ‘T∆S’ term large enough to

counter-balance the ∆H term▶ At any temperature below a certain point, the reaction will be

spontaneous◻ This does not mean that the reaction will be fast, just that it will (in

theory) happen

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Summarising the Effect Of Temperature Changes

∆G = ∆H - T∆SUse the equation to help yourself reason these through

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Free Energy and Equilibrium

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Favors Reactants to Products

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Favors Products to Reactants

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Summarising▶ ∆G = ∆H - T∆S

▶ The mathematics means that:▶ Some reactions are spontaneous at all temperatures▶ Some reactions are only spontaneous above certain

temperatures▶ Some reactions are only spontaneous below certain

temperatures

▶ ∆S can be calculated using Hess cycles in much the same way as ∆G and ∆H

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Your Turn…1. Draw a mind-map to summarise the topic

2. Draw a Venn-Diagram to compare and contrast enthalpy, entropy and Gibb’s free energy

3. Draw a flow-chart to explain how to build Hess cycles or Haber cycles