Upload
dale-dawson
View
220
Download
3
Embed Size (px)
Citation preview
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)1
ThermochemistryThermochemistry- Energy of Chemical Reactions
ContentsContents:• heat, work, forms of energy• specific heat and energies of phase changes • enthalpy changes in chemical reactions• standard enthalpies of formation • Hess’s law • estimating enthalpies of reaction from Bond Energies
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)2
CHEMICAL ENERGYCHEMICAL ENERGY
Chemical bonds are a source of energy• BOND BREAKING - requires energy• BOND MAKING - releases energy
In a chemical reaction:• if more energy is released in forming bonds than is used in breaking bonds then . . . reaction is EXOTHERMICEXOTHERMIC
• if more energy is used in breaking bonds than is released in
forming bonds then . . . reaction is ENDOTHERMIC
Energy is released as HEAT, LIGHT, SOUND, WORK
Energy can be provided by - LIGHT - photochemistry- WORK - electrochemistry- COOLING of surroundings
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)3
Energy and ChemistryEnergy and ChemistryENERGYENERGY is the capacity to do is the capacity to do workwork or transfer or transfer heatheat..
HEATHEAT is the form of energy that flows between 2 is the form of energy that flows between 2 samples because of a difference in temperature.samples because of a difference in temperature.
WORKWORK is the form of energy that results in a is the form of energy that results in a macroscopic displacement of matter such as gas macroscopic displacement of matter such as gas expansion or motion of an object expansion or motion of an object (force x distance)(force x distance)
Other forms of energy —Other forms of energy —• light light • electricalelectrical• kinetickinetic
• ChemicalChemical
• gravitational potentialgravitational potential• electrostatic potentialelectrostatic potential
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)4
Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat CapacityThermochemistry is the science of heat (energy) flow.
A difference in temperature leads to energy transfer.
The heat “lost” or “gained” is related to a) sample massb) change in T, andc) specific heat capacity by
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g) (T change, K)
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)5
Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity
SubstanceSubstance Spec. Heat (J/g•K)Spec. Heat (J/g•K)
HH22OO 4.1844.184
AlAl 0.9020.902
glassglass 0.840.84
AluminumAluminum
WaterWater
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)6
Specific Heat Capacity - an exampleSpecific Heat Capacity - an exampleSpecific Heat Capacity - an exampleSpecific Heat Capacity - an example
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
where T = Tfinal - Tinitial = 37 - 310 = -273 Kq = (0.902 J/g•K)(25.0 g)(-273 K)q = -6160 J
negative sign of q heat is “lost by” or transferred from Alnegative sign of q heat is “lost by” or transferred from Al
Specific heat capacity =
heat lost or gained by substance (J)
(mass, g)(T change, K)
0.902 J/g.K
=
heat gain/lost = q = (specific heat)(mass)(T)heat gain/lost = q = (specific heat)(mass)(T)
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)7
Heat Transfer and Heat Transfer and Changes of StateChanges of State
Heat Transfer and Heat Transfer and Changes of StateChanges of State
Changes of state involve energyChanges of state involve energy
Ice Water333 J/g333 J/g
(Heat of FusionHeat of Fusion)
Water Vapor 2260 J/g2260 J/g
(Heat of vaporization)(Heat of vaporization)
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)8
Heating/Cooling Curve for WaterHeating/Cooling Curve for Water
11 22
33 44
Heat waterHeat water
Evaporate waterEvaporate water
Melt iceMelt ice
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)9
CHEMICALCHEMICAL REACTIVITYREACTIVITYCHEMICALCHEMICAL REACTIVITYREACTIVITY
• What drives chemical reactions? How do they occur?
The first is answered by THERMODYNAMICS and the second by KINETICS.
• In Ch. 4 we saw a number of “driving forces” for reactions that are PRODUCT-FAVORED.
• formation of a precipitate
• gas formation
• H2O formation (acid-base reaction)
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)10
• Energy transfer also allows us to predict reactivity.
• In general, reactions that transfer energy to their surroundings are “product-favored”.
• How do we describe heat transfer in chemical processes ?
CHEMICALCHEMICAL REACTIVITYREACTIVITYCHEMICALCHEMICAL REACTIVITYREACTIVITY
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)11
Heat Energy Transfer in Heat Energy Transfer in Physical & Chemical ProcessesPhysical & Chemical Processes
• COCO2 2 (s, -78 (s, -78 ooC) ---> COC) ---> CO2 2 (g, -78 (g, -78 ooC)C)
Heat flows into the Heat flows into the SYSTEMSYSTEM (solid CO (solid CO22) from the ) from the
SURROUNDINGSSURROUNDINGS in an E in an ENDOTHERMICNDOTHERMIC process.process.
heatheat
SurroundingsSurroundings
SystemSystem
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)12
• ENERGY is the capacity to do work or transfer heat.• HEAT is the form of energy that flows between 2 samples because of a difference in temperature.• WORK is the form of energy that results in a macroscopic displacement of matter such as gas expansion or motion of an object (force x distance)
In CO2 sublimation & expansion, the same amount the same amount
of of ENERGY flows from flows from surroundingssurroundings to systemsystem
If expanding gas is enclosed, part of the energy transfer appears in the form of WORK OF EXPANSION
wexp = - PV (for an ideal gas)
If expanding gas is not enclosed, the energy transfer appearsonly as HEATHEAT (CO2 gas gets warm).
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)13
FIRST LAW OF THERMODYNAMICSFIRST LAW OF THERMODYNAMICS
q = q = E - wE - w
heat energy transferred
Energy change
work doneby the system
Energy is conserved!Energy is conserved!
OR E = q + wE = q + w
NB - q and w positive when they are
transferred FROM surroundings
TO system
SurroundingsSurroundings
Heat
qsys > 0
System
Workwsys > 0
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)14
ENTHALPYENTHALPYENTHALPYENTHALPYMost chemical reactions occur at constant P, so
Heat transferred at constant P is called qp with
qp = H = E - w = E + P V = E+PV)
where H = enthalpy H is defined as E + PV)
H = heat transferred at constant PH = change in heat content of the system
H = HH = Hfinalfinal - H - Hinitialinitial
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)15
H = HH = Hfinalfinal - H - HinitialinitialH = HH = Hfinalfinal - H - Hinitialinitial
If Hfinal > Hinitial then H is positive
Process is ENDOTHERMIC
If Hfinal > Hinitial then H is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then H is negative
Process is EXOTHERMIC
If Hfinal < Hinitial then H is negative
Process is EXOTHERMIC
ENTHALPYENTHALPYENTHALPYENTHALPY
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)16
Endo- and ExothermicEndo- and ExothermicEndo- and ExothermicEndo- and Exothermic
SurroundingsSurroundings
Heatqsys > 0
System
ENDOTHERMICENDOTHERMICENDOTHERMICENDOTHERMIC
HeatHeat qqsyssys < 0 < 0
SurroundingsSurroundings
System
EXOTHERMICEXOTHERMIC
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)17
But the reverse reaction, the decomposition of water :
H2O(g) + 242 kJ ---> H2(g) + 1/2 O2(g)Endothermic reaction — heat is a “reactant”, H = +242 kJ. This does not occur spontaneously.
Consider the combustion of H2 to form water . .
H2(g) + 1/2 O2(g) ---> H2O(g) 242 kJExothermic reaction — heat is a “product”. H = -242 kJ. This is spontaneous and proceeds readily once initiated.
USING ENTHALPYUSING ENTHALPYUSING ENTHALPYUSING ENTHALPY
BUT . . . Decomposition of water can be made to occur by coupling to another, spontaneous process . . .
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)18
How can we How can we make Hmake H22 gas ? gas ?
N. Lewis, N. Lewis,
American Scientist,American Scientist,
Nov. 1995, page 534.Nov. 1995, page 534.
H2O
O2H2
LIGHT
S e m ic o nduc t o r Me t a l
w i r e
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)19
Making H2 from liquidliquid H2O involves two steps.
H2O(liq) + 44 kJ H2O(g)
H2O(g) + 242 kJ H2(g) + 1/2 O2(g)---------------------------------------------------H2O(liq) + 286 kJ H2(g) + 1/2 O2(g)
This is an example of HESS’S LAW —
If a reaction is the sum of 2 or more others, the net H is the sum of the H’s of the other rxns.
If a reaction is the sum of 2 or more others, the net H is the sum of the H’s of the other rxns.
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)20
Calc. Calc. HHrxnrxn for S(s) + 3/2 O for S(s) + 3/2 O22(g) --> SO(g) --> SO33(g)(g)
knowing thatknowing that
S(s) + OS(s) + O22(g) --> SO(g) --> SO22(g) (g) HH11 = -320.5 kJ = -320.5 kJ
SOSO22(g) + 1/2 O(g) + 1/2 O22(g) --> SO(g) --> SO33(g) (g) HH22 = -75.2 kJ = -75.2 kJ
Hess’s Law - a second example :Hess’s Law - a second example :
The two rxns. add to give the desired rxn.,
S(s) + 3/2 OS(s) + 3/2 O22(g) --> SO(g) --> SO33(g)(g)
so Hrxn = H1 + H2 = -395.7 kJ
HH33 = -395.7 kJ = -395.7 kJ
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)21
energy
2
S solid
SO3 gas
direct path
+ 3/2 O
H3 = -395.7 kJ SO2 gas
+O2H1 = -320.5 kJ
+ 1/2 O2H2 = -75.2 kJ
HH33 = -395.7 = -395.7 HH(2+3)(2+3) = -320.5 + -75.2 = -395.7 = -320.5 + -75.2 = -395.7
H along one path = H along one path = H along another pathH along another path
HH33 = -395.7 = -395.7 HH(2+3)(2+3) = -320.5 + -75.2 = -395.7 = -320.5 + -75.2 = -395.7
H along one path = H along one path = H along another pathH along another path
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)22
• This equation is valid because This equation is valid because H is a H is a STATE FUNCTIONSTATE FUNCTION
• These depend only on the These depend only on the state of state of the systemthe system and not how it got there. and not how it got there.
• Other Other state functionsstate functions include: include:
V, T, P, energy . . V, T, P, energy . .
H along one path =H along one path =
H along another pathH along another path
H along one path =H along one path =
H along another pathH along another path
— and your bank account!
• Unlike V, T, and P, one cannot Unlike V, T, and P, one cannot
measure absolute H. Can only measure measure absolute H. Can only measure H.H.
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)23
Standard Enthalpy ValuesStandard Enthalpy ValuesStandard Enthalpy ValuesStandard Enthalpy ValuesMost H values are labeled Ho
• P = 1 atmosphere ( = 760 torr = 101.3 kPa)
• Concentration = 1 mol/L
• T = usually 25 oC
• with all species in standard states
e.g., C = graphite and O2 = gas
o means measured under standard conditions
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)24
- the enthalpy change when 1 mol of compound is formed from elements under standard conditions.
Values: Kotz, Table 6.2 and Appendix KValues: Kotz, Table 6.2 and Appendix K
By definition, Hof = 0 for elements in
their standard states.
Hof = standard molar enthalpy of formationHo
f = standard molar enthalpy of formation
H2(g) + 1/2 O2(g) --> H2O(g)
Hof = -241.8 kJ/mol
H2(g) + 1/2 O2(g) --> H2O(g)
Hof = -241.8 kJ/mol
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)25
Using Standard Enthalpy ValuesUsing Standard Enthalpy ValuesUsing Standard Enthalpy ValuesUsing Standard Enthalpy Values
In general, when ALL
enthalpies of formation are known,
Horxn =
Hof (products)
- Hof (reactants)
Calculate Calculate H H of reaction?of reaction?
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)26
Horxn = Ho
f (prod) - Hof (react)Ho
rxn = Hof (prod) - Ho
f (react)
Example: Calculate the heat of combustion of ethanol, i.e., Ho
rxn for
C2H5OH(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(g)
Horxn = { 2 Ho
f (CO2) + 3 Hof (H2O) }
- {7/2 Hof (O2) + Ho
f (C2H5OH)} = { 2 (-393.5 kJ) + 3 (-241.8 kJ) } - {7/2 (0 kJ) + (-235.1 kJ)}
Horxn = -1035.5 kJ per mol of ethanol
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)27
• Given by D - the bond dissociation energy
D = energy required to break a bond in a gas phase molecule under standard conditions
e.g. CH4 (g) C (g) + 4 H (g) Hrxn = -1664 kJ = 4 * D(C-H)D(C-H) = 416 kJ per mole of C-H bonds
• D (C-H) (kJ/mol) varies slightly among compounds :
CH4 416 C2H6 392 C3H8 380
C2H4 432 C2H2 445 C6H6 448
Bond Energies Bond Energies (Kotz, sect. 9.4, pp 418-422)(Kotz, sect. 9.4, pp 418-422)
H
H
H
H
• D can be derived from Hrxn for atomization . . .
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)28
The GREATER the number of bonds (bond order) the HIGHER the bond dissociation energy
The GREATER the number of bonds (bond order) the HIGHER the bond dissociation energy
BOND D (kJ/mol) (Bond Energy) H—H 436 C—C 347 C=C 611 CC 837
N—N 159 NN 946see table 9.5 for Dissociation Energies of other bonds.
• D is similar for same bond in different molecules• Average values over many compounds are tabulated• Bond energy depends on bond order
Bond Energies Bond Energies
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)29
Using Bond EnergiesUsing Bond Energies
• Estimate the energy of the reaction
H—H + Cl—Cl ----> 2 H—ClH—H + Cl—Cl ----> 2 H—ClH—H + Cl—Cl ----> 2 H—ClH—H + Cl—Cl ----> 2 H—Cl
Net energy = Hrxn = energy required to break bonds - energy evolved when bonds are made
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)30
H—H = 436 kJ/molH—H = 436 kJ/molCl—Cl = 243 kJ/molCl—Cl = 243 kJ/mol H—Cl = 431 kJ/molH—Cl = 431 kJ/mol
H—H = 436 kJ/molH—H = 436 kJ/molCl—Cl = 243 kJ/molCl—Cl = 243 kJ/mol H—Cl = 431 kJ/molH—Cl = 431 kJ/mol
• Sum of H-H + Cl-Cl bond energies = 436 kJ + 243 kJ = +679 kJ
• 2 mol H-Cl bond energies = 862 kJ
• Net = H = +679 kJ - 862 kJ = -183 kJ
THEREFORE, , Hf for H-Cl is ? ? ?? ? ?
Estimating Hrxn for H—H + Cl—Cl 2 H—Cl
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)31
• Is the reaction exo- or endothermic?
Energy for bond breaking:
4 mol O—H bonds = 4 (464 kJ)
2 mol O—O bonds = 2 (138 kJ)
TOTAL = 2132 kJ
Energy from bond making :
1 mol O=O bonds = 498 kJ
4 mol O—H bonds = 4 (464 kJ)
TOTAL = 2354 kJ
EXAMPLE 2: Estimate the energy of the reaction
2 H—O—O—H ----> O=O + 2 H—O—H
• Which is larger: energy req’d to break bonds . . .
or energy evolved on making bonds?
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)32
2 H—O—O—H ---->
O=O + 2 H—O—H
More energy is evolved on More energy is evolved on making bonds than is making bonds than is expended in breaking expended in breaking bonds.bonds.
More energy is evolved on More energy is evolved on making bonds than is making bonds than is expended in breaking expended in breaking bonds.bonds.
The reaction is exothermic!The reaction is exothermic!The reaction is exothermic!The reaction is exothermic!
Net energy = +2132 kJ - 2354 kJ = - 222 kJNet energy = +2132 kJ - 2354 kJ = - 222 kJ
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)33
Enthalpies of Reaction from Bond EnergiesEnthalpies of Reaction from Bond Energies
REACTANTS
GaseousAtoms
PRODUCTS
ENDOTHERMICENDOTHERMIC
Bond Breaking costs morethan is gained by Bond Making
Bond Making releases more Ethan required for Bond Breaking
PRODUCTS
REACTANTS
EXOTHERMICEXOTHERMIC
GaseousAtoms
22 September, 1997 Chem 1A03E/1E03E
THERMOCHEMISTRY (Ch. 6)34
Key Concepts from Chapter 6: ThermochemistryKey Concepts from Chapter 6: Thermochemistry
• heat transfer - specific heat• phase transitions - heats of fusion, vaporization, etc
• First law of thermodynamics E = q - w
• endothermic versus exothermic reactions• enthalpy change in chemical reactions• Hess’s law
• standard molar enthalpies of formationHrxn = Hf(products) - Hf (reactants)
• bond energies Hrxn = D(bonds broken) - D(bonds made)