ENTHALPY CHANGESA guide for A level students

KNOCKHARDY PUBLISHING

ENTHALPY CHANGESINTRODUCTIONThis Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available.Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...www.argonet.co.uk/users/hoptonj/sci.htm

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ENTHALPY CHANGESCONTENTS Thermodynamics Enthalpy changes Standard enthalpy values Standard enthalpy of formation Standard enthalpy of combustion Enthalpy of neutralisation Bond dissociation enthalpy Hesss law Calculating enthalpy changes using bond enthalpies Calculating enthalpy changes using enthalpy of formation values Calculating enthalpy changes using enthalpy of combustion values Practical measurement Check list

Before you start it would be helpful to

be able to balance equations be confident with simple arithmetical operationsENTHALPY CHANGES

THERMODYNAMICSFirst LawEnergy can be neither created nor destroyed but It can be converted from one form to another

Energychanges all chemical reactions are accompanied by some form of energy change changes can be very obvious (e.g. coal burning) but can go unnoticed

ExothermicEnergy is given out

Endothermic Energy is absorbed

ExamplesExothermiccombustion of fuelsrespiration (oxidation of carbohydrates)

Endothermicphotosynthesisthermal decomposition of calcium carbonate

THERMODYNAMICS - ENTHALPY CHANGESEnthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH , delta H

Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants

Enthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH , delta H

Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants

Enthalpy of reactants > products DH = - ive EXOTHERMIC Heat given outTHERMODYNAMICS - ENTHALPY CHANGES

Enthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH , delta H

Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants

Enthalpy of reactants > products Enthalpy of reactants < products DH = - ive DH = + iveEXOTHERMIC Heat given out ENDOTHERMIC Heat absorbedREACTION CO-ORDINATEENTHALPYTHERMODYNAMICS - ENTHALPY CHANGES

Why astandard? enthalpy values vary according to the conditions a substance under these conditions is said to be in its standard state ...

Pressure:- 100 kPa (1 atmosphere)

A stated temperature usually 298K (25C)

as a guide, just think of how a substance would be under normal lab conditions assign the correct subscript [(g), (l) or (s) ] to indicate which state it is in any solutions are of concentration 1 mol dm-3 to tell if standard conditions are used we modify the symbol for DH .

Enthalpy Change Standard Enthalpy Change(at 298K)STANDARD ENTHALPY CHANGES

STANDARD ENTHALPY OF FORMATIONDefinition The enthalpy change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states.

Symbol DHf

Values Usually, but not exclusively, exothermic

Example(s)C(graphite) + O2(g) > CO2(g)

H2(g) + O2(g) > H2O(l)

2C(graphite) + O2(g) + 3H2(g) > C2H5OH(l)

Notes Only ONE MOLE of product on the RHS of the equation

Elements In their standard states have zero enthalpy of formation.

Carbon is usually taken as the graphite allotrope.

Definition The enthalpy change when ONE MOLE of a substance undergoes complete combustion under standard conditions. All reactants and products are in their standard states.

Symbol DHc

Values Always exothermic

Example(s)C(graphite) + O2(g) > CO2(g)

H2(g) + O2(g) > H2O(l)

C2H5OH(l) + 3O2(g) > 2CO2(g) + 3H2O(l)

NotesAlways only ONE MOLE of what you are burning on the LHS of the equationTo aid balancing the equation, remember...you get one carbon dioxide molecule for every carbon atom in the original and one water molecule for every two hydrogen atomsWhen you have done this, go back and balance the oxygen.STANDARD ENTHALPY OF COMBUSTION

Definition The enthalpy change when ONE MOLE of water is formed from its ions in dilute solution.

Values Exothermic

Equation H+(aq) + OH(aq) > H2O(l)

Notes A value of -57kJ mol-1 is obtained when strong acids react with strong alkalis.

See later slides for practical details of measurementENTHALPY OF NEUTRALISATION

Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms.

Values Endothermic - Energy must be put in to break any chemical bond

Examples Cl2(g) > 2Cl(g) O-H(g) > O(g) + H(g)

Notes strength of bonds also depends on environment; MEAN values quoted making bonds is exothermic as it is the opposite of breaking a bond for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation smaller bond enthalpy = weaker bond = easier to break

BOND DISSOCIATION ENTHALPY

Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms.

Values Endothermic - Energy must be put in to break any chemical bond

Examples Cl2(g) > 2Cl(g) O-H(g) > O(g) + H(g)

Notes strength of bonds also depends on environment; MEAN values quoted making bonds is exothermic as it is the opposite of breaking a bond for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation smaller bond enthalpy = weaker bond = easier to break

Mean Values H-H 436 H-F562 N-N163 C-C 346 H-Cl431 N=N409 C=C 611 H-Br366 NN944 CC 837 H-I299 P-P172 C-O 360 H-N388 F-F158 C=O 743 H-O463 Cl-Cl242 C-H 413 H-S338 Br-Br193 C-N 305 H-Si318 I-I151 C-F 484 P-H322 S-S264 C-Cl 338 O-O146 Si-Si176BOND DISSOCIATION ENTHALPYAverage (mean) values are quoted because the actual value depends on the environment of the bond i.e. where it is in the moleculeUNITS = kJ mol-1

The enthalpy change is independent of the path taken

HowThe enthalpy change going fromA to B can be found by adding thevalues of the enthalpy changes forthe reactions A to X, X to Y and Y to B.

DHr = DH1 + DH2 + DH3

HESSS LAW

The enthalpy change is independent of the path taken

HowThe enthalpy change going fromA to B can be found by adding thevalues of the enthalpy changes forthe reactions A to X, X to Y and Y to B.

DHr = DH1 + DH2 + DH3

If you go in the opposite direction of an arrow, you subtract the value ofthe enthalpy change.

e.g. DH2 = - DH1 + DHr - DH3

The values of DH1 and DH3 have been subtracted because the routeinvolves going in the opposite direction to their definition.HESSS LAW

The enthalpy change is independent of the path taken

Useapplying Hesss Law enables one to calculate enthalpy changes fromother data, such as...

changes which cannot be measured directly e.g. Lattice Enthalpy

enthalpy change of reaction from bond enthalpy

enthalpy change of reaction from DHc

enthalpy change of formation from DHfHESSS LAW

TheoryImagine that, during a reaction, all the bonds of reacting species are brokenand the individual atoms join up again but in the form of products. Theoverall energy change will depend on the difference between the energyrequired to break the bonds and that released as bonds are made.

energy released making bonds > energy used to break bonds ... EXOTHERMIC energy used to break bonds > energy released making bonds ... ENDOTHERMICEnthalpy of reaction from bond enthalpiesStep 1Energy is put in to break bonds to form separate, gaseous atoms

Step 2The gaseous atoms then combine to form bonds and energy is releasedits value will be equal and opposite to that of breaking the bondsApplying Hesss Law DHr = Step 1 + Step 2

Enthalpy of reaction from bond enthalpies DH = bond enthalpies bond enthalpies of reactants of productsAlternative viewStep 1Energy is put in to break bonds to form separate, gaseous atoms.

Step 2Gaseous atoms then combineto form bonds and energy is released; its value will be equaland opposite to that of breaking the bondsDHr = Step 1 - Step 2

Because, in Step 2 the route involvesgoing in the OPPOSITE DIRECTIONto the defined change of bond enthalpy,its value is subtracted.

Calculate the enthalpy change for the hydrogenation of etheneEnthalpy of reaction from bond enthalpies

Calculate the enthalpy change for the hydrogenation of etheneEnthalpy of reaction from bond enthalpiesDH21 x C=C bond @ 611= 611 kJ4 x C-H bonds @ 413= 1652 kJ1 x H-H bond @ 436= 436 kJTotal energy to break bonds of reactants= 2699 kJ

DH31 x C-C bond @ 346= 346 kJ6 x C-H bonds @ 413= 2478 kJTotal energy to break bonds of products= 2824 kJ

Applying Hesss Law DH1 = DH2 DH3 = (2699 2824) = 125 kJ

If you formed the products from their elements you should need the same amounts of every substance as if you formed the reactants from their elements.

Enthalpy of formation tends to be an exothermic processEnthalpy of reaction from enthalpies of formation

Step 1Energy is released as reactantsare formed from their elements.

Step 2Energy is released as productsare formed from their elements.DHr = - Step 1 + Step 2

orStep 2 - Step 1

In Step 1 the route involves going in the OPPOSITE DIRECTION to the defined enthalpy change, its value is subtracted.Enthalpy of reaction from enthalpies of formation DH = DHf of products DHf of reactants

Sample calculationCalculate the standard enthalpy change for the following reaction, given that the standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286, +33 and -173 kJ mol-1 respectively; the value for oxygen is ZERO as it is an element

2H2O(l) + 4NO2(g) + O2(g) > 4HNO3(l)By applying Hesss Law ... The Standard Enthalpy of Reaction DHr will be...

PRODUCTS REACTANTS [ 4 x DHf of HNO3 ] minus [ (2 x DHf of H2O) + (4 x DHf of NO2) + (1 x DHf of O2) ]

DHr = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0

ANSWER = - 252 kJEnthalpy of reaction from enthalpies of formation DH = DHf of products DHf of reactants

Enthalpy of reaction from enthalpies of combustionIf you burned all the products you should get the same amounts of oxidation products such a CO2 and H2O as if you burned the reactants.

Enthalpy of combustion is an exothermic process

DH = DHc of reactants DHc of products Enthalpy of reaction from enthalpies of combustionStep 1Energy is released as reactantsundergo combustion.

Step 2Energy is released as productsundergo combustion.DHr = Step 1 - Step 2

Because, in Step 2 the route involves going in the OPPOSITE DIRECTION to the defined change of Enthalpy of Combustion, its value is subtracted.

Enthalpy of reaction from enthalpies of combustionSample calculationCalculate the standard enthalpy of formation of methane; the standard enthalpies of combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1 .

C(graphite) + 2H2(g) > CH4(g)By applying Hesss Law ... The Standard Enthalpy of Reaction DHr will be...

REACTANTS PRODUCTS [ (1 x DHc of C) + (2 x DHc of H2) ] minus [ 1 x DHc of CH4O ]

DHr = 1 x (-394) + 2 x (-286) - 1 x (-890)

ANSWER = - 74 kJ mol-1 DH = DHc of reactants DHc of products

Calorimetryinvolves the practical determination of enthalpy changesusually involves heating (or cooling) known amounts of waterwater is heated upreaction is EXOTHERMICwater cools downreaction is ENDOTHERMIC

CalculationThe energy required to change the temperatureof a substance can be calculated using...q = m x c x DT

where q= heat energykJ m= masskg c= Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] DT= change in temperatureKMEASURING ENTHALPY CHANGES

Calorimetryinvolves the practical determination of enthalpy changesusually involves heating (or cooling) known amounts of waterwater is heated upreaction is EXOTHERMICwater cools downreaction is ENDOTHERMIC

CalculationThe energy required to change the temperatureof a substance can be calculated using...q = m x c x DT

where q= heat energykJ m= masskg c= Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] DT= change in temperatureK

Example On complete combustion, 0.18g of hexane raised the temperature of100g water from 22C to 47C. Calculate its enthalpy of combustion.

Heat absorbed by the water (q) = 0.1 x 4.18 x 25 = 10.45 kJMoles of hexane burned = mass / Mr = 0.18 / 86 = 0.00209Enthalpy change = heat energy / moles= 10.45 / 0.00209ANS= 5000 kJ mol -1MEASURING ENTHALPY CHANGES

Example 1 - graphicalThe temperature is taken every halfminute before mixing the reactants.

Reactants are mixed after three minutes.

Further readings are taken every halfminute as the reaction mixture cools.

Extrapolate the lines as shown andcalculate the value of DT.

MEASURING ENTHALPY CHANGES

Example 1 - graphicalThe temperature is taken every halfminute before mixing the reactants.

Reactants are mixed after three minutes.

Further readings are taken every halfminute as the reaction mixture cools.

Extrapolate the lines as shown andcalculate the value of DT.

Example calculationWhen 0.18g of hexane underwent complete combustion, it raised the temperature of100g (0.1kg) water from 22C to 47C. Calculate its enthalpy of combustion.

Heat absorbed by the water (q) = m C DT = 0.1 x 4.18 x 25 = 10.45 kJ

Moles of hexane burned= mass / Mr= 0.18 / 86= 0.00209Enthalpy change = heat energy / moles= 10.45 / 0.00209= 5000 kJ mol -1MEASURING ENTHALPY CHANGES

Example 225cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20C. The highest temperature reached by the solution was 33C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1]

NaOH + HCl > NaCl + H2O

MEASURING ENTHALPY CHANGES

Example 225cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20C. The highest temperature reached by the solution was 33C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1]

NaOH + HCl > NaCl + H2O

Temperature rise (DT) = 306K 293K= 13KVolume of resulting solution= 25 + 25 = 50cm3 = 0.05 dm3Equivalent mass of water= 50g= 0.05 kg(density is 1g per cm3) Heat absorbed by the water (q)= m x c x DT = 0.05 x 4.18 x 13= 2.717 kJ

MEASURING ENTHALPY CHANGES

Example 225cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20C. The highest temperature reached by the solution was 33C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1]

NaOH + HCl > NaCl + H2O

Temperature rise (DT) = 306K 293K= 13KVolume of resulting solution= 25 + 25 = 50cm3 = 0.05 dm3Equivalent mass of water= 50g= 0.05 kg(density is 1g per cm3) Heat absorbed by the water (q)= m x c x DT = 0.05 x 4.18 x 13= 2.717 kJ

Moles of HCl reacting= 2 x 25/1000= 0.05 molMoles of NaOH reacting= 2 x 25/1000= 0.05 molMoles of water produced= 0.05 mol

Enthalpy change per mol (DH)= heat energy / moles of water = 2.717 / 0.05= 54.34 kJ mol -1MEASURING ENTHALPY CHANGES

REVISION CHECKWhat should you be able to do?Explain the difference between an endothermic and exothermic reactionUnderstand the reasons for using standard enthalpy changesRecall the definitions of enthalpy of formation and combustionWrite equations representing enthalpy of combustion and formationRecall and apply Hesss lawRecall the definition of bond dissociation enthalpy Calculate standard enthalpy changes using bond enthalpy valuesCalculate standard enthalpy changes using enthalpies of formation and combustionKnow simple calorimetry methods for measuring enthalpy changesCalculate enthalpy changes from calorimetry measurementsCAN YOU DO ALL OF THESE? YES NO

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2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHINGENTHALPY CHANGES

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