5 Energetics Year 11 DP Chemistry R Slider. 5.1 Exothermic and Endothermic.

5 Energetics5 Energetics

Year 11 DP ChemistryYear 11 DP ChemistryR SliderR Slider

5.1 Exothermic and Endothermic5.1 Exothermic and Endothermic

Chemical Energy-EnthalpyChemical Energy-Enthalpy

Kinetic energy Kinetic energy is the energy of motionis the energy of motion

•Particles that make up matter are Particles that make up matter are constantly in motionconstantly in motion

Potential energyPotential energy is stored energy (no motion) is stored energy (no motion)

•Every bond between atoms and ions has Every bond between atoms and ions has stored energy within the bondsstored energy within the bonds•This stored energy is known as This stored energy is known as chemical chemical potential energypotential energy

Total energy = Kinetic energy +Potential Energy Total energy = Kinetic energy +Potential Energy (known as (known as EnthalpyEnthalpy or heat or heat content–symbol H)content–symbol H)

Change in Enthalpy (Change in Enthalpy (ΔΔH)H)

Because it is difficult to measure the amount of energy in individual Because it is difficult to measure the amount of energy in individual reactants or products, Chemists measure the reactants or products, Chemists measure the change in enthalpy (change in enthalpy (ΔΔH)H) that occurs when a reaction takes place. that occurs when a reaction takes place.

ΔΔH is the difference in the enthalpy of the products and the enthalpy of H is the difference in the enthalpy of the products and the enthalpy of the reactants.the reactants.

Enthalpy changes result from bonds breaking Enthalpy changes result from bonds breaking and new bonds reforming. and new bonds reforming. This breaking and reforming of bonds results in This breaking and reforming of bonds results in energy changes within the system which are energy changes within the system which are measured as temperature changesmeasured as temperature changes

Note: Standard enthalpy changes are Note: Standard enthalpy changes are measured under standard temperature measured under standard temperature (298K) and pressure (101.3kPa)(298K) and pressure (101.3kPa)

Endothermic or Exothermic?Endothermic or Exothermic?

EndothermicEndothermicSome chemical reactions absorb heat Some chemical reactions absorb heat from the surroundings so the products from the surroundings so the products of the reaction contain more heat.of the reaction contain more heat.

The surroundings The surroundings feel colder, feel colder, ΔΔH = +H = +

Examples: photosynthesis, cold packsExamples: photosynthesis, cold packs

ExothermicExothermicSome chemical reactions release heat Some chemical reactions release heat to the surroundings so the products to the surroundings so the products have less heat than the reactantshave less heat than the reactants

The surroundings The surroundings feel hotter, feel hotter, ΔΔH = -H = -

Examples: combustion, neutralisation Examples: combustion, neutralisation reactionsreactions

Heat exchangeHeat exchangeAnother way to visualise Another way to visualise ΔΔH is to think about energy being exchanged H is to think about energy being exchanged between the chemical system and the surroundingsbetween the chemical system and the surroundings

Source diagrams: Source diagrams: http://www.chemhume.co.uk/ASCHEM/Unit%203/13%20Enthalpy/13%20Enthalpyc.htm http://www.chemhume.co.uk/ASCHEM/Unit%203/13%20Enthalpy/13%20Enthalpyc.htm

Heat loss in a chemical systemHeat loss in a chemical system == Heat gain to the surroundingsHeat gain to the surroundings ( (Temperature increases)Temperature increases) Heat gain in a chemical systemHeat gain in a chemical system == Heat loss to the surroundingsHeat loss to the surroundings ( (Temperature decreases)Temperature decreases)

Neutralisation is exothermicNeutralisation is exothermic

AcidAcid BaseBase SaltSalt WaterWater

Acid + base reactions are called neutralisation reactions because the Acid + base reactions are called neutralisation reactions because the low pH (<7) of an acid and the high pH (>7) of a base, results in a pH low pH (<7) of an acid and the high pH (>7) of a base, results in a pH closer to neutral (7)closer to neutral (7)

These reactions release heat and are exothermic. For example:These reactions release heat and are exothermic. For example:

HCl + NaOH HCl + NaOH NaCl + H NaCl + H220 + 58 kJ/mol0 + 58 kJ/mol

Combustion is exothermicCombustion is exothermic

Combustion is the process of burningCombustion is the process of burning

Most often, the combustion of a Most often, the combustion of a material involves its combination material involves its combination with oxygen gas.with oxygen gas.

Because combustion reactions Because combustion reactions release energy in the form of heat release energy in the form of heat and light, they are and light, they are exothermicexothermic..

Fire in the Penola Forest (SA), Ash Wednesday 1983 Fire in the Penola Forest (SA), Ash Wednesday 1983

(Source: (Source: www.austehc.unimelb.edu.au/ fam/1611_image.html)www.austehc.unimelb.edu.au/ fam/1611_image.html)

Example:Example:The combustion of methane The combustion of methane (natural gas) in oxygen:(natural gas) in oxygen:

CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22OO

∆ ∆ HHrxnrxn = -832 kJ/mol (exothermic) = -832 kJ/mol (exothermic)

Combustion: why exothermic?Combustion: why exothermic?

• Energy is required to break bonds(endothermic)Energy is required to break bonds(endothermic)• Energy is released when bonds form (exothermic)Energy is released when bonds form (exothermic)

In summary:In summary:

∆∆ HHrxnrxn = H (products) – H (reactants) = H (products) – H (reactants)

This reaction releases more energy than it absorbs resulting in a negative value for This reaction releases more energy than it absorbs resulting in a negative value for ∆∆ HrxnHrxn

H reactants > H productsH reactants > H products

Breaking bondsBreaking bondsAbsorbs energyAbsorbs energy

Forming bondsForming bondsReleases energyReleases energy

Energy requiredEnergy requiredto break bondsto break bonds

Energy releasedEnergy releasedwhen bonds formedwhen bonds formed

In the previous example involving the combustion of methane:In the previous example involving the combustion of methane:

CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22O + 832kJO + 832kJ A thermochemical A thermochemical equationequation

Energy changes in chemical reactions – Energy changes in chemical reactions – Enthalpy diagramsEnthalpy diagrams

Exothermic reactionsExothermic reactions

The enthalpy of the reactants is The enthalpy of the reactants is higher than the products.higher than the products.

ReactantsReactants

ProductsProducts

Endothermic reactionsEndothermic reactions

The enthalpy of the reactants is The enthalpy of the reactants is lower than the products.lower than the products.

ProductsProducts

ReactantsReactants

Enth

alp

y (

H)

Enth

alp

y (

H)

Enth

alp

y (

H)

Enth

alp

y (

H)

∆∆ H is -ve ∆∆ H is +veHeat releasedHeat released Heat absorbedHeat absorbed

Reaction progressionReaction progression Reaction progressionReaction progression

ExamplesExamples

The combustion of methane is The combustion of methane is exothermic and releases exothermic and releases energy to the surroundingsenergy to the surroundings

The decomposition of calcium The decomposition of calcium carbonate is endothermic and carbonate is endothermic and absorbs energy from the absorbs energy from the surroundingssurroundings


Activation EnergyActivation Energy

Reaction progressReaction progress

ReactantsReactants

ProductsProducts

enth

alp

yen

thal

py

ActivationActivationenergyenergy

∆∆ H

The activation energy EThe activation energy Eaa is the minimum is the minimum

amount of energy that is required by the amount of energy that is required by the reactants for bonds to be broken and the reactants for bonds to be broken and the reaction to proceed to the products.reaction to proceed to the products.

High EHigh Eaa means strong means strong

bonds in the reactantsbonds in the reactants

Chemical stabilityChemical stability

Source: http://www.teachmetuition.co.uk/Energetics/chemical_energetics1.htmSource: http://www.teachmetuition.co.uk/Energetics/chemical_energetics1.htm

Exothermic reactionsExothermic reactions•Products are more stable than Products are more stable than reactants due to lower enthalpy reactants due to lower enthalpy contentcontent•The lower the enthalpy the more The lower the enthalpy the more stablestable•The lower the enthalpy, the The lower the enthalpy, the stronger the bondsstronger the bonds

Endothermic reactionsEndothermic reactions•Products are less stable than Products are less stable than reactants due to a higher enthalpy reactants due to a higher enthalpy contentcontent•The higher the enthalpy the less The higher the enthalpy the less stablestable•The higher the enthalpy, the The higher the enthalpy, the weaker the bondsweaker the bonds

SummarySummaryThermochemical Reaction Exothermic Endothermic

Enthalpy Change (ΔH = HP – HR)

HP < HR HP > HR

Sign ΔH Negative Positive

ΔT of surroundings Increases (warmer) Decreases (colder)

Enthalpy Diagram


ExercisesExercises1.1. Classify each of the reactions below as endothermic or exothermicClassify each of the reactions below as endothermic or exothermic

a)a) CC66HH1212OO6(aq)6(aq) + 6O + 6O2(g)2(g) 6CO 6CO2(g)2(g) + 6H + 6H22OO(l)(l) ΔΔH = -2801 kJ molH = -2801 kJ mol-1-1

b)b) 6CO6CO2(g)2(g) + 6H + 6H22OO(l)(l) C C66HH1212OO6(aq)6(aq) + 6O + 6O2(g)2(g) ΔΔH = +2801 kJ molH = +2801 kJ mol-1-1

c)c) HH2(g)2(g) + 1/2 O + 1/2 O2(g)2(g) H H22OO(l)(l) + 286 kJmol + 286 kJmol-1-1

2.2. When a spark is introduced into a vessel containing a mixture of hydrogen and oxygen gases, When a spark is introduced into a vessel containing a mixture of hydrogen and oxygen gases, they react explosively.they react explosively.

a)a) What is the role of the spark in this reaction?What is the role of the spark in this reaction?b)b) What bonds are broken when this reaction is initiated?What bonds are broken when this reaction is initiated?c)c) What bonds are generated when the products are formed?What bonds are generated when the products are formed?d)d) Is the reaction endothermic or exothermic? Justify your answer.Is the reaction endothermic or exothermic? Justify your answer.

3.3. 10.0g of ammonium nitrate is dissolved in 100cm10.0g of ammonium nitrate is dissolved in 100cm33 of water and the temperature of the solution of water and the temperature of the solution decreases from 19.0decreases from 19.000C to 10.50C to 10.5000C.C.

a)a) Which is greater, the enthalpy of the reactants or the products?Which is greater, the enthalpy of the reactants or the products?b)b) Is the reaction endothermic or exothermic?Is the reaction endothermic or exothermic?c)c) What chemical bonds were broken when the ammonium nitrate dissolved in water?What chemical bonds were broken when the ammonium nitrate dissolved in water?

4.4. Consider the reaction represented by the following equation: A + 2B Consider the reaction represented by the following equation: A + 2B C CIn this reaction, the total enthalpy of the reactants is 80kJmolIn this reaction, the total enthalpy of the reactants is 80kJmol -1-1, the total enthalpy for the , the total enthalpy for the products is -90kJ.molproducts is -90kJ.mol-1-1and the activation energy for the forward reaction is 120kJ.moland the activation energy for the forward reaction is 120kJ.mol -1-1..

a)a) Draw a diagram for the energy profile for this reaction. Label the diagram clearly to show Draw a diagram for the energy profile for this reaction. Label the diagram clearly to show ΔΔH and EH and Eaa..

b)b) State whether the forward reaction is endothermic or exothermicState whether the forward reaction is endothermic or exothermicc)c) Calculate the change in enthalpy Calculate the change in enthalpy d)d) Calculate the amount of energy released as the products are formedCalculate the amount of energy released as the products are formede)e) Calculate the activation energy of the reverse reaction: C Calculate the activation energy of the reverse reaction: C A 2B A 2B

Exercises - AnswersExercises - Answers

1.1. a) exothermic b) endothermic c) exothermica) exothermic b) endothermic c) exothermic2.2. a)spark is to overcome the activation energya)spark is to overcome the activation energy

b) H-H and O=O covalent bonds broken b) H-H and O=O covalent bonds broken c) O-H bonds are made during the reactionc) O-H bonds are made during the reaction d) exothermic as energy is released d) exothermic as energy is released

3.3. a) products>reactantsa) products>reactants b) b) endothermicendothermic c) c) ionic bonds between NHionic bonds between NH44

++ and NO and NO33--

4.4. a) diagram should show exothermic forward rxna) diagram should show exothermic forward rxn b) b) exothermicexothermic c) c) ΔΔH H = -90 – 80 kJ mol= -90 – 80 kJ mol-1-1 = -170 kJ mol = -170 kJ mol-1-1 d) 290 d) 290 kJ mol kJ mol -1-1 e) 290 e) 290 kJ molkJ mol-1-1

5.2 Calculating Enthalpy5.2 Calculating Enthalpy

Measuring Enthalpy ChangesMeasuring Enthalpy Changes

Recall that changes in temperature are related to changes in the amount of heat energy that is being absorbed or released in a system. This quantity of heat energy, qq in joules (J), can be calculated using the following equation and often involves water being heated by reactions:

qq = = mm C C ∆T∆T

Heat energy Heat energy released or absorbed released or absorbed in Joules (J)in Joules (J) Mass (g)Mass (g)

Specific heatSpecific heatCapacity (4.18 Capacity (4.18 J gJ g-1-1KK-1 -1 for water) for water)

Change in Change in TemperatureTemperature(final – initial)(final – initial)

Example:Example:What quantity of energy is required to raise the temperature of 0.5L of What quantity of energy is required to raise the temperature of 0.5L of water from 20water from 2000C to 100C to 10000C?C?q = q = mm C ∆T C ∆T = 500g X 4.18 J g= 500g X 4.18 J g-1-1KK-1-1 X 80 X 8000C or KC or K = 167,200 J= 167,200 J = 167 kJ= 167 kJ NB: NB: 00C = K - 273C = K - 273

Heat Capacity ValuesHeat Capacity Values

SubstanceSubstance

Specific Specific Heat Heat CapacityCapacity

J KJ K-1-1gg-1-1 SubstanceSubstance

Specific Specific Heat Heat CapacityCapacity

J KJ K-1-1gg-1-1

WaterWater 4.184.18 EthanolEthanol 2.412.41

Ethylene Ethylene glycolglycol

2.392.39 HexaneHexane 2.262.26

50/50 50/50 water/E. water/E. glycolglycol

2.862.86 ChloroformChloroform 0.960.96

AcetoneAcetone 2.172.17 AluminiumAluminium 0.900.90

IronIron 0.4480.448 CopperCopper 0.3860.386Specific heat capacitySpecific heat capacity (C), is the amount of heat energy in Joules (J), (C), is the amount of heat energy in Joules (J), required to raise the temperature of 1g of a substance by 1 Kelvin (K).required to raise the temperature of 1g of a substance by 1 Kelvin (K).

Heat capacity is a measure of how much energy a substance can absorb without Heat capacity is a measure of how much energy a substance can absorb without changing temperature. These values vary greatly among substances.changing temperature. These values vary greatly among substances.

Practice QuestionsPractice Questions

1.1. A Bunsen burner was used to heat 100 cmA Bunsen burner was used to heat 100 cm33 of water for 10 min. of water for 10 min. The temperature of the water increased from 15.0The temperature of the water increased from 15.000C to 80.0C to 80.000C. C. Determine the heat energy change of the water. (note: the density Determine the heat energy change of the water. (note: the density of water is 1.0 g cmof water is 1.0 g cm-3-3))

2.2. The same Bunsen burner was used to heat a 500g block of copper The same Bunsen burner was used to heat a 500g block of copper for 10 minutes. Assuming the same amount of energy is transferred for 10 minutes. Assuming the same amount of energy is transferred from the Bunsen burner to the block as in question 1, determine the from the Bunsen burner to the block as in question 1, determine the highest temperature that the block could reach if the starting highest temperature that the block could reach if the starting temperature was 15.0temperature was 15.000C.C.

3.3. Explain the difference in the temperatures considering the same Explain the difference in the temperatures considering the same amount of energy was put into the water and the copperamount of energy was put into the water and the copper

Solutions:Solutions:1)27.2kJ1)27.2kJ 2) 1562) 15600CC 3) The water has a much higher heat capacity than the copper 3) The water has a much higher heat capacity than the copper which means the water can absorb much more heat energy before changing temperature.which means the water can absorb much more heat energy before changing temperature.

CalorimetryCalorimetry

What is calorimetryWhat is calorimetryWater can be used to measure the change in heat Water can be used to measure the change in heat

energy in a chemical reaction due to its ability energy in a chemical reaction due to its ability to absorb heat. This is known as to absorb heat. This is known as calorimetrycalorimetry..

A A calorimetercalorimeter is a device that is used to measure is a device that is used to measure the enthalpy change that occurs during a the enthalpy change that occurs during a chemical reaction. chemical reaction.

Measuring heat in the laboratoryMeasuring heat in the laboratory

If a known quantity of water is placed in the If a known quantity of water is placed in the calorimeter and a reaction is carried out, the calorimeter and a reaction is carried out, the change in temperature due to the reaction is change in temperature due to the reaction is transferred to the water and is measured using transferred to the water and is measured using a thermometer placed in a hole in the lid.a thermometer placed in a hole in the lid.

The heat energy change is: The heat energy change is:

q = q = mm C ∆T C ∆T

The amount of heat lost or gained in the reaction is The amount of heat lost or gained in the reaction is equal in size but opposite in sign to the amount equal in size but opposite in sign to the amount of heat lost or gained by the water. of heat lost or gained by the water.

Coffee cup calorimeterCoffee cup calorimeterThe The simplest calorimetersimplest calorimeter makes use of two makes use of two polystyrene (Styrofoam) cups, one inside the other polystyrene (Styrofoam) cups, one inside the other with a lid on top. This minimises the amount of with a lid on top. This minimises the amount of heat lost to the surroundings. This is the most heat lost to the surroundings. This is the most significant source of error.significant source of error.

Calorimetry exampleCalorimetry exampleWhen a student added excess Mg is to 100cmWhen a student added excess Mg is to 100cm33 of 2.00 mol dm of 2.00 mol dm-3-3 CuSO CuSO44, the temperature rose , the temperature rose

from 20.0from 20.0ooC to 65C to 65ooC.C.

i) The energy change:i) The energy change:QQ == mm xx cc xx ΔΔT T QQ == 100100 xx 4.184.18 xx (65 - 20)(65 - 20) QQ == 100100 xx 4.184.18 xx 4545 QQ == 18810 J18810 J QQ == 18.81 kJ18.81 kJ Convert the energy calculated into KJ (divide by 1000)Convert the energy calculated into KJ (divide by 1000) ii) Calculate the number of moles used:ii) Calculate the number of moles used: No moles = No moles = c x Vc x V 10001000 No moles = No moles = 2 x 1002 x 100 10001000 No moles = 0.2No moles = 0.2 iii) Calculate the amount of energy exchanged per mole , this is the enthalpy:iii) Calculate the amount of energy exchanged per mole , this is the enthalpy: Enthalpy = Enthalpy = EnergyEnergy molesmoles Enthalpy = Enthalpy = - 18.81- 18.81 0.20.2 Enthalpy = - 94.05 kJ molEnthalpy = - 94.05 kJ mol-1-1

iv) Finally write the equation with the enthalpy change:iv) Finally write the equation with the enthalpy change:

MgMg(s)(s) ++ CuSOCuSO4(aq)4(aq) MgSOMgSO4(aq)4(aq) ++ CuCu(s)(s) ΔΔHH == - 94.05 kjmol- 94.05 kjmol-1 -1 It is exothermic, therefore negativeIt is exothermic, therefore negative

Heat of solutionHeat of solution

In order for a solute to dissolve in a In order for a solute to dissolve in a solvent, the solute particles and the solvent, the solute particles and the solvent particles must separate, which solvent particles must separate, which involves energy changes.involves energy changes.

In addition, solute particles interact with In addition, solute particles interact with solvent particles, which also can result solvent particles, which also can result in energy changes.in energy changes.

Overall:Overall:ΔHΔHsolnsoln = = ΔΔHH11 + + ΔΔHH22 + + ΔΔHH33

The energy profile diagrams on the right The energy profile diagrams on the right show net exothermic and endothermic show net exothermic and endothermic dissolutionsdissolutions

Question: How are these processes Question: How are these processes similar to a chemical reaction? How are similar to a chemical reaction? How are they different?they different?

Diagrams Source: Diagrams Source: http://wps.prenhall.com/wps/media/objects/3312/3391718/blb1301.html http://wps.prenhall.com/wps/media/objects/3312/3391718/blb1301.html

Heat of Combustion (Heat of Combustion (ΔΔHHcc))

• Heat of Combustion of a substance is the heat Heat of Combustion of a substance is the heat liberated when 1 mole of the substance undergoes liberated when 1 mole of the substance undergoes complete combustion with oxygen at constant complete combustion with oxygen at constant pressure. pressure.

• Combustion is always exothermic, Combustion is always exothermic, ΔΔH is negative. H is negative. • By definition, the heat of combustion is minus the By definition, the heat of combustion is minus the

enthalpy change for the combustion reaction. enthalpy change for the combustion reaction. • By definition, the heat of combustion is a positive By definition, the heat of combustion is a positive

value. value. • Heat of Combustion can be measured experimentally.Heat of Combustion can be measured experimentally.

Heat of Combustion (Heat of Combustion (ΔΔHHcc))

The diagram to the right shows a typical school The diagram to the right shows a typical school laboratory setup to test the heat of combustion of laboratory setup to test the heat of combustion of a fuel such as ethanol.a fuel such as ethanol.

1.1. A known quantity of water is placed in a flask or A known quantity of water is placed in a flask or beaker (calorimeter) beaker (calorimeter)

2.2. A thermometer is positioned with bulb near the A thermometer is positioned with bulb near the middle of the volume of watermiddle of the volume of water

3.3. A known quantity of fuel, such as an alcohol A known quantity of fuel, such as an alcohol (alkanol), is placed in the spirit burner (alkanol), is placed in the spirit burner

4.4. The initial temperature of the water is measured The initial temperature of the water is measured and recorded (Tand recorded (Tii))

5.5. The wick on the spirit burner is lit, burning the The wick on the spirit burner is lit, burning the fuel, and heating the waterfuel, and heating the water

6.6. When the temperature has risen an appreciable When the temperature has risen an appreciable amount, the spirit burner is extinguished and the amount, the spirit burner is extinguished and the final temperature recorded (Tfinal temperature recorded (Tff))

7.7. The final quantity of fuel is measured and The final quantity of fuel is measured and recorded recorded

Source: http://www.ausetute.com.au/heatcomb.html Source: http://www.ausetute.com.au/heatcomb.html

Heat of Combustion ExampleHeat of Combustion Example

initial water temperature (Ti) = 20oC intial mass burner + ethanol = 37.25 g

final water temperature (Tf)= 75oC final mass burner + ethanol = 35.50 g

change in temperature = Tf - Ti = 55oCmass ethanol used = 1.75g

A student used the apparatus on the previous slide to determine the A student used the apparatus on the previous slide to determine the heat of combustion of ethanol. Their results for one trial are below:heat of combustion of ethanol. Their results for one trial are below:

Use the results above to determine the heat of combustion for Use the results above to determine the heat of combustion for ethanol in kJ/mol. (Solution on the next slide)ethanol in kJ/mol. (Solution on the next slide)

Heat of Combustion ExampleHeat of Combustion Example

1.1. Calculate moles (n) of fuel used Calculate moles (n) of fuel used molecular mass (MM) of ethanol = 46.1g/mol molecular mass (MM) of ethanol = 46.1g/mol

mass ethanol used = 1.75g mass ethanol used = 1.75g n = mass ÷ MM = 1.75 ÷ 46.1 = 0.0380 mol n = mass ÷ MM = 1.75 ÷ 46.1 = 0.0380 mol

2. Calculate energy required to change temperature of water 2. Calculate energy required to change temperature of water energy = mass of water x specific heat capacity of water x change in water energy = mass of water x specific heat capacity of water x change in water temp (mCtemp (mCΔΔT)T)energy = 200g x 4.184 JKenergy = 200g x 4.184 JK-1-1gg-1-1 x 55 x 55ooC = 46024 J = 46.024 kJ C = 46024 J = 46.024 kJ

3. Calculate the heat of combustion of ethanol 3. Calculate the heat of combustion of ethanol Assume all the heat produced from burning ethanol has gone into heating the Assume all the heat produced from burning ethanol has gone into heating the water, ie, no heat has been wasted. water, ie, no heat has been wasted. 0.0380 mole ethanol produced 46.024 kJ of heat. 0.0380 mole ethanol produced 46.024 kJ of heat. Therefore 1 mole of ethanol would produce 46.024 kJ ÷ 0.0380 mol = 1211 Therefore 1 mole of ethanol would produce 46.024 kJ ÷ 0.0380 mol = 1211 kJ/mol kJ/mol The heat of combustion of ethanol is 1211 kJ/mol The heat of combustion of ethanol is 1211 kJ/mol

The experimentally determined value for the heat of combustion of ethanol is usually less The experimentally determined value for the heat of combustion of ethanol is usually less than the accepted value of 1368 kJ/mol because some heat is always lost to the than the accepted value of 1368 kJ/mol because some heat is always lost to the atmosphere and in heating the vessel. atmosphere and in heating the vessel.

Heat loss compensationHeat loss compensation

During exothermic reactions using a During exothermic reactions using a calorimeter in the laboratory, there calorimeter in the laboratory, there are many errors associated with the are many errors associated with the loss of heat.loss of heat.

Because reactions are not Because reactions are not instantaneous, heat loss is often instantaneous, heat loss is often gradual.gradual.

We can compensate for this heat loss We can compensate for this heat loss by graphing temperature against by graphing temperature against time. Then, we can extrapolate back time. Then, we can extrapolate back to what would have been the final to what would have been the final temperature if no heat had been lost temperature if no heat had been lost to the surroundings.to the surroundings.

5.3 Hess’s Law5.3 Hess’s Law

Hess’s LawHess’s LawAccording to the Law of Conservation of Energy, the energy exchanges that take place in According to the Law of Conservation of Energy, the energy exchanges that take place in reactions must be conserved. reactions must be conserved.

Hess’s Law states that the enthalpy of a chemical process is the same, whether the process Hess’s Law states that the enthalpy of a chemical process is the same, whether the process takes place in one step or several steps. takes place in one step or several steps. H is the same regardless of how the reactants H is the same regardless of how the reactants change into productschange into products

The diagram above shows an The diagram above shows an enthalpy cycle enthalpy cycle and reflects Hess’s Law:and reflects Hess’s Law:

ΔHΔHxx = = ΔΔHH11 + + ΔΔHH22

So, if we cannot measure the enthalpy for the overall reaction (x), but can measure the reactions (1) and (2), we So, if we cannot measure the enthalpy for the overall reaction (x), but can measure the reactions (1) and (2), we can determine the enthalpy of reaction mathematically. can determine the enthalpy of reaction mathematically.

Hess’s Law AnalogyHess’s Law Analogy

Little Red Riding Hood wants to visit her Little Red Riding Hood wants to visit her grandmother, who lives at the top of the grandmother, who lives at the top of the mountain.mountain.

Her grandmother’s house is 2000 meters Her grandmother’s house is 2000 meters above Red’s house.above Red’s house.

Regardless of how Red climbs the Regardless of how Red climbs the mountain, her change in altitude will be the mountain, her change in altitude will be the same: 2000 meters.same: 2000 meters.

Red’s HouseRed’s House

Grandma’s HouseGrandma’s HouseThe pathway The pathway

doesn’t matter; doesn’t matter; the altitude the altitude

change is the change is the same.same.

Hess’s Law ProblemsHess’s Law ProblemsHelpful HintsHelpful Hints

1. If you reverse a reaction, the sign of H changes.– 2H2O2(l) 2H2O(l) + O2(g)

H = -196.4 kJ

– 2H2O(l) + O2(g) 2H2O2(l) H = +196.4 kJ


2. If you multiply the coefficients of a thermochemical equation by some number, you must multiply H by the same number.– 2H2O2(l) 2H2O(l) + O2(g)

H = -196.4 kJ

– 4H2O2(l) 4H2O(l) + 2O2(g) H = -392.8 kJ

– H2O2(l) H2O(l) + ½O2(g) H = -98.2 kJ


3 Thermochemical equations can be combined 3 Thermochemical equations can be combined to give to give newnew thermochemical equations. thermochemical equations.– NN22 + 2O + 2O22 2NO2NO22

H = 67.8 kJH = 67.8 kJ

– NN22OO44 NN22 + 2O + 2O22

H = -9.6 kJH = -9.6 kJ

– NN22 + 2O + 2O22 + + NN22OO44 2NO2NO22 + + NN22 + 2O + 2O22

– NN22OO44 2NO2NO22

H = (67.8 kJ) + (-9.6 kJ) = 58.2 kJH = (67.8 kJ) + (-9.6 kJ) = 58.2 kJ

Hess’s Law Problems - summaryBelow are the steps to follow when performing Hess’s Law problems:Below are the steps to follow when performing Hess’s Law problems:

1.1. Rearrange the balanced chemical equations so that the reactants and Rearrange the balanced chemical equations so that the reactants and products are on the correct sides. If you reverse a chemical equation, then products are on the correct sides. If you reverse a chemical equation, then the sign of the enthalpy also changes.the sign of the enthalpy also changes.

2.2. Check that you have the correct states, enthalpy changes will be different Check that you have the correct states, enthalpy changes will be different for species in the solid, liquid and gas states.for species in the solid, liquid and gas states.

3.3. Assign each rearranged equation the correct Assign each rearranged equation the correct ΔΔH value.H value.Remember, if you need to multiply each species in the chemical equation Remember, if you need to multiply each species in the chemical equation by 2, then the enthalpy change must also be multiplied by 2.by 2, then the enthalpy change must also be multiplied by 2.

4.4. Add the rearranged equations together to give the overall equation for the Add the rearranged equations together to give the overall equation for the reaction. Add the reaction. Add the ΔΔH values for these equations to calculate H values for these equations to calculate ΔΔH for the H for the overall reaction.overall reaction.

Example 1Example 1When carbon combusts in an excess of oxygen, carbon dioxide is formed and When carbon combusts in an excess of oxygen, carbon dioxide is formed and 393.5 kJ of heat is released per mole of carbon. 393.5 kJ of heat is released per mole of carbon.

CC(s)(s) + O + O2(g)2(g) -----> CO -----> CO2(g)2(g) ΔΔH = -393.5 kJ H = -393.5 kJ

When carbon combusts in an excess of oxygen, carbon dioxide is formed and When carbon combusts in an excess of oxygen, carbon dioxide is formed and 393.5 kJ of heat is released per mole of carbon. 393.5 kJ of heat is released per mole of carbon.

CC(s)(s) + O + O2(g)2(g) -----> CO -----> CO2(g)2(g) ΔΔH = -393.5 kJ H = -393.5 kJ

This overall reaction can also be produced as a two stage process: This overall reaction can also be produced as a two stage process: Carbon combusts in limited oxygen producing carbon monoxide: Carbon combusts in limited oxygen producing carbon monoxide: CC(s)(s) + ½O + ½O2(g)2(g) CO CO(g)(g) ΔΔH = -110.5 kJ H = -110.5 kJ

Carbon monoxide then combusts in additional oxygen: Carbon monoxide then combusts in additional oxygen: COCO(g)(g) + ½O + ½O2(g)2(g) CO CO2(g)2(g) ΔΔH = -283.0 kJH = -283.0 kJ

This overall reaction can also be produced as a two stage process: This overall reaction can also be produced as a two stage process: Carbon combusts in limited oxygen producing carbon monoxide: Carbon combusts in limited oxygen producing carbon monoxide: CC(s)(s) + ½O + ½O2(g)2(g) CO CO(g)(g) ΔΔH = -110.5 kJ H = -110.5 kJ

Carbon monoxide then combusts in additional oxygen: Carbon monoxide then combusts in additional oxygen: COCO(g)(g) + ½O + ½O2(g)2(g) CO CO2(g)2(g) ΔΔH = -283.0 kJH = -283.0 kJ

Example 1 - GraphExample 1 - Graph

This shows a graphical This shows a graphical representation of the representation of the previous example.previous example.

Using one step or two, Using one step or two, the overall enthalpy the overall enthalpy change is the same.change is the same.

This follows Hess’s This follows Hess’s LawLaw

Source: http://www.ausetute.com.au/hesslaw.html Source: http://www.ausetute.com.au/hesslaw.html

Example 2 – you tryExample 2 – you try

Calculate Calculate ΔΔH for the reaction: H for the reaction: NHNH3(g)3(g) + + HClHCl(g)(g) NHNH44ClCl(s)(s)

Given that: Given that: ½N½N2(g)2(g) + 1½H + 1½H2(g)2(g) NHNH3(g)3(g) ΔΔH = -46.1 kJH = -46.1 kJ

½H½H2(g)2(g) + ½Cl + ½Cl2(g)2(g) HClHCl(g)(g) ΔΔH = -92.3 kJ H = -92.3 kJ

½N½N2(g)2(g) + 2H + 2H2(g)2(g) + ½Cl + ½Cl2(g)2(g) NHNH44ClCl(s)(s) ΔΔH = -314.4 kJH = -314.4 kJ

Example 2 - SolutionExample 2 - Solution

ExerciseExercise

Calculate the enthalpy change (Calculate the enthalpy change (Δ Δ H°), in kJ, of the H°), in kJ, of the reaction:reaction:NN22HH44 + 2 H + 2 H22OO22 N N22 + 4 H + 4 H22OO

using the following enthalpy of combustion data:using the following enthalpy of combustion data:a. Na. N22HH44 + O + O22 N N22 + 2 H + 2 H22O O ΔΔH° = -622 kJH° = -622 kJ

b. Hb. H22 + O + O22 H H22OO22 Δ Δ H° = -188 kJH° = -188 kJ

c. Hc. H22 + ½ O + ½ O22 H H22O O Δ Δ H° = -286 kJH° = -286 kJ

To solve this, leave equation (a) alone. Reverse equation (b) and double the To solve this, leave equation (a) alone. Reverse equation (b) and double the coefficients. Then, doublecoefficients. Then, doublethe coefficients for equation (c). When that is complete, equations a, b, and c the coefficients for equation (c). When that is complete, equations a, b, and c add up to the desiredadd up to the desiredequation and the overall DH° value equals -818 kJ.equation and the overall DH° value equals -818 kJ.

Finding Enthalpy changes using Hess’s Finding Enthalpy changes using Hess’s Law cyclesLaw cycles

Standard enthalpy change of Standard enthalpy change of formation formation ΔΔHH00

ff is the energy is the energy

change when 1 mole of a change when 1 mole of a compound forms from its compound forms from its elements at under standard elements at under standard conditions and constant conditions and constant pressure.pressure.

If we want to determine If we want to determine ΔΔHH00ff the for a the for a

substance such as benzene Csubstance such as benzene C66HH66, we can , we can

use the standard enthalpy changes of use the standard enthalpy changes of combustion and Hess’s Law cycles to help combustion and Hess’s Law cycles to help us. You can find the data below in your us. You can find the data below in your data booklet.data booklet.

Standard enthalpy changes of combustionStandard enthalpy changes of combustion

ΔH°c (kJ mol-1)

C6H6(l) -3267

C(s) -394

H2(g) -286

Drawing the cycleDrawing the cycle

1. The formation of benzene is drawn 1. The formation of benzene is drawn horizontally horizontally

1. The formation of benzene is drawn 1. The formation of benzene is drawn horizontally horizontally

2. The Hess’s Law cycle 2. The Hess’s Law cycle is completed showing is completed showing the combustion the combustion products. (Oproducts. (O22 has been has been

left off for convenience left off for convenience only - it would be added only - it would be added in excess anyway)in excess anyway)

2. The Hess’s Law cycle 2. The Hess’s Law cycle is completed showing is completed showing the combustion the combustion products. (Oproducts. (O22 has been has been

left off for convenience left off for convenience only - it would be added only - it would be added in excess anyway)in excess anyway)

3. There are two possible routes 3. There are two possible routes to the same products. These to the same products. These routes are shown in blue.routes are shown in blue.

3. There are two possible routes 3. There are two possible routes to the same products. These to the same products. These routes are shown in blue.routes are shown in blue.

4. Construct and solve the equation:4. Construct and solve the equation:ΔH - 3267 = 6(-394) + 3(-286)ΔH - 3267 = 6(-394) + 3(-286)ΔΔH = 3267 + 6(-394) + 3(-286)H = 3267 + 6(-394) + 3(-286)ΔΔH = +45 kJ molH = +45 kJ mol-1-1

4. Construct and solve the equation:4. Construct and solve the equation:ΔH - 3267 = 6(-394) + 3(-286)ΔH - 3267 = 6(-394) + 3(-286)ΔΔH = 3267 + 6(-394) + 3(-286)H = 3267 + 6(-394) + 3(-286)ΔΔH = +45 kJ molH = +45 kJ mol-1-1

Hess’s Law cycle – example 2Hess’s Law cycle – example 2

Standard enthalpy changes of formation

ΔH°f (kJ mol-1)

C2H4(g) +52

HCl(g) -92

C2H5Cl(g) -137

Calculate the enthalpy change of the reaction between ethene and hydrogen Calculate the enthalpy change of the reaction between ethene and hydrogen chloride gases to form chloroethane gas using a Hess’s Law cycle.chloride gases to form chloroethane gas using a Hess’s Law cycle.

-137-137

These standard enthalpy changes of formation can These standard enthalpy changes of formation can be found in your data booklet (except HCl)be found in your data booklet (except HCl)These standard enthalpy changes of formation can These standard enthalpy changes of formation can be found in your data booklet (except HCl)be found in your data booklet (except HCl)

ΔΔH for this reaction can easily be found by the same method as in the previous question H for this reaction can easily be found by the same method as in the previous question once the cycle has been drawn. Find the two pathways that lead to the same product and once the cycle has been drawn. Find the two pathways that lead to the same product and equate them. Then solve for the unknown.equate them. Then solve for the unknown.+52 - 92 + Δ+52 - 92 + ΔH = -137H = -137ΔΔH = -52 + 92 - 137H = -52 + 92 - 137ΔΔH = -97 kJ molH = -97 kJ mol-1-1

ΔΔH for this reaction can easily be found by the same method as in the previous question H for this reaction can easily be found by the same method as in the previous question once the cycle has been drawn. Find the two pathways that lead to the same product and once the cycle has been drawn. Find the two pathways that lead to the same product and equate them. Then solve for the unknown.equate them. Then solve for the unknown.+52 - 92 + Δ+52 - 92 + ΔH = -137H = -137ΔΔH = -52 + 92 - 137H = -52 + 92 - 137ΔΔH = -97 kJ molH = -97 kJ mol-1-1

5.4 Bond Enthalpies5.4 Bond Enthalpies

Bond enthalpyBond enthalpy

Also known as bond energies, this is the enthalpy change required to break a Also known as bond energies, this is the enthalpy change required to break a covalent bond when all species are in the gaseous state. These are positive covalent bond when all species are in the gaseous state. These are positive values as breaking bonds is an exothermic process that requires energy.values as breaking bonds is an exothermic process that requires energy.

We can use bond enthalpy We can use bond enthalpy values to estimate the values to estimate the overall enthalpy changes in overall enthalpy changes in reactions. reactions.

We simply take the sum of the individual bond energies for reactants and We simply take the sum of the individual bond energies for reactants and subtract the sum of the bond enthalpies for the products (negative because subtract the sum of the bond enthalpies for the products (negative because forming bonds releases energy – exothermic)forming bonds releases energy – exothermic)

Consider this example:Consider this example:

COCO(g) (g) + H+ H22OO(g) (g) CO CO2(g)2(g) + H + H2(g)2(g)

Bond Bond energy, kJ/mol

C–C 347

C=C 612

C≡C 838

C–O 358

C=O 746

C≡O 1077

F–F 158

Cl–Cl 243

C–H 413

H–H 436

H–O 464

O=O 498

AVERAGE BOND ENERGIES OF COMMON BONDSAVERAGE BOND ENERGIES OF COMMON BONDSThe amount of energy required to break these bondsThe amount of energy required to break these bonds AVERAGE BOND ENERGIES OF COMMON BONDSAVERAGE BOND ENERGIES OF COMMON BONDSThe amount of energy required to break these bondsThe amount of energy required to break these bonds

Use these bond energy values to determine the ∆ Hrxn of methane combustionUse these bond energy values to determine the ∆ Hrxn of methane combustion∆∆ HHrxnrxn = ∑ E (bonds broken) - ∑ E (bonds formed) = ∑ E (bonds broken) - ∑ E (bonds formed)

Example - SolutionExample - Solution

COCO(g) (g) + H+ H22OO(g) (g) CO CO2(g)2(g) + H + H2(g)2(g)

bond enthalpy (kJ mol-1)

C-O in carbon monoxide

+1077

C-O in carbon dioxide

+746

O-H +464

H-H +436

ΔΔH = 1077 + 2(464) – 2(746) – 436H = 1077 + 2(464) – 2(746) – 436

ΔH = + 77 kJ mol ΔH = + 77 kJ mol -1-1

5 Energetics5 EnergeticsCompiled by: Robert Slider (2011)Compiled by: Robert Slider (2011)Please share this resource with othersPlease share this resource with others

5 Energetics Year 11 DP Chemistry R Slider. 5.1 Exothermic and Endothermic.

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