Thermochemistry

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Thermochemistry
The study of heat transfer in chemical rxns
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ReadingChapter 13 pages 498 - 506Chapter 15
page 591- 602HW due Friday November
10thChapter 13 p 535 43, 47, 49, 51Chapter 15
p 640 59

• HW For tonight

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System
That part of nature upon which attention is
focused
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Surroundings
That part of nature around the part upon which we
focus
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• System Surroundings The Universe!

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Reaction Coordinate
graph of energy change vs. time in a chemical
reaction
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Exothermic reaction
Releases Gives off Loses heat
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Endothermic Reaction
Absorbs Takes in Gains heat
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EnthalpyDH q mcDT
m is for mass!
Heat flow/change in a system
c is for specific heat!
?T is for change in temp!
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Grammar of Thermochemistry

• Exothermic condensation reaction
• H2O (g) ? H2O (l) 44kJ
• H2O (g) ? H2O (l) ?Ho -44 kJ
• Endothermic evaporation reaction
• 2 H2O (l) 88 kJ ? 2 H2O (g)
• 2 H2O (l) ? 2 H2O (g) ?Ho 88 kJ

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Specific Heat (capacity) c
Ability of a specific quantity (1g) of a
substance to store heat as its temp rises by
1oC units ? J g oC
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Heat Capacity C
Ability of a thing to store heat as its
temperature rises units ? J oC
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Calorimeter

• Device that measures ? heat
• It tries to be an adiabatic system
• In real life, gives experimental yield

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Adiabatic System
Does not lose heat to or take heat from
surroundings DHsystem 0
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Calorimetry
DHsystem 0 DHsys DHcal DHrxn DHrxn
-DHcal DHrxn mcDTcal
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• 3.358 kJ of heat added to the 50.0 g water inside
  a calorimeter. Twater increases from 22.34oC to
  36.74oC. What is the heat capacity of the
  calorimeter in J/oC?
• cwater 4.180 J/g oC
• ?T (36.74oC 22.34oC) 14.40oC
• 50.00g (4.184 J/g oC) 14.40oC
  3.012 x 103 J
• 3.012 kJ goes into water
• 3.358 kJ 3.012 kJ .346 kJ absorbed by
  calorimeter
• .346 kJ 346 J 14.40oC 24.0 J/oC

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• 100.0 g of water at 50.0oC is added to a
  calorimeter that already contains 100.0g of water
  at 30.0oC. The final temperature is 39oC. What
  is the heat capacity of the calorimeter?
• cwater 4.184 J/g oC
• ?Tadded water (50.0oC 39.0oC) 11oC
• 100.0g (4.184 J/g oC) 11oC 4.60 x
  103 J
• ?Tcalorimeter water (39.0oC 30.0oC) 9oC
• 100.0g (4.184 J/g oC) 9oC 3.76 x
  103 J
• 4.60 kJ 3.76 kJ .834 kJ absorbed by
  calorimeter
• .834 kJ 834 J 9.0oC 93 J/oC

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Thermodynamics and ? of state
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Heat of Fusion Hf
Heat reqd to melt 1g of a substance at its
MP units ? J/g or J/kg
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Heat of VaporizationHv
Heat reqd to boil 1g of a substance at its
normal BP units ? J/g or J/kg
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• Calculate the amount of heat that must be
  absorbed by 50.0 grams of ice at -12.0oC to
  convert it to water at 20.0oC.
• cice 2.09 J/g oC Hf for ice
  334 J/g
• cwater 4.184 J/g oC
• Step 1 warm the ice to 0oC requires
• (50.0 g) (2.09 J/g oC) (0oC (-12oC)) 0.125
  x 104 J
• Step 2 melt the ice with no ? in temp
• 50.0 g 334J/g 1.67 x 104 J
• Step 3 warm the liquid to 20.0oC requires
• 50.0 g 4.18 J/g oC (20 oC - 0 oC) .418 x
  104 J

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• Answer 2.21 x 104 J 22.1 kJ must be absorbed

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Homework

• Extra Credit Homework AssignmentDue Monday
  November 13th
• 55 page 535, chapter 13
• Homework due Tuesday November 14th
• Chapter 15, page 641 61, 63, 67, 69

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• A certain calorimeter absorbs 20 J/oC. If 50.0 g
  of 50oC water is mixed with 50.0 g of 20oC water
  inside the calorimeter, what will be the final
  temperature of the mixture?
• Heat lost by the hot water will be gained by the
  cold water and the calorimeter
• ?Hhot water ?Hcool water ?Hcalorimeter
• ?Hhot water (50.0 g) (4.180 J/oCg) (50oC x)
• 209J/oC (50oC x)
• ?Hcool water (50.0 g) (4.180 J/oCg) (x 20oC)
• 209 J/oC (x 20oC)
• ?Hcalorimeter 20 J/oC (x 20oC)

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Solve algebraically

• 209 (50 x) 209 (x 20) 24 (x 20)
• 209 (50 x) 235 (x 20)
• 0.889 (50 x) x 20
• 44 0.889x x 20
• 64 1.889x
• x 33.9oC 30oC

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• A certain calorimeter absorbs 24 J/oC. If 50.0 g
  of 52.7oC water is mixed with 50.0 g of 22.3oC
  water inside the calorimeter, what will be the
  final temperature of the mixture?
• Heat lost by the hot water will be gained by the
  cold water and the calorimeter
• ?Hhot water ?Hcool water ?Hcalorimeter
• ?Hhot water (50.0 g) (4.180 J/oCg) (52.7oC
  x)
• 209J/oC (52.7oC x)
• ?Hcool water (50.0 g) (4.180 J/oCg) (x
  22.3oC)
• 209 J/oC (x 22.3oC)
• ?Hcalorimeter 24 J/oC (x 22.3oC)

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Solve algebraically

• 209 (52.7 x) 209 (x 22.3) 24 (x 22.3)
• 209 (52.7 x) 235 (x 22.3)
• 0.889 (52.7 x) x 22.3
• 46.87 0.889x x 22.3
• 69.17 1.889x
• x 36.6oC 37oC

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Heat of ReactionDHrxn
Heat/enthalpy change of a chemical reaction Units
? J or kJ Sometimes, units ? J/mol rxn
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Mole of reaction

• Depends on how it is given in the problem (or how
  you balance your reaction)
• Can say that
• O2 (g) 2 H2 (g) ? 2 H2O (g) 45 kJ
• ?Hrxn 45 kJ/mol rxn
• You can use the following conversion factors
• 1 mol O2 2 mol H2 2 mol H2O 1 mol rxn
• 45 kJ 45 kJ 45 kJ
  45 kJ

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When X reacts with water the temp in a 1.5 kg
calorimeter containing 2.5 kg water went from
22.5oC to 26.5oC. Calculate DHrxn. cwater 4.18
J/g oC ccalorimeter 2.00 J/g oC
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• ?Hrxn ?Hwater ?Hcalorimeter
• ? T 26.5oC 22.5oC 4oC
• Heat absorbed by water ? Hwater mc ?T
• 2.5 kg 2,500 g
• (2,500 g)(4.18J/goC)(4oC) 41,800 J 41.8 kJ
• Heat absorbed by calorimeter ? Hcalorimeter mc
  ?T
• 1.5 kg 1,500 g
• (1,500 g)(2.00 J/goC)(4oC) 12,000 J 12 kJ
• Total heat added to system 41.8 12 53.8 kJ
• 54 kJ

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Heat of SolutionDHsoln

• The heat or enthalpy change when a substance is
  dissolved

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• 80 g NaOH is dissolved with 1.40 L of 0.7 M HCl
  in a calorimeter. HCl solution has a mass of 1.4
  kg or 1,400g.
• Ccalorimeter 20 J/oC DTwater 10oC
• cHCl same as cwater 4.18 J/goC
• What is the heat released by the solution
• What is the DHsolution for the reaction
• NaOH (s) HCl (aq) ? NaCl (aq) H2O (l)

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• Heat absorbed by calorimeter
• 20 J/oC 10oC 200 J
• Heat absorbed by HCl solution
• 1,400 g (4.18 J/goC) (10oC) 58,520 J
• Hsolution Hcalorimeter HHCl solution
• 200 J 58,520 J 58,720 J
• Heat released by solution 58,720 J 59 kJ
• Go back and see how many moles of NaOH HCl
  reacted
• 80 g NaOH is 2 moles therefore you have 2 moles
  rxn
• DHsolution 59 kJ/2 mol rxn 30 kJ/mol rxn

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Change! To the HW Due Wednesday

• Chapter 15, page 637 13 15
• Due Thursday November 16th
• Chapter 15, page 637 8 25, 27, 29, 31

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• When 2.61 g of C2H6O is burned at constant
  pressure,82.5 kJ of heat is given off. What is
  ?H for the reaction
• C2H6O (l) O2 (g) ? 2 CO2 (g) 3 H2O (l)
• 82.5 kJ 46.0 g C2H6O 1 mol
  C2H6O
• 2.61 g C2H6O mol C2H6O mol
  rxn
• ?H for the reaction -1450 kJ/mol rxn

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• When Al metal is exposed to O2 it is oxidized to
  form Al2O3. How much heat is released by the
  complete oxidation of 24.2 g of Al at 25oC and 1
  atm?
• 4 Al (s) 3 O2 (g) ? 2 Al2O3 (s) ?H -3352
  kJ/mol rxn
• 24.2 g Al 1 mol Al 1 mol rxn -3352
  kJ
• 27 g Al 4 mol Al
  mol rxn
• -751 kJ 751 kJ of heat are released

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Heat of Combustion DHcombustion

• The heat or enthalpy change when a substance is
  burned

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Heat of Formation DHfo

• The heat reqd to form 1 mol of a compound from
  pure elements
• units ? kJ/mole

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Gibbs Free EnergyDGo

• Energy of a system that can be converted to work
• Determines spontaneity

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Energy of Formation DGfo
The energy reqd to form 1 mol of a compound from
pure elements units ? kJ/mole
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Exergonic Reaction

• A reaction in which free energy is given off
• DG lt 0

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Endergonic Reaction

• A reaction in which free energy is absorbed
• DG gt 0

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Reaction at Equilibrium
DG 0
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Entropy

• A measure of disorder
• DSo

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Entropy of Formation

• The entropy change when one mole of a substance
  is formed
• Sfo (J/moleoK)

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Thermochemical Equation

• An equation that shows changes in heat, energy,
  etc

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Thermochemical Equation

• DHorxn
• SDHfoproducts -
• SDHforeactants

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Thermochemical Equation

• DGorxn
• SDGfoproducts -
• SDGforeactants

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Thermochemical Equation
DSorxn SSfoproducts - SSforeactants
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Thermochemical Equation

• Stoichiometry of heat change
• Solves theoretical yield

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Interrelating Equation

• DG DH - TDS

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Calculate DH, DG, DS when 19.7 kg of BaCO3 is
decomposed into BaO CO2Cmpd BaCO3 CO2
. BaODHfo -1216.3 -393.5 -553.5 DGfo
-1137.6 -394.4 -525.1 Sfo 112.1
213.6 70.4
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Calculate DH, DG, DS when 13.6 g of CaSO4 is
changed into CaO SO2 O2 at 27oCCmpd CaSO4
SO2 CaO DHfo -1434.1 -296.8
-635.1 DGfo -1321.8 -300.2 -604.0
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Lab Results Cup H2O NaOH Thermo 5.0
g 50.0 g 4.0 g 15.0 g Ti 22.0oC Tf
27.0oCCmpd NaOH Na OH-DHfo
-425.6 -240.1 -230.0Determine theoretical and
experimental heat changes,
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Calculate the potential DH, DG, DS for the
reaction Sfo for O2 when burning 8.8 kg of C3H8
Cpd C3H8 CO2 H2O DHfo -103.8
-393.5 -241.8DGfo - 23.5 -394.4 -228.6Sfo 269
.9 213.6 188.7
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Bond Energy

• The energy change when one mole of bonds are
  broken
• DHobond

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Bond Equation

• DHbondorxn
• SDHbondoproducts
• - SDHbondoreactants

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Bond Energies (kJ/mole)

• C-C 347
• C-H 414
• O-H 464
• CO 715

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Calculate DH, DG, DS in the production of 831ML
ammonia at 227oC under 125 kPa pressureCompd
NH3DHfo -46.1 DGfo -16.5
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1st Law Thermodynamics

• Total energy change heat work
• DE q W

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Work

• W Fd
• P F/A
• V Ad
• W PDV DnRT

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2nd Law Thermodynamics

• Total entropy in a system always increases
  assuming no energy is added to the system

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Thermodynamic Rxns are State Rxns
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State Reaction

• Reactions that are independent of the path thus
  not dependent on intermediates

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Calculate DHo, DGo, DS when A BC ? AC2 Bat
-23oC Teq Compd BC AC2DHfo(kJ/mole)
-150 -250DGfo(kJ/mole) -125 -225
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Write TE for the process

• 2 A B C D
• C A H
• D B 2 K
• H K M B
• K M Product

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Write TE for the process
2 A 2 B C D C A 2 H D B 2 K H
K P B
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Write TE for the process
2 A 2 B C D C A 3 H D B 2 K H
K P Q
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Review
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Calculate DHtotal, when 40.0 g of H2O is changed
from - 25oC to 125oC. FPw 0.0oC BPw 100.0
oC Hv 2260 J/g Cice 2.06 (J/g K) Hf
330 J/g Cwater 4.18 (J/g K) Csteam 2.02 (J/g
K)
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Calculate DHo, DGo, DS for AD2 BC AC2 BD
at (-23oC)Cpd BC AD2 AC2 BD
DHfo -150 -250 -300 -175 DGfo -125
-225 -250 -150 Sfo 75 50
80 ? Determine SfoBD
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Calculate DHo, DGo, DSo for PbO2 CO CO2
Pb Cpd PbO2 CO CO2 DHfo -277.4
-110.5 -393.5 DGfo -217.4 -137.2
-394.4 Calculate Teq DH of 48 g PbO2