Thermochem

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1. 1THERMOCHEMISTRYTHERMOCHEMISTRY
ThermodynamicsThermodynamics
The study of Heat and Work The study of Heat and Work
and State Functionsand State Functions

2. 2
Energy & ChemistryEnergy & ChemistryEnergy & ChemistryEnergy & Chemistry
ENERGYENERGY is the capacity to is the capacity to
do work or transfer heat.do work or transfer heat.
HEATHEAT is the form of energy is the form of energy
that flows between 2 that flows between 2
objects because of their objects because of their
difference in temperature.difference in temperature.
Other forms of energy —Other forms of energy —
•lightlight
•electricalelectrical
•kinetic and potentialkinetic and potential

3. 3
Energy & ChemistryEnergy & Chemistry
•Burning peanuts peanuts
supply sufficient supply sufficient
energy to boil a cup energy to boil a cup
of water.of water.
•Burning sugar Burning sugar
(sugar reacts with (sugar reacts with
KClOKClO
33, a strong , a strong
oxidizing agent)oxidizing agent)

4. 4
Energy & ChemistryEnergy & Chemistry
•These reactions are These reactions are PRODUCT PRODUCT
FAVOREDFAVORED
•They proceed almost completely They proceed almost completely
from reactants to products, perhaps from reactants to products, perhaps
with some outside assistance.with some outside assistance.

5. 5
Energy & ChemistryEnergy & Chemistry
2 H2 H
22(g) + O(g) + O
22(g) --> (g) -->
2 H2 H
22O(g) + heat and lightO(g) + heat and light
This can be set up to provide This can be set up to provide
ELECTRIC ENERGYELECTRIC ENERGY in a in a
fuel cellfuel cell..
Oxidation:Oxidation:
2 H2 H
22 ---> 4 H ---> 4 H
++
+ 4 e + 4 e
--
Reduction: Reduction:
4 e4 e
--
+ O + O
22 + 2 H + 2 H
22O ---> 4 OHO ---> 4 OH
--
CCR, page 845

6. 6
Potential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy
Potential Potential
energy energy — —
energy a energy a
motionless motionless
body has by body has by
virtue of its virtue of its
position.position.

7. 7
•Positive and
negative particles
(ions) attract one
another.
•Two atoms can
bond
•As the particles
attract they have a
lower potential
energy
Potential EnergyPotential Energy
on the Atomic Scaleon the Atomic Scale
NaCl — composed of NaCl — composed of
NaNa
++
and Cl and Cl
--
ions. ions.

8. 8
•Positive and
negative particles
(ions) attract one
another.
•Two atoms can
bond
•As the particles
attract they have a
lower potential
energy
Potential EnergyPotential Energy
on the Atomic Scaleon the Atomic Scale

9. 9
Potential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy
Kinetic energy Kinetic energy
— energy of — energy of
motionmotion
• • TranslationTranslation

10. 10
Potential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic EnergyPotential & Kinetic Energy
Kinetic energy Kinetic energy
— energy of — energy of
motion.motion.
translate
rotate
vibrate
translate
rotate
vibrate

11. 11
Internal Energy (E)Internal Energy (E)
•PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U)
•Int. E of a chemical system Int. E of a chemical system
depends ondepends on
•number of particlesnumber of particles
•type of particlestype of particles
•temperaturetemperature

12. 12
Internal Energy (E)Internal Energy (E)
•PE + KE = Internal energy (E or U)PE + KE = Internal energy (E or U)
QuickTime™ and a
Graphics decompressor
are needed to see this picture.

13. 13
Internal Energy (E)Internal Energy (E)
•The higher the T The higher the T
the higher the the higher the
internal energyinternal energy
•So, use changes So, use changes
in T (∆T) to in T (∆T) to
monitor changes monitor changes
in E (∆E).in E (∆E).

14. 14
ThermodynamicsThermodynamics
•Thermodynamics is the science of heat
(energy) transfer.
Heat energy is associated Heat energy is associated
with molecular motions.with molecular motions.
Heat transfers until thermal equilibrium is
established.

15. 15
Directionality of Heat TransferDirectionality of Heat Transfer
•Heat always transfer from hotter object to
cooler one.
•EXOthermic: heat transfers from SYSTEM to
SURROUNDINGS .
T(system) goes downT(system) goes down
T(surr) goes upT(surr) goes up

16. 16
Directionality of Heat TransferDirectionality of Heat Transfer
•Heat always transfer from hotter object to
cooler one.
•ENDOthermic: heat transfers from
SURROUNDINGS to the SYSTEM.
T(system) goes upT(system) goes up
T (surr) goes downT (surr) goes down

17. 17
Energy & ChemistryEnergy & Chemistry
All of thermodynamics depends All of thermodynamics depends
on the law of on the law of
CONSERVATION OF ENERGYCONSERVATION OF ENERGY ..
•The total energy is unchanged The total energy is unchanged
in a chemical reaction.in a chemical reaction.
•If PE of products is less than If PE of products is less than
reactants, the difference must reactants, the difference must
be released as KE.be released as KE.

18. 18
Energy Change in Energy Change in
Chemical ProcessesChemical Processes
Reactants
Products
Kinetic
Energy
PE
PE of system dropped. KE increased. Therefore, PE of system dropped. KE increased. Therefore,
you often feel a T increase.you often feel a T increase.

19. 19
UNITS OF ENERGYUNITS OF ENERGYUNITS OF ENERGYUNITS OF ENERGY
1 calorie = heat required to 1 calorie = heat required to
raise temp. of 1.00 g of raise temp. of 1.00 g of
HH
22O by 1.0 O by 1.0
oo
C.C.
1000 cal = 1 kilocalorie = 1 1000 cal = 1 kilocalorie = 1
kcalkcal
1 kcal = 1 Calorie (a food 1 kcal = 1 Calorie (a food
“calorie”)“calorie”)
But we use the unit called But we use the unit called
the the JOULEJOULE
1 cal = 4.184 joules1 cal = 4.184 joules
James JouleJames Joule
1818-18891818-1889

20. 20
HEAT CAPACITYHEAT CAPACITY
The heat required to raise an
object’s T by 1 ˚C.
Which has the larger heat capacity?Which has the larger heat capacity?

21. 21
SpecificSpecific Heat Capacity Heat CapacitySpecificSpecific Heat Capacity Heat Capacity
How much energy is How much energy is
transferred due to T transferred due to T
difference?difference?
The heat The heat (q)(q) “lost” or “gained” “lost” or “gained”
is related to is related to
a)a) sample masssample mass
b) b) change in T andchange in T and
c) c) specific heat capacityspecific heat capacity
Specific heat capacity =

heat lost or gained by substance (J)
(mass, g)(T change, K)

22. 22
Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity
SubstanceSubstance Spec. Heat (J/g•K)Spec. Heat (J/g•K)
HH
22OO 4.1844.184
Ethylene glycolEthylene glycol2.392.39
AlAl 0.8970.897
glassglass 0.840.84
AluminumAluminum

23. 23
Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity
If 25.0 g of Al cool from If 25.0 g of Al cool from
310 310
oo
C to 37 C to 37
oo
C, how C, how
many joules of heat many joules of heat
energy are lost by energy are lost by
the Al?the Al?
Specific heat capacity =

heat lost or gained by substance (J)
(mass, g)(T change, K)

24. 24
Specific Heat CapacitySpecific Heat CapacitySpecific Heat CapacitySpecific Heat Capacity
If 25.0 g of Al cool from 310 If 25.0 g of Al cool from 310
oo
C to 37 C to 37
oo
C, how many C, how many
joules of heat energy are lost by the Al?joules of heat energy are lost by the Al?

heat gain/lose = q = (sp. ht.)(mass)(∆T)
where ∆T = Twhere ∆T = T
finalfinal - T - T
initialinitial
q = (0.897 J/g•K)(25.0 g)(37 - 310)Kq = (0.897 J/g•K)(25.0 g)(37 - 310)K
q = - 6120 Jq = - 6120 J
Notice that the negative sign on q signals Notice that the negative sign on q signals
heat “lost by” or transferred OUT of Al.heat “lost by” or transferred OUT of Al.
Notice that the negative sign on q signals Notice that the negative sign on q signals
heat “lost by” or transferred OUT of Al.heat “lost by” or transferred OUT of Al.

25. 25
Heat TransferHeat Transfer
No Change in StateNo Change in State
q transferred = (sp. ht.)(mass)(∆T)

26. 26Heat Transfer with Heat Transfer with
Change of StateChange of State
Heat Transfer with Heat Transfer with
Change of StateChange of State
Changes of state involve energy Changes of state involve energy (at constant T)(at constant T)
Ice + 333 J/g (heat of fusion) -----> Liquid waterIce + 333 J/g (heat of fusion) -----> Liquid water
q = (heat of fusion)(mass)q = (heat of fusion)(mass)

27. 27
Heat Transfer and Heat Transfer and
Changes of StateChanges of State
Heat Transfer and Heat Transfer and
Changes of StateChanges of State
Requires energy Requires energy
(heat).(heat).
This is the reasonThis is the reason
a)a)you cool down after you cool down after
swimming swimming
b)b)you use water to put you use water to put
out a fire.out a fire.
+ energy
Liquid ---> VaporLiquid ---> Vapor

28. 28
Heat waterHeat water
Evaporate waterEvaporate water
Melt iceMelt ice
Heating/Cooling Curve for WaterHeating/Cooling Curve for Water
Note that T is Note that T is
constant as ice meltsconstant as ice melts
Note that T is Note that T is
constant as ice meltsconstant as ice melts

29. 29
Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/g
Specific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•K
Heat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g
Heat of fusion of ice = 333 J/gHeat of fusion of ice = 333 J/g
Specific heat of water = 4.2 J/g•KSpecific heat of water = 4.2 J/g•K
Heat of vaporization = 2260 J/gHeat of vaporization = 2260 J/g
What quantity of heat is required to What quantity of heat is required to
melt 500. g of melt 500. g of iceice and heat the and heat the
water to water to steamsteam at 100 at 100
oo
C?C?
Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State
+333 J/g+333 J/g
+2260 J/g+2260 J/g

30. 30
How much heat is required to melt 500. g of How much heat is required to melt 500. g of
ice and heat the water to steam at 100 ice and heat the water to steam at 100
oo
C?C?
1. 1. To melt iceTo melt ice
q = (500. g)(333 J/g) = 1.67 x 10q = (500. g)(333 J/g) = 1.67 x 10
55
J J
2.2.To raise water from 0 To raise water from 0
oo
C to 100 C to 100
oo
CC
q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x
1010
55
J J
3.3.To evaporate water at 100 To evaporate water at 100
oo
CC
q = (500. g)(2260 J/g) = 1.13 x 10q = (500. g)(2260 J/g) = 1.13 x 10
66
J J
4. 4. Total heat energy = 1.51 x 10Total heat energy = 1.51 x 10
66
J = J =
1510 kJ1510 kJ
Heat & Changes of StateHeat & Changes of StateHeat & Changes of StateHeat & Changes of State

31. 31
ChemicalChemical Reactivity Reactivity
What drives chemical reactions? How do they What drives chemical reactions? How do they
occur?occur?
The first is answered by The first is answered by THERMODYNAMICSTHERMODYNAMICS
and the second by and the second by KINETICSKINETICS..
Have already seen a number of “driving Have already seen a number of “driving
forces” for reactions that are forces” for reactions that are PRODUCT-PRODUCT-
FAVOREDFAVORED..
••formation of a precipitateformation of a precipitate
••gas formationgas formation
••HH
22O formation (acid-base reaction)O formation (acid-base reaction)
••electron transfer in a batteryelectron transfer in a battery

32. 32
ChemicalChemical Reactivity Reactivity
But energy transfer also allows us to predict But energy transfer also allows us to predict
reactivity.reactivity.
In general, reactions that transfer energy In general, reactions that transfer energy
to their surroundings are product-to their surroundings are product-
favored.favored.
So, let us consider heat transfer in chemical processes.So, let us consider heat transfer in chemical processes.

33. 33
Heat Energy Transfer in Heat Energy Transfer in
a Physical Processa Physical Process
COCO
2 2 (s, -78 (s, -78
oo
C) ---> COC) ---> CO
2 2 (g, -78 (g, -78
oo
C)C)
Heat transfers from surroundings to system in endothermic process.

34. 34
Heat Energy Transfer in Heat Energy Transfer in
a Physical Processa Physical Process
•COCO
2 2 (s, -78 (s, -78
oo
C) ---> C) --->
CO CO
2 2 (g, -78 (g, -78
oo
C)C)
•A regular array of A regular array of
molecules in a molecules in a
solid solid
-----> gas phase -----> gas phase
molecules. molecules.
•Gas molecules Gas molecules
have higher kinetic have higher kinetic
energy.energy.

35. 35
Energy Level Diagram Energy Level Diagram
for Heat Energy for Heat Energy
TransferTransfer
∆E = E(final) - E(initial)
= E(gas) - E(solid)
COCO
22 solid solid
COCO
22 gas gas

36. 36
Heat Energy Transfer in Heat Energy Transfer in
Physical ChangePhysical Change
•Gas molecules have higher Gas molecules have higher
kinetic energy.kinetic energy.
•Also, Also, WORKWORK is done by the is done by the
system in pushing aside the system in pushing aside the
atmosphere.atmosphere.
COCO
2 2 (s, -78 (s, -78
oo
C) ---> COC) ---> CO
2 2 (g, -78 (g, -78
oo
C)C)
Two things have happened!Two things have happened!

37. 37
FIRST LAW OF FIRST LAW OF
THERMODYNAMICSTHERMODYNAMICS
∆∆E = q + wE = q + w
heat energy transferredheat energy transferred
energyenergy
changechange
work donework done
by the by the
systemsystem
Energy is conserved!Energy is conserved!

38. 38
heat transfer outheat transfer out
(exothermic), -q(exothermic), -q
heat transfer inheat transfer in
(endothermic), +q(endothermic), +q
SYSTEMSYSTEM
∆E = q + w
w transfer inw transfer in
(+w)(+w)
w transfer outw transfer out
(-w)(-w)

39. 39
ENTHALPYENTHALPY
Most chemical reactions occur at constant P, soMost chemical reactions occur at constant P, so
and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small)
∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E
∆∆H = change in H = change in heat content heat content of the systemof the system
∆∆H = HH = H
finalfinal - H - H
initialinitial
and so ∆E = ∆H + w (and w is usually small)and so ∆E = ∆H + w (and w is usually small)
∆∆H = heat transferred at constant P ≈ ∆EH = heat transferred at constant P ≈ ∆E
∆∆H = change in H = change in heat content heat content of the systemof the system
∆∆H = HH = H
finalfinal - H - H
initialinitial
Heat transferred at constant P = qHeat transferred at constant P = q
pp
qq
pp = = H
∆
H
∆
where where H = enthalpyH = enthalpy
Heat transferred at constant P = qHeat transferred at constant P = q
pp
qq
pp = = H
∆
H
∆
where where H = enthalpyH = enthalpy

40. 40
If If HH
finalfinal < H < H
initialinitial then H is negative
∆
then H is negative
∆
Process is Process is EXOTHERMICEXOTHERMIC
If If HH
finalfinal < H < H
initialinitial then H is negative
∆
then H is negative
∆
Process is Process is EXOTHERMICEXOTHERMIC
If If HH
finalfinal > H > H
initialinitial then H is positive
∆
then H is positive
∆
Process is Process is ENDOTHERMICENDOTHERMIC
If If HH
finalfinal > H > H
initialinitial then H is positive
∆
then H is positive
∆
Process is Process is ENDOTHERMICENDOTHERMIC
ENTHALPYENTHALPY
∆∆H = HH = H
fina lfina l - H - H
initia linitia l

41. 41
Consider the formation of waterConsider the formation of water
HH
22(g) + 1/2 O(g) + 1/2 O
22(g) --> H(g) --> H
22O(g) + O(g) + 241.8 241.8
kJkJ
USING ENTHALPYUSING ENTHALPY
Exothermic reaction — heat is a “product” Exothermic reaction — heat is a “product”
and H = – 241.8 kJ
∆and H = – 241.8 kJ
∆

42. 42
Making Making liquidliquid H H
22O from HO from H
22
+ O+ O
22 involves involves twotwo
exoexothermic steps. thermic steps.
USING ENTHALPYUSING ENTHALPY
H
2
+ O
2
gas
Liquid H
2
OH
2
O vapor

43. 43
Making HMaking H
22O from HO from H
22 involves two steps. involves two steps.
HH
22(g) + 1/2 O(g) + 1/2 O
22(g) ---> H(g) ---> H
22O(g) + 242 kJO(g) + 242 kJ
HH
22O(g) ---> HO(g) ---> H
22O(liq) + 44 kJ O(liq) + 44 kJ
------------------------------------------------------------------
-----
HH
22(g) + 1/2 O(g) + 1/2 O
22(g) --> H(g) --> H
22O(liq) + 286 kJO(liq) + 286 kJ
Example of Example of HESS’S LAWHESS’S LAW——
If a rxn. is the sum of 2 or more If a rxn. is the sum of 2 or more
others, the net ∆H is the sum of the others, the net ∆H is the sum of the
∆H’s of the other rxns.∆H’s of the other rxns.
USING ENTHALPYUSING ENTHALPY

44. 44
Hess’s Law Hess’s Law
& Energy Level Diagrams& Energy Level Diagrams
Forming H
2
O can occur in a
single step or in a two
steps.
∆H
total
is the same no matter
which path is followed.

45. 45
Hess’s Law Hess’s Law
& Energy Level Diagrams& Energy Level Diagrams
Forming CO
2
can occur in a
single step or in a two steps.

∆H
total
is the same no matter
which path is followed.

46. 46
•This equation is valid because This equation is valid because
∆H is a ∆H is a STATE FUNCTIONSTATE FUNCTION
•These depend only on the state These depend only on the state
of the system and of the system and notnot on how on how
the system got there.the system got there.
•V, T, P, energy — and your bank V, T, P, energy — and your bank
account!account!
•Unlike V, T, and P, one cannot Unlike V, T, and P, one cannot
measure absolute H. Can only measure absolute H. Can only
measure ∆H.measure ∆H.
S S ∆∆H along one path =H along one path =
S S ∆∆H along another pathH along another path
S S ∆∆H along one path =H along one path =
S S ∆∆H along another pathH along another path

47. 47
Standard Enthalpy ValuesStandard Enthalpy Values
Most H values are labeled
∆
Most H values are labeled
∆
H
∆
H
∆
oo

Measured under Measured under standard conditionsstandard conditions
P = 1 bar = 10P = 1 bar = 10
55
Pa = 1 atm / Pa = 1 atm /
1.01325 1.01325 Concentration = 1 Concentration = 1
mol/Lmol/L
T = usually 25 T = usually 25
oo
CC
with all species in standard stateswith all species in standard states
e.g., C = graphite and Oe.g., C = graphite and O
22 = gas = gas

48. 48
Enthalpy ValuesEnthalpy Values
HH
22(g) + 1/2 O(g) + 1/2 O
22(g) --> H(g) --> H
22O(g)O(g)
∆∆H˚ = -242 kJH˚ = -242 kJ
2 H2 H
22(g) + O(g) + O
22(g) --> 2 H(g) --> 2 H
22O(g)O(g)
∆∆H˚ = -484 kJH˚ = -484 kJ
HH
22O(g) ---> HO(g) ---> H
22(g) + 1/2 O(g) + 1/2 O
22(g) (g)
∆∆H˚ = +242 kJH˚ = +242 kJ
HH
22(g) + 1/2 O(g) + 1/2 O
22(g) --> H(g) --> H
22O(liquid)O(liquid)
∆∆H˚ = -286 kJH˚ = -286 kJ
Depend on Depend on how the reaction is writtenhow the reaction is written and on phases and on phases
of reactants and productsof reactants and products
Depend on Depend on how the reaction is writtenhow the reaction is written and on phases and on phases
of reactants and productsof reactants and products

49. 49
Standard Enthalpy ValuesStandard Enthalpy Values
NIST (Nat’l Institute for Standards and NIST (Nat’l Institute for Standards and
Technology) gives values ofTechnology) gives values of
∆∆HH
ff
oo
= standard molar enthalpy of = standard molar enthalpy of
formationformation
— — the enthalpy change when 1 mol of the enthalpy change when 1 mol of
compound is formed from elements compound is formed from elements
under standard conditions.under standard conditions.
See Table 6.2See Table 6.2

50. 50
∆∆HH
ff
oo
, standard molar , standard molar
enthalpy of formationenthalpy of formation
Enthalpy change when 1 mol of Enthalpy change when 1 mol of
compound is formed from the compound is formed from the
corresponding elements under corresponding elements under
standard conditionsstandard conditions
HH
22(g) + 1/2 O(g) + 1/2 O
22(g) --> H(g) --> H
22O(g)O(g)
∆∆HH
ff
oo
(H (H
22O, g)= -241.8 kJ/molO, g)= -241.8 kJ/mol
By definition, By definition,
∆∆HH
ff
oo
= 0 for elements in their standard = 0 for elements in their standard
states.states.

51. 51
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
Use H ’s to calculate
∆ ˚
Use H ’s to calculate
∆ ˚
enthalpy changeenthalpy change
forfor

HH
22O(g) + C(graphite) --> HO(g) + C(graphite) --> H
22(g) + CO(g)(g) + CO(g)
(product is called “(product is called “water gaswater gas”)”)

52. 52
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
HH
22O(g) + C(graphite) --> HO(g) + C(graphite) --> H
22(g) + CO(g)(g) + CO(g)
From reference books we findFrom reference books we find
•HH
22(g) + 1/2 O(g) + 1/2 O
22(g) --> H(g) --> H
22O(g) H
∆
O(g) H
∆
ff = - 242
˚
= - 242
˚
kJ/molkJ/mol
•C(s) + 1/2 OC(s) + 1/2 O
22(g) --> CO(g)(g) --> CO(g) H
∆
H
∆
ff = - 111
˚
= - 111
˚
kJ/molkJ/mol

53. 53
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
HH
22O(g) --> HO(g) --> H
22(g) + 1/2 O(g) + 1/2 O
22(g) H
∆
(g) H
∆
oo
= +242 = +242
kJkJ
C(s) + 1/2 OC(s) + 1/2 O
22(g) --> CO(g)(g) --> CO(g) H
∆
H
∆
oo
= -111 = -111
kJkJ
-----------------------------------------------------------------------
---------
To convert 1 mol of water to 1 mol each To convert 1 mol of water to 1 mol each
of Hof H
22 and CO and CO requiresrequires 131 kJ of energy. 131 kJ of energy.
The “water gas” reaction is The “water gas” reaction is ENDOENDOthermic.thermic.
HH
22O(g) + C(graphite) --> HO(g) + C(graphite) --> H
22(g) + CO(g)(g) + CO(g)
∆∆HH
oo
netnet = +131 kJ = +131 kJ
HH
22O(g) + C(graphite) --> HO(g) + C(graphite) --> H
22(g) + CO(g)(g) + CO(g)
∆∆HH
oo
netnet = +131 kJ = +131 kJ

54. 54
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
In general, when In general, when ALLALL
enthalpies of formation are enthalpies of formation are
known: known:
Calculate ∆H of Calculate ∆H of
reaction?reaction?
∆∆HH
oo
rxnrxn = = SS H
∆
H
∆
ff
oo
(products) - (products) - SS H
∆
H
∆
ff
oo
(reactants)(reactants)∆∆HH
oo
rxnrxn = = SS H
∆
H
∆
ff
oo
(products) - (products) - SS H
∆
H
∆
ff
oo
(reactants)(reactants)
Remember that ∆ always = final – initial

55. 55
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
Calculate the heat of combustion of Calculate the heat of combustion of
methanol, i.e., ∆Hmethanol, i.e., ∆H
oo
rxnrxn for for
CHCH
33OH(g) + 3/2 OOH(g) + 3/2 O
22(g) --> CO(g) --> CO
22(g) + 2 H(g) + 2 H
22O(g)O(g)
∆∆HH
oo
rxnrxn = = SS ∆H ∆H
ff
oo
(prod) - (prod) - SS ∆H ∆H
ff
oo
(react)(react)

56. 56
Using Standard Enthalpy ValuesUsing Standard Enthalpy Values
∆∆HH
oo
rxnrxn = ∆H = ∆H
ff
oo
(CO(CO
22) + 2 ∆H) + 2 ∆H
ff
oo
(H(H
22O) O)
- {3/2 ∆H- {3/2 ∆H
ff
oo
(O(O
22) + ∆H) + ∆H
ff
oo
(CH(CH
33OH)} OH)}
= (-393.5 kJ) + 2 (-241.8 kJ) = (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}- {0 + (-201.5 kJ)}
∆∆HH
oo
rxnrxn = -675.6 kJ per mol of methanol = -675.6 kJ per mol of methanol
CHCH
33OH(g) + 3/2 OOH(g) + 3/2 O
22(g) --> CO(g) --> CO
22(g) + 2 H(g) + 2 H
22O(g)O(g)
∆∆HH
oo
rxnrxn = = SS H
∆
H
∆
ff
oo
(prod) - (prod) - SS H
∆
H
∆
ff
oo
(react)(react)

57. 57
Measuring Heats of Reaction
CALORIMETRYCALORIMETRY
Constant Volume
“Bomb” Calorimeter
•Burn combustible
sample.
•Measure heat evolved
in a reaction.
•Derive ∆E for
reaction.

58. 58
CalorimetryCalorimetry
Some heat from reaction warms
water
q
water
= (sp. ht.)(water mass)(∆T)
Some heat from reaction warms
“bomb”
q
bomb
= (heat capacity, J/K)(∆T)
Total heat evolved = q
total
= q
water
+ q
bomb

59. 59
Calculate heat of combustion of octane. Calculate heat of combustion of octane.
CC
88HH
1818 + 25/2 O + 25/2 O
22 --> 8 CO --> 8 CO
22 + 9 H + 9 H
22OO
••Burn 1.00 g of octaneBurn 1.00 g of octane
•Temp rises from 25.00 to 33.20 Temp rises from 25.00 to 33.20
oo
CC
•Calorimeter contains 1200 g waterCalorimeter contains 1200 g water
•Heat capacity of bomb = 837 J/KHeat capacity of bomb = 837 J/K
Measuring Heats of ReactionMeasuring Heats of Reaction
CALORIMETRYCALORIMETRY
Measuring Heats of ReactionMeasuring Heats of Reaction
CALORIMETRYCALORIMETRY

60. 60
Step 1Step 1 Calc. heat transferred from reaction to Calc. heat transferred from reaction to
water.water.
q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 Jq = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J
Step 2Step 2 Calc. heat transferred from reaction to Calc. heat transferred from reaction to
bomb.bomb.
q = (bomb heat capacity)( T)
∆
q = (bomb heat capacity)( T)
∆
= (837 J/K)(8.20 K) = 6860 J= (837 J/K)(8.20 K) = 6860 J
Step 3Step 3 Total heat evolvedTotal heat evolved
41,170 J + 6860 J = 48,030 J41,170 J + 6860 J = 48,030 J
Heat of combustion of 1.00 g of octane = - 48.0 kJHeat of combustion of 1.00 g of octane = - 48.0 kJ
Measuring Heats of ReactionMeasuring Heats of Reaction
CALORIMETRYCALORIMETRY
Measuring Heats of ReactionMeasuring Heats of Reaction
CALORIMETRYCALORIMETRY