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
5.1 Thermodynamics The science of heat and work of energy transfer therme heat dynamis power (Greek)
ENERGY Energy - Capacity to Do Work (w) or Transfer Heat (q) Kinetic Energy - Energy of Motion Potential Energy - Stored Energy
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
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
Law of Conservation of Energy The total energy of the universe is constant Energy can not be created or destroyed, only transformed
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 = 4.184 Joule Note Dietary Calorie (Cal) = 1 Kcal
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 (1818-1889) 1J 1 kg m /sec 2 2
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
Spontaneous Thermal Energy Transfer Heat (q) Flows From Hot to Cold
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
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.
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 )
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).
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.
What Is the Specific Heat of Water? c(h 2 O) = 1cal/(g o C) (original definition of the calorie)
Specific Heat of Common Materials Substance c(j/g. K) Aluminum 0.902 Graphite 0.72 glass 0.888 gold 0.128 water(l) 4.184 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
Specific Heat of Common Materials Substance c(j/g. K) Aluminum 0.902 Graphite 0.72 glass 0.888 gold 0.128 water(l) 4.184 water(s) 2.06 water(g) 2.01 wood 1.76 The Heat Capacity of a Substance Depends on it s Phase
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 0.902 J/(g-K).
Calculating Energy Requirements How much Energy is Required to raise 1.000 gallon of water from 0.000 deg C to 100.0 deg C? d water = 1 g/ml, 1 gal = 3.7854 L Q=mcDT Q 3.7854l 1000ml 1g 1cal 1gal( )( )( )( )(100deg C 0deg C) gal l ml g deg C Q=379 kcal
Chapter 6 Interactive Quizzes 1-3 http://www.ualr.edu/rebelford/chem1402/q1402/chem1402qp.htm 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 0.902 i i f i J i g 0 C
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.
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
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)
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)
Heat & Changes of State
6 Phase Changes V L ENDO S
6 Phase Changes V EXO L S
6 Phase Changes V EXO L ENDO S
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
For a Given Substance, Which Is Greater, the Molar Heat of Vaporization or the Molar Heat of Fusion?
Heating Diagram
Calculating Energy Changes How much energy is required raise 100. g of ice from - 10 0 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
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 2 100 2.06 0 10 100 6.02 gc J o o o 2 100g 4.186 100 C 0 C 100g 40.6 gc J o o o 100g 2.01 110 C 100 C gc mol H O 18g KJ mol KJ mol
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
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?
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
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?
Sign Conventions of Energy Transfer Endothermic Reactions - absorb heat from the surroundings (DH > 0) Exothermic Reactions - release heat to the surroundings (DH < 0)
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
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
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?
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
Calorimetry What happens when a hot object comes into thermal contact with a cold object? Insulated Jacket
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)
-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 )(30.0 2 2 O O )( 4.184J g C C) (2g Au)( ) (2g Au)( 0.128J g C 0.128J g C )(100 ) C) T F = 30.17 o C
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)
-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 4.184 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 4.184 J J 0.128J (25 ml H2O)( )( ) 36 o (2 g Au)( ) 2 ml H O g C C g C
-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 4.184 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 4.184 J J 0.128J (25 ml H2O)( )( ) 36 o (2 g Au)( ) 2 ml H O g C C g C T F = 30.12 o C
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
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
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
Bomb Calorimeter & Combustion Reactions
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= -197.0kJ Note, to cancel SO 2, multiply first eq by 2
S(s) + O 2 (g) --> SO 2 (g) DH= -296.8kJ 2SO 2 (g) + O 2 (g) --> 2SO 3 (g) DH= -197.0kJ DH (kj) 2S(s) + 2O 2 (g) --> 2SO 2 (g) 2SO 2 (g) + O 2 (g) --> 2SO 3 (g) -593.6-197.0 2S(s) + 3O 2 (g) --> 2SO 3 (g) -790.6 Interactive Quiz 6-11 http://www.ualr.edu/rebelford/chem1402/q1402/chem1402qp.htm
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
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
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 = - 285.8kJ/mol DH o f Listed in Appendix L of Text
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)
Enthalpy of Formation Calculate the Enthalpy of Formation of Acetylene (C 2 H 2 ) from the following combustion data 1. 2. 3. 2C 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)
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
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
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 ) = -1118.4 kj/mol DH o f (Al 2 O 3 ) = -1675.7 kj/mol 3Fe(s) + 2O 2 (g) Fe 3 O 4 (s) 2Al(s) + 3/2O 2 (g) Al 3 O 3 (s) -1118.4kJ -1675.5kJ
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(-1118.4kJ) 4(-1675.5kJ) 8Al(s) + 3Fe 3 O 4 4Al 2 O 3 (s) + 9Fe(s) -3363.3kJ In 1 Step, Using the Eq: DH o rxn o o n[ DH ( products )] m[ DH ( reactants ) f f
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 ) = -1118.4 kj/mol DH o f (Al 2 O 3 ) = -1675.7 kj/mol DH o rxn o o n[ DH ( products )] m[ DH ( reactants ) f f DH o rxn =[4(-1675.7 kj/mol) + 8(0)] -[3(-1118.4 kj/mol) + 9(0)] = -3363.6 kj
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 http://www.ualr.edu/rebelford/chem1402/q1402/chem1402qp.htm
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?
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
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
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 = -3363.6 kj
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 = -3363.6 kj mol Al 1 15gAl 0.069 26.98g 8mol Al mol Fe O 1 3 4 30gFe3O 4 0.043 231.54g 3mol Fe3O 4 Fe 3 O 4 is Limiting Reagent
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 = -3363.6 kj mol Fe O 3363kJ 3 4 30gFe3O 4 145 231.54g 3mol Fe3O 4 kj Key Step Correlates Ratio of the Enthalpy of Reaction to the Coefficient of a Chemical Species