Chapter 5: Thermochemistry
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1 Chapter 5: Thermochemistry 1. Thermodynamics 2. Energy 3. Specific Heat 4. Enthalpy 5. Enthalpies of Reactions 6. Hess s Law 7. State Functions 8. Standard Enthalpies of Formation 9. Determining Enthalpies of Reactions
2 5.1 Thermodynamics The science of heat and work of energy transfer therme heat dynamis power (Greek)
3 ENERGY Energy - Capacity to Do Work (w) or Transfer Heat (q) Kinetic Energy - Energy of Motion Potential Energy - Stored Energy
4 Molecular Kinetic Energy -Translational energy E k, translational = 1/2mv 2 -Rotational energy E k, rotaional = 1/2Iw 2 I = moment of inertia w = rotational frequency -Vibrational energy E k, vibrational = 1/2Kx 2 K = Hooke s constant x = displacement from equilbrium
5 Molecular Potential Energy E atom coulombic attraction of e - to nucleus E bond coulombic forces of covalent and ionic bonds E nucleus strong and weak nuclear forces holding nucleus together
6 Law of Conservation of Energy The total energy of the universe is constant Energy can not be created or destroyed, only transformed
7 Units of Energy Calorie (cal) - Amount of Energy (Heat) Required to Raise 1 Gram of Water 1 Degree Celsius (14.5 to 15.5 o C) Modern Definition is Based on The Joule 1 Calorie = Joule Note Dietary Calorie (Cal) = 1 Kcal
8 Units of Energy Two Common Units of Energy Joule work based definition Calorie heat based definition Joule SI derived unit, the amount of work required to accelerate a 1 kg object 1 m/s 2 a distance of 1 meter JP Joule ( ) 1J 1 kg m /sec 2 2
9 Thermal Energy Depends on the Temperature, # and Type of Atoms Present The Hotter an Object Is, the More Kinetic Energy It s Constituent Atoms Posses
10 Spontaneous Thermal Energy Transfer Heat (q) Flows From Hot to Cold
11 5.2 Heat Capacity (C) q = CDT -The heat required to induce a temperature change in a substance C = q/dt -The capacity of a substance to absorb heat as its temperature changes *Note: Heat capacity describes an object and has units in terms of energy per degree temperature
12 Is Heat an Intensive or Extensive Property? Extensive: It requires twice as much heat to raise 2 grams of water 1 o C than it does 1 gram.
13 Specific Heat Capacity -the heat capacity per gram of substance q = mcdt q = Heat or Energy (cal or joule) m = Mass (in grams) c = Specific heat, the heat required to raise one gram of a substance by one degree Celsius DT = Change in Temperature (T final - T initial )
14 Specific Heat Capacity c = q mdt Specific heat capacity describes a type of substance and not a specific object. Has Units of J/g o C or cal/ g o C To know how much heat capacity an object has you need to know how much of it you have (it s mass times it s specific heat capacity).
15 Is Specific Heat Capacity an Intensive or Extensive Property? Intensive: The amount of heat required to raise the temperature of substance 1 o C on a per gram basis is the same, no matter how many grams the substance has.
16 What Is the Specific Heat of Water? c(h 2 O) = 1cal/(g o C) (original definition of the calorie)
17 Specific Heat of Common Materials Substance c(j/g. K) Aluminum Graphite 0.72 glass gold water(l) water(s) 2.06 water(g) 2.01 wood 1.76 Metals have small specific heat capacities, so it does not take much heat to raise the temperature of a metal
18 Specific Heat of Common Materials Substance c(j/g. K) Aluminum Graphite 0.72 glass gold water(l) water(s) 2.06 water(g) 2.01 wood 1.76 The Heat Capacity of a Substance Depends on it s Phase
19 Difference between C and c: C: Heat Capacity describes an object, like a 10 g piece of aluminum has a heat capacity of 9.02 J/K. c: Specific Heat Capacity describes a material, aluminum has a specific heat capacity of J/(g-K).
20 Calculating Energy Requirements How much Energy is Required to raise gallon of water from deg C to deg C? d water = 1 g/ml, 1 gal = L Q=mcDT Q l 1000ml 1g 1cal 1gal( )( )( )( )(100deg C 0deg C) gal l ml g deg C Q=379 kcal
21 Chapter 6 Interactive Quizzes Heat Problem What would be the final temperature if a 250 g piece of aluminum at 20. o C absorbed 1.5 kj of energy? q mcdt mc T T f f f q mct mct mct q mct T q mc T 1,500J o 0 Tf 20. C 27 C 250g i i f i J i g 0 C
22 First Law of Thermodynamics -Total amount of energy of the universe is constant DE = q + w The change in the internal energy of a system is equal to the heat transfer and work done on/by the system with the surroundings.
23 Internal Energy Q system (reaction) surroundings W DE = E f - E i = Q + W DE > 0, the system gains energy DE < 0, the system loses energy
24 Heat of Reaction Endothermic Reaction - heat is added to the system (q is positive and the surroundings lose the heat) Exothermic Reaction - heat is lost from the system (q is negative and surroundings gain the heat)
25 5.3 Energy & Changes of State What Are the Three Effects That Adding or Subtracting Heat Can Have on a Substance. 1. Decomposition the destruction of intramolecular bonds (chemical change) 2. Temperature Change within a phase (physical change) 3. Phase Change - Changes in the intermolecular forces (physical change)
26 Heat & Changes of State
27 6 Phase Changes V L ENDO S
28 6 Phase Changes V EXO L S
29 6 Phase Changes V EXO L ENDO S
30 How Do We Quantitatively Determine the Heat Associated With Vaporization or Fusion Change? With the Molar Heat of Fusion or Vaporization Define the Molar Heat of Fusion: DH f (kj/mol) -Heat required to melt one mole of a substance at it s melting pt Define the Molar Heat of Vaporization: DH v (kj/mol) -Heat required to vaporize one mole of a substance at it s boiling pt
31 For a Given Substance, Which Is Greater, the Molar Heat of Vaporization or the Molar Heat of Fusion?
32 Heating Diagram
33 Calculating Energy Changes How much energy is required raise 100. g of ice from C to 110 o C? Given: DH f =6.02 kj/mol DH v =40.6 kj/mol c l =4.184 J/g.o C c v =2.01 J/g.o C c s =2.06 J/g.o C T e m p e r a t u r e ( o C) Heat Added
34 Calculating Energy Changes q q q q q q total DT melt ice DT boil water DT ice water steam qtotal mici DTi ndh fus mwcwdt w ndh vap mscsdt s o o mol H O J qtotal g o C C g 18g gc J o o o 2 100g C 0 C 100g 40.6 gc J o o o 100g C 100 C gc mol H O 18g KJ mol KJ mol
35 Calculating Energy Changes q q q q q q total DT melt ice DT boil water DT ice water steam qtotal mici DTi ndh fus mwcwdt w ndh vap mscsdt s q 2.06kJ 33.4KJ 41.86KJ 225kJ 2.01kJ total 304kJ
36 Work -Work occurs when something moves against an opposing force - Lets investigate the work of expansion at constant pressure W = FDX F = Force, X = distance P = F/A, F = P. A V = X. A, X = V/A W = (P. A)(DV/A) W = -PDV Why is this negative?
37 Enthalpy DH = Enthalpy change, the heat transfer in/out of a system at constant P DH = q p DE = q + w w = -PDV DE = q -PDV DH = q P = DE + PDV
38 State Functions Enthalpy & Internal Energy are State Functions - their values are path independent and only depend on their current states, not how they were attained Is Work a State Function?
39 Sign Conventions of Energy Transfer Endothermic Reactions - absorb heat from the surroundings (DH > 0) Exothermic Reactions - release heat to the surroundings (DH < 0)
40 5.5 DH and Chemical Reactions Reactants + Heat --> Products -endothermic, heat was added to the reaction Reactants --> Products + Heat -exothermic, heat was released by the reaction
41 DH rxn is associated with a chemical equation C(s) + 2H 2 (g) --> CH 4 (g) DH = -74.8kJ What do the units of the DH in the above reaction mean? 74.8 kj are released for every mole of carbon or for every 2 moles of hydrogen consumed
42 Is the Following Reaction Endothermic or Exothermic? CaO(s) + CO 2 (g) --> CaCO 3 (s) DH = -178 kj DH Is Negative and Energy Is Lost From the System So It Is Exothermic Would this reaction heat the surroundings?
43 DH Forward = - DH reverse CaO(s) + CO 2 (g) --> CaCO 3 (s) DH = -178 kj What is DH For the reaction: CaCO 3 (s) --> CaO(s) + CO 2 (g) DH =+178 kj It is endothermic by the same order of magnitude the first was exothermic
44 Calorimetry What happens when a hot object comes into thermal contact with a cold object? Insulated Jacket
45 Calorimetry What is the final temperature if 2g of gold at 100 o C is dropped into 25 ml of water at 30 o C in an ideal calorimeter? c Au = 0.128J/g.o C. Heat lost by Au = Heat gained by water -Q hot = Q cold (First Law)
46 -Q hot = Q cold -m H c H (T f - T H ) = m C c C (T f - T C ) T F = m C c C T C + m H c H T H m C c C + m H c H T F (25ml H 2 O)( 1g H ml H 2 2 (25ml H O O 2 )( O)( 4.184J g C 1g H ml H )( O O )( 4.184J g C C) (2g Au)( ) (2g Au)( 0.128J g C 0.128J g C )(100 ) C) T F = o C
47 Calorimetry What is the final temperature if 2g of gold at 100 o C is dropped into 25 ml of water at 30 o C in a real calorimeter? c Au = 0.128J/g.o C, C cal = 36J/ o C Heat lost by Au = Heat gained by water + calorimeter -Q hot = Q cold (First Law)
48 -Q hot = Q cold -Q Au = Q water +Q calorimeter -m H c H (T f - T H ) = m C c C (T f - T C ) + C cal (T f - T C ) -m H c H (T f - T H ) = (m C c C + C cal )(T f - T C ) T F = (m C c C + C cal ) T C + m H c H T H (m C c C + C cal ) + m H c H T F 1g H2O J J 0.128J (25 ml H2O)( )( ) 36 o (30.0 C) (2 g Au)( )(100 C) 2 ml H O g C C g C 1g H2O J J 0.128J (25 ml H2O)( )( ) 36 o (2 g Au)( ) 2 ml H O g C C g C
49 -Q hot = Q cold -Q Au = Q water +Q calorimeter T F T F = (m C c C + C cal ) T C + m H c H T H (m C c C + C cal ) + m H c H 1g H2O J J 0.128J (25 ml H2O)( )( ) 36 o (30.0 C) (2 g Au)( )(100 C) 2 ml H O g C C g C 1g H2O J J 0.128J (25 ml H2O)( )( ) 36 o (2 g Au)( ) 2 ml H O g C C g C T F = o C
50 6.7 Hess s Law Conservation of Energy If the products of one reaction are consumed by another, the two equations can be coupled into a third equation. The energy of the third reaction will be the sum of the energies of the first 2 reactions
51 Hess s Law Calculate DH for the reaction 2C(s) + O 2 (g) --> 2CO(g) From the reactions 2C(s) + 2O 2 (g) --> 2CO 2 (g) 2CO 2 (g) --> 2CO (g) + O 2 (g) 2C(s) + O 2 (g) --> 2CO(g) DH= -787kJ DH= 566kJ DH= -221kJ
52 Hess s Law 2C(s) + 2O 2 (g) E DH= -221kJ 2CO (g) + O 2 (g) DH= 566kJ 2 2CO 2 (g) DH= -787kJ 1 2C(s) + 2O 2 (g) --> 2CO 2 (g) 2CO 2 (g) --> 2CO (g) + O 2 (g) 2C(s) + O 2 (g) --> 2CO(g) DH= -787kJ DH= 566kJ DH= -221kJ 2 Step Path
53 Bomb Calorimeter & Combustion Reactions
54 Hess s Law Calculate DH for the reaction 2S(s) + 3O 2 (g) --> 2SO 3 (g) From the reactions S(s) + O 2 (g) --> SO 2 (g) DH=-296.8kJ 2SO 2 (g) + O 2 (g) --> 2SO 3 (g) DH= kJ Note, to cancel SO 2, multiply first eq by 2
55 S(s) + O 2 (g) --> SO 2 (g) DH= kJ 2SO 2 (g) + O 2 (g) --> 2SO 3 (g) DH= kJ DH (kj) 2S(s) + 2O 2 (g) --> 2SO 2 (g) 2SO 2 (g) + O 2 (g) --> 2SO 3 (g) S(s) + 3O 2 (g) --> 2SO 3 (g) Interactive Quiz
56 Why are Enthalpies Additive? -Enthalpy is a state function -Its value is only dependent on the state of the system, not the path -It s a consequence of the conservation of energy For a Chemical Reaction DH = H products -H reactants Is Not Dependent on the Steps the Reaction Involves
57 Standard State Enthalpies DH o = Standard State Enthalpy Change of Reaction Standard State is the most stable form of a substance as it exists at 1 atm and 25 o C. For a Solution this is at a Concentration of 1M
58 Standard Molar Enthalpy of Formation DH o f = Standard Molar Enthalpy of Formation The Enthalpy Change Associated With the Formation of 1 Mol of a Substance From It s Elements in Their Standard States H 2 (g) +1/2O 2 (g) --> H 2 O(l) DH o f = kJ/mol DH o f Listed in Appendix L of Text
59 Exercises Write the DH o f equation for the following 1. NH 3 (g) 2. H 2 O(l) 3. O 2 (l) 4. O 2 (g) 5. O 3 (g) 1/2N 2 (g) + 3/2H 2 (g) --> NH 4 (g) 1/2O 2 (g) + H 2 (g) --> H 2 O(l) O 2 (g) --> O 2 (l) No reaction 3/2O 2 (g) --> O 3 (g)
60 Enthalpy of Formation Calculate the Enthalpy of Formation of Acetylene (C 2 H 2 ) from the following combustion data C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(l) C(s) + O 2 (g) CO 2 (g) 2H 2 (g) + O 2 (g) 2H 2 O(l) -2600KJ -394KJ -572KJ 2C(s) + H 2 (g) C 2 H 2 (g)
61 Enthalpy of Formation Calculate the Enthalpy of Formation of Acetylene (C 2 H 2 ) from the following combustion data 2CO 2 (g) + H 2 O(l) C 2 H 2 (g) + 5/2O 2 (g) 2C(s) + 2O 2 (g) 2CO 2 (g) H 2 (g) + 1/2O 2 (g) H 2 O(l) 2C(s) + H 2 (g) C 2 H 2 (g) - ½ (-2600KJ) 2(-394KJ) ½(-572KJ) 226KJ
62 Standard Enthalpies of Reaction DH o rxn DH o rxn can be determined from DH o f o o n[ DH ( products )] m[ DH ( reactants ) f Where n & m are the stoichiometric coefficients of the products & reactants Explain This in Terms of Enthalpy as a State Function and Hess s Law for the following Reaction: 8Al(s) + 3Fe 3 O 4 --> 4Al 2 O 3 (s) + 9Fe(s) f
63 DH o rxn Exercises 8Al(s) + 3Fe 3 O 4 --> 4Al 2 O 3 (s) + 9Fe(s) From Appendix L: DH o f (Fe 3 O 4 ) = kj/mol DH o f (Al 2 O 3 ) = kj/mol 3Fe(s) + 2O 2 (g) Fe 3 O 4 (s) 2Al(s) + 3/2O 2 (g) Al 3 O 3 (s) kJ kJ
64 DH o rxn Exercises 8Al(s) + 3Fe 3 O 4 --> 4Al 2 O 3 (s) + 9Fe(s) 3(Fe 3 O 4 (s) 3Fe(s) + 2O 2 (g)) 4(2Al(s) +3/2O 2 (g) Al 2 O 3 (s)) -3( kJ) 4( kJ) 8Al(s) + 3Fe 3 O 4 4Al 2 O 3 (s) + 9Fe(s) kJ In 1 Step, Using the Eq: DH o rxn o o n[ DH ( products )] m[ DH ( reactants ) f f
65 DH o rxn Exercises 8Al(s) + 3Fe 3 O 4 4Al 2 O 3 (s) + 9Fe(s) From Appendix L: DH o f (Fe 3 O 4 ) = kj/mol DH o f (Al 2 O 3 ) = kj/mol DH o rxn o o n[ DH ( products )] m[ DH ( reactants ) f f DH o rxn =[4( kj/mol) + 8(0)] -[3( kj/mol) + 9(0)] = kj
66 Standard Enthalpies of Reaction DH o rxn can be determined from DH o f DH o rxn o o n[ DH ( products )] m[ DH ( reactants ) f f Interactive Quiz 6-1 & 6-2
67 Enthalpy of Vaporization for Water from Enthalpies of Formation Calculate DH Vap (H 2 O) From Appendix 4: DH o f [H 2 O(l)] = -286 kj/mol DH o f [H 2 O(g)] = -242 kj/mol What is the Equation that Describes Vaporization of Water?
68 Enthalpy of Vaporization for Water from Enthalpies of Formation From Appendix 4: DH o f [H 2 O(l)] = -286 kj/mol DH o f [H 2 O(g)] = -242 kj/mol DH o rxn H 2 O(l) H 2 O(g) o o n[ DH ( products )] m[ DH ( reactants ) f f -242kJ/mol [-286kJ/mol] = 44 kj/mol
69 Enthalpies of Reaction and Stoichiometric Problems 8Al(s) + 3Fe 3 O 4 4Al 2 O 3 (s) + 9Fe(s) Determine the standard state enthalpy change when 15 g of Al react with 30.0 g of Fe 3 O 4. 3 Steps Balance Eq. & Identify Limiting Reagent Calculate DH o rxn for Balanced Eq. Calculate DH o rxn for this reaction based on complete consumption of the limiting reagent
70 Enthalpies of Reaction and Stoichiometric Problems 8Al(s) + 3Fe 3 O 4 4Al 2 O 3 (s) + 9Fe(s) 1. Calculate DH o rxn.for balanced Equation This was done in previous problem, DH o rxn = kj
71 Enthalpies of Reaction and Stoichiometric Problems 8Al(s) + 3Fe 3 O 4 4Al 2 O 3 (s) + 9Fe(s) 2. Determine Limiting Reagent DH o rxn = kj mol Al 1 15gAl g 8mol Al mol Fe O gFe3O g 3mol Fe3O 4 Fe 3 O 4 is Limiting Reagent
72 Enthalpies of Reaction and Stoichiometric Problems 8Al(s) + 3Fe 3 O 4 4Al 2 O 3 (s) + 9Fe(s) 3. Determine DH o rxn based on complete consumption of limiting reagent DH o rxn = kj mol Fe O 3363kJ gFe3O g 3mol Fe3O 4 kj Key Step Correlates Ratio of the Enthalpy of Reaction to the Coefficient of a Chemical Species
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