PPP Enthalpy Changes

8/10/2019 PPP Enthalpy Changes

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ENTH LPY

CH NGES

A guide for A level students


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ENTHALPY CHANGES

INTRODUCTIONThis Powerpointshow 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.knockhardy.org.uk/sci.htm

Navigationis achieved by...

either clicking on the grey arrows at the foot of each page

or using the left and right arrow keys on the keyboard


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Before you start it would be helpful to

be able to balance equations

be confident with simple arithmetical operations

ENTHALPY CHANGES


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THERMODYNAMICS

First Law Energy can be neither created nor destroyed butIt 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

Exothermic Energy is given out

Endothermic Energy is absorbed

Examples Exothermic combustion of fuels

respiration (oxidation of carbohydrates)

Endothermic photosynthesis

thermal decomposition of calcium carbonate


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THERMODYNAMICS - 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 CHANGEwritten as the symbol DH, delta H

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


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Enthalpy of reactants > products

DH= - ive

EXOTHERMIC Heat given out

REACTION CO-ORDINATE

ENTHALPY



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Enthalpy of reactants > products Enthalpy of reactants < products

DH= - ive DH= + ive

EXOTHERMIC Heat given out ENDOTHERMIC Heat absorbed


ENTHALPY


ENTHALPY



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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 forDH.

Enthalpy Change Standard Enthalpy Change(at 298K)

STANDARD ENTHALPY CHANGES


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STANDARD ENTHALPY OF FORMATION

Definition The enthalpy change when ONE MOLE of a compound is formed in itsstandard 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

ElementsIn their standard states have zero enthalpy of formation.

Carbon is usually taken as the graphite allotrope.


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

Values Exothermic

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

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

See later slides for practical details of measurement

ENTHALPY OF NEUTRALISATION


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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


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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-F 562 N-N 163C-C 346 H-Cl 431 N=N 409

C=C 611 H-Br 366 NN 944

CC 837 H-I 299 P-P 172

C-O 360 H-N 388 F-F 158

C=O 743 H-O 463 Cl-Cl 242

C-H 413 H-S 338 Br-Br 193

C-N 305 H-Si 318 I-I 151

C-F 484 P-H 322 S-S 264

C-Cl 338 O-O 146 Si-Si 176

BOND DISSOCIATION ENTHALPY

Average (mean)values are quotedbecause the actual

value depends onthe environment ofthe bond i.e. whereit is in the molecule

UNITS = kJ mol-1


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The enthalpy change is independent of the path taken

How The enthalpy change going from

A to B can be found by adding thevalues of the enthalpy changes for

the reactions A to X, X to YandY to B.

DHr = DH + DH2+ DH3

HESSS LAW


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How The enthalpy change going from

A to B can be found by adding thevalues of the enthalpy changes for

the reactions A to X, X to YandY to B.

DHr = DH + DH2+ DH3

If you go in the opposite direction of an arrow, you subtract the value of

the enthalpy change.

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

The values of DH

and

DH3 have been subtracted because the routeinvolves going in the opposite direction to their definition.

HESSS LAW


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Use applying Hesss Law enables one to calculate enthalpy changes from

other data, such as...

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

enthalpy change of reaction from bond enthalpy

enthalpy change of reaction fromDH

c

enthalpy change of formation from DHf

HESSS LAW


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Theory Imagine 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 energy

required 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 ... ENDOTHERMIC

Enthalpy of reaction from bond enthalpies

Step 1 Energy is put in to break bonds to form separate, gaseous atoms

Step 2 The gaseous atoms then combine to form bonds and energy is releasedits value will be equal and opposite to that of breaking the bonds

Applying Hesss Law DHr = Step 1 + Step 2


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DH = bond enthalpies bond enthalpiesof reactants of products

Alternative view

Step 1 Energy is put in to break bonds

to form separate, gaseous atoms.

Step 2 Gaseous atoms then combineto form bonds and energy isreleased; its value will be equaland opposite to that of breakingthe bonds

DHr = 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.

SUM OFTHE BONDENTHALPIES OF

THE REACTANTS

REACTANTS

PRODUCTS

ATOMS

DH

SUM OFTHEBONDENTHALPIES OFTHE PRODUCTS


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Calculate the enthalpy change for the hydrogenation of ethene



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DH2 1 x C=C bond @ 611 = 611 kJ4 x C-H bonds @ 413 = 1652 kJ1 x H-H bond @ 436 = 436 kJ

Total energy to break bonds of reactants = 2699 kJ


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DH2 1 x C=C bond @ 611 = 611 kJ4 x C-H bonds @ 413 = 1652 kJ1 x H-H bond @ 436 = 436 kJ

Total energy to break bonds of reactants = 2699 kJ

DH3 1 x C-C bond @ 346 = 346 kJ6 x C-H bonds @ 413 = 2478 kJ

Total energy to break bonds of products = 2824 kJ

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


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

Enthalpy of formation tends to be an exothermic process

Enthalpy of reaction from enthalpies of formation


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

Step 2 Energy is released as productsare formed from their elements.

DHr= -Step 1 + Step 2

or Step 2 - Step 1

In Step 1 the route involves going in theOPPOSITE DIRECTIONto the definedenthalpy change, its value is subtracted.

SUM OFTHEENTHALPIES OFFORMATION OF

THE REACTANTS

REACTANTS

PRODUCTS

ELEMENTS

DH

SUM OFTHEENTHALPIESOFFORMATIONOF THEPRODUCTS


DH = DHfof products DHfof reactants


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Sample calculationCalculate the standard enthalpy change for the following reaction, given that thestandard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286,+33 and -173 kJ mol-1respectively; 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 ReactionDH

rwill be...

PRODUCTS REACTANTS

[ 4 x DHfof HNO3 ] minus [ (2 x DHfof H2O) + (4 x DHfof NO2) + (1 x DHfof O2) ]

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

ANSWER = - 252 kJ


DH =

DHfof products

DHfof reactants


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Enthalpy of reaction from enthalpies of combustion

If you burned all the products you should get the same amounts of oxidation productssuch a CO2and H2O as if you burned the reactants.

Enthalpy of combustion is an exothermic process


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DH = DHcof reactants DHcof products


Step 1 Energy is released as reactantsundergo combustion.

Step 2 Energy is released as products

undergo combustion.

DHr = Step 1 - Step2

Because, in Step 2 the route involvesgoing in the OPPOSITE DIRECTION to thedefined change of Enthalpy ofCombustion, its value is subtracted.

SUM OFTHEENTHALPIES OFCOMBUSTIONOF THEREACTANTS

REACTANTS

PRODUCTS

OXIDATIONPRODUCTS

DH

SUM OFTHEENTHALPIES OF

COMBUSTION OFTHE PRODUCTS


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Sample calculationCalculate the standard enthalpy of formation of methane; the standard enthalpies ofcombustion 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 DHcof C)

+ (2 x DH

cof H

2) ] minus [ 1 x DH

cof CH

4]

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

ANSWER = - 76 kJ mol-1

DH

= DHcof reactants DHcof products


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Calorimetry involves the practical determination of enthalpy changesusually involves heating (or cooling) known amounts of water

water is heated up reaction is EXOTHERMIC

water cools down reaction is ENDOTHERMIC

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

where q = heat energy kJ

m = mass kgc = Specific Heat Capacity kJ K -1kg -1 [ water is 4.18 ]

DT = change in temperature K

MEASURING ENTHALPY CHANGES


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Calorimetry involves the practical determination of enthalpy changesusually involves heating (or cooling) known amounts of water

water is heated up reaction is EXOTHERMIC

water cools down reaction is ENDOTHERMIC

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

where q = heat energy kJ

m = mass kgc = Specific Heat Capacity kJ K -1kg -1 [ water is 4.18 ]

DT = change in temperature K

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 kJ

Moles of hexane burned = mass / Mr = 0.18 / 86= 0.00209

Enthalpy change = heat energy / moles = 10.45 / 0.00209

ANS = 5000 kJ mol -1



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Example 1 - graphical

The 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 ofDT.



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Example 1 - graphical

The 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 ofDT.

Example calculation

When 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 = 0.00209

Enthalpy change = heat energy / moles = 10.45 / 0.00209 = 5000 kJ mol -1



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Example 2

25cm3 of 2.0M HCl was added to 25cm3of 2.0M NaOH in an insulated beaker. The initialtemperature of both solutions was 20C. The highest temperature reached by the

solution was 33C. Calculate the Molar Enthalpy of Neutralisation. [The specific heatcapacity (c) of water is 4.18 kJ K -1kg -1]

NaOH + HCl > NaCl + H2O



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Example 2




Temperature rise (DT) = 306K293K = 13K

Volume of resulting solution = 25 + 25 = 50cm3 = 0.05 dm3

Equivalent 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



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Example 2




Temperature rise (DT) = 306K293K = 13K

Volume of resulting solution = 25 + 25 = 50cm3 = 0.05 dm3

Equivalent 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 mol

Moles 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 -1



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REVISION CHECK

What should you be able to do?

Explainthe difference between an endothermic and exothermic reaction

Understandthe reasons for using standard enthalpy changes

Recallthe definitions of enthalpy of formation and combustion

Writeequations representing enthalpy of combustion and formation

Recall and applyHesss law

Recallthe definition of bond dissociation enthalpy

Calculatestandard enthalpy changes using bond enthalpy values

Calculatestandard enthalpy changes using enthalpies of formation and combustion

Knowsimple calorimetry methods for measuring enthalpy changes

Calculateenthalpy changes from calorimetry measurements

CAN YOU DO ALL OF THESE? YES NO


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ENTH LPY

CH NGES

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