PPP Enthalpy Changes
Transcript of 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 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
Enthalpy of reactants > products
DH= - ive
EXOTHERMIC Heat given out
REACTION CO-ORDINATE
ENTHALPY
THERMODYNAMICS - ENTHALPY CHANGES
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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
Enthalpy of reactants > products Enthalpy of reactants < products
DH= - ive DH= + ive
EXOTHERMIC Heat given out ENDOTHERMIC Heat absorbed
REACTION CO-ORDINATE
ENTHALPY
REACTION CO-ORDINATE
ENTHALPY
THERMODYNAMICS - ENTHALPY CHANGES
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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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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
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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The enthalpy change is independent of the path taken
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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Enthalpy of reaction from bond enthalpies
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
Enthalpy of reaction from bond enthalpies
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Calculate the enthalpy change for the hydrogenation of ethene
Enthalpy of reaction from bond enthalpies
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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Calculate the enthalpy change for the hydrogenation of ethene
Enthalpy of reaction from bond enthalpies
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
Enthalpy of reaction from enthalpies of formation
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
Enthalpy of reaction from enthalpies of formation
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
Enthalpy of reaction from enthalpies of combustion
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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Enthalpy of reaction from enthalpies of combustion
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
MEASURING ENTHALPY CHANGES
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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.
MEASURING ENTHALPY CHANGES
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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
MEASURING ENTHALPY CHANGES
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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
MEASURING ENTHALPY CHANGES
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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
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
MEASURING ENTHALPY CHANGES
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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
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
MEASURING ENTHALPY CHANGES
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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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WELL DONE!Try some past paper questions
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ENTH LPY
CH NGES
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