Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 1 of 58 Philip Dutton University of Windsor, Canada Prentice-Hall © 2007 CHEMISTRY Ninth Edition GENERAL Principles and Modern Applications Petrucci Harwood Herring Madura Chapter 7: Thermochemistry

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 2 of 58 Contents 7-1Getting Started: Some Terminology 7-2Heat 7-3Heats of Reaction and Calorimetry 7-4Work 7-5The First Law of Thermodynamics 7-6Heats of Reaction:  U and  H 7-7The Indirect Determination of  H: Hess’s Law 7-8Standard Enthalpies of Formation 7-9Fuels as Sources of Energy  Focus On Fats, Carbohydrates and Energy Storage

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 3 of Getting Started: Some Terminology  System  Surroundings

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 4 of 58 Terminology  Energy, U  The capacity to do work.  Work  Force acting through a distance.  Kinetic Energy  The energy of motion.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 5 of 58 Energy  Kinetic Energy ek =ek = 1 2 mv 2 [e k ] = m s = J kg 2 w = Fd [w ] = kg m s2s2 = J m  Work

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 6 of 58  Potential Energy  Energy due to condition, position, or composition.  Associated with forces of attraction or repulsion between objects.  Energy can change from potential to kinetic. Energy

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 7 of 58 Potential and Kinetic Energy Slide 7 of 58General Chemistry: Chapter 7 Prentice-Hall © 2007

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 8 of 58 Energy and Temperature  Thermal Energy  Kinetic energy associated with random molecular motion.  In general proportional to temperature.  An intensive property.  Heat and Work  q and w.  Energy changes.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 9 of 58 Heat Energy transferred between a system and its surroundings as a result of a temperature difference.  Heat flows from hotter to colder.  Temperature may change.  Phase may change (an isothermal process).

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 10 of 58 Units of Heat  Calorie (cal)  The quantity of heat required to change the temperature of one gram of water by one degree Celsius.  Joule (J)  SI unit for heat 1 cal = J

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 11 of 58 Heat Capacity  The quantity of heat required to change the temperature of a system by one degree.  Molar heat capacity. ◦System is one mole of substance.  Specific heat capacity, c. ◦System is one gram of substance  Heat capacity ◦Mass  specific heat. q = mc  T q = CTCT

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 12 of 58 Conservation of Energy  In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed. q system + q surroundings = 0 q system = -q surroundings

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 13 of 58 Determination of Specific Heat

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 14 of 58 Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. q lead = -q water q water = mc  T = (50.0 g)(4.184 J/g °C)( )°C q water = 1.4  10 3 J q lead = -1.4  10 3 J = mc  T = (150.0 g)(c lead )( )°C c lead = 0.13 Jg -1 °C -1 EXAMPLE 7-2

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 15 of Heats of Reaction and Calorimetry  Chemical energy.  Contributes to the internal energy of a system.  Heat of reaction, q rxn.  The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system, at constant temperature.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 16 of 58  Exothermic reactions.  Produces heat, q rxn < 0.  Endothermic reactions.  Consumes heat, q rxn > 0.  Calorimeter  A device for measuring quantities of heat. Heats of Reaction Add water Slide 16 of 58General Chemistry: Chapter 7 Prentice-Hall © 2007

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 17 of 58 Bomb Calorimeter q rxn = -q cal q cal = q bomb + q water + q wires +… Define the heat capacity of the calorimeter: q cal =  m i c i  T = C  T all i heat

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 18 of 58 Using Bomb Calorimetry Data to Determine a Heat of Reaction. The combustion of g sucrose, in a bomb calorimeter, causes the temperature to rise from to 28.33°C. The heat capacity of the calorimeter assembly is 4.90 kJ/°C. (a) What is the heat of combustion of sucrose, expressed in kJ/mol C 12 H 22 O 11 ? (b) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only 19 calories. EXAMPLE 7-3

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 19 of 58 Calculate q calorimeter : q cal = C  T = (4.90 kJ/°C)( )°C = (4.90)(3.41) kJ = 16.7 kJ Calculate q rxn : q rxn = -q cal = kJ per g EXAMPLE 7-3

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 20 of 58 Calculate q rxn in the required units: q rxn = -q cal = kJ g = kJ/g g 1.00 mol = kJ/g =  10 3 kJ/mol q rxn (a) Calculate q rxn for one teaspoon: 4.8 g 1 tsp = (-16.5 kJ/g)(q rxn (b) )( )= -19 kcal/tsp 1.00 cal J EXAMPLE 7-3

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 21 of 58 Coffee Cup Calorimeter  A simple calorimeter.  Well insulated and therefore isolated.  Measure temperature change. q rxn = -q cal See example 7-4 for a sample calculation.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 22 of Work  In addition to heat effects chemical reactions may also do work.  Gas formed pushes against the atmosphere.  Volume changes.  Pressure-volume work.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 23 of 58 Pressure Volume Work w = F  d = (m  g) (m  g) = P  V w = -P ext  V A  h  h =  A  h  h

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 24 of 58 Assume an ideal gas and calculate the volume change: V i = nRT/P = (0.100 mol)( L atm mol -1 K -1 )(298K)/(2.40 atm) = 1.02 L V f = 2.04 L Calculating Pressure-Volume Work. Suppose the gas in the previous figure is mol He at 298 K and the each mass in the figure corresponds to an external pressure of 1.20 atm. How much work, in Joules, is associated with its expansion at constant pressure.  V = 2.04 L-1.02 L = 1.02 L EXAMPLE 7-5

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 25 of 58 Calculate the work done by the system: w = -P  V = -(1.20 atm)(1.02 L)( =  10 2 J ) -101 J 1 L atm Where did the conversion factor come from? Compare two versions of the gas constant and calculate J/mol K ≡ L atm/mol K 1 ≡ J/L atm Hint: If you use pressure in kPa you get Joules directly. A negative value signifies that work is done ON the surroundings EXAMPLE 7-5

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 26 of The First Law of Thermodynamics  Internal Energy, U.  Total energy (potential and kinetic) in a system. Translational kinetic energy. Molecular rotation. Bond vibration. Intermolecular attractions. Chemical bonds. Electrons.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 27 of 58 First Law of Thermodynamics  A system contains only internal energy.  A system does not contain heat or work.  These only occur during a change in the system.  Law of Conservation of Energy  The energy of an isolated system is constant  U = q + w

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 28 of 58 First Law of Thermodynamics

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 29 of 58 Functions of State  Any property that has a unique value for a specified state of a system is said to be a State Function. ◦Water at K and 1.00 atm is in a specified state. ◦d = g/mL ◦This density is a unique function of the state. ◦It does not matter how the state was established.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 30 of 58 Functions of State  U is a function of state.  Not easily measured.  U has a unique value between two states.  Is easily measured.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 31 of 58 Path Dependent Functions  Changes in heat and work are not functions of state.  Remember example 7-5, w =  10 2 J in a one step expansion of gas:  Consider 2.40 atm to 1.80 atm and finally to 1.20 atm.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 32 of 58 Path Dependent Functions w = (-1.80 atm)( )L – (1.30 atm)( )L = L atm – 0.82 L atm = L atm =  10 2 J Compared  10 2 J for the two stage process

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 33 of Heats of Reaction:  U and  H Reactants → Products U i U f  U = U f - U i  = q rxn + w In a system at constant volume:  U = q rxn + 0 = q rxn = q v But we live in a constant pressure world! How does q p relate to q v ?

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 34 of 58 Heats of Reaction

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 35 of 58 Heats of Reaction q V = q P + w We know that w = - P  V and  U = q P, therefore:  U = q P - P  V q P =  U + P  V These are all state functions, so define a new function. Let H = U + PV Then  H = H f – H i =  U +  PV If we work at constant pressure and temperature:  H =  U + P  V = q P

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 36 of 58 Comparing Heats of Reaction q P = -566 kJ/mol =  H P  V = P(V f – V i ) = RT(n f – n i ) = -2.5 kJ  U =  H - P  V = kJ/mol = q V

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 37 of 58 Changes of State of Matter H 2 O (l) → H 2 O(g)  H = 44.0 kJ at 298 K Molar enthalpy of vaporization: Molar enthalpy of fusion: H 2 O (s) → H 2 O(l)  H = 6.01 kJ at K

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 38 of 58 Break the problem into two steps: Raise the temperature of the liquid first then completely vaporize it. The total enthalpy change is the sum of the changes in each step. Enthalpy Changes Accompanying Changes in States of Matter. Calculate  H for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 25.0°C. = (50.0 g)(4.184 J/g °C)( )°C g 18.0 g/mol 44.0 kJ/mol Set up the equation and calculate: q P = mc H 2 O  T + n  H vap = 3.14 kJ kJ = 125 kJ EXAMPLE 7-8

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 39 of 58 Standard States and Standard Enthalpy Changes  Define a particular state as a standard state.  Standard enthalpy of reaction,  H °  The enthalpy change of a reaction in which all reactants and products are in their standard states.  Standard State  The pure element or compound at a pressure of 1 bar and at the temperature of interest.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 40 of 58 Enthalpy Diagrams

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 41 of Indirect Determination of  H: Hess’s Law  H is an extensive property.  Enthalpy change is directly proportional to the amount of substance in a system. N 2 (g) + O 2 (g) → 2 NO(g)  H = kJ ½N 2 (g) + ½O 2 (g) → NO(g)  H = kJ  H changes sign when a process is reversed NO(g) → ½N 2 (g) + ½O 2 (g)  H = kJ

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 42 of 58 Hess’s Law  Hess’s law of constant heat summation  If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. ½N 2 (g) + O 2 (g) → NO 2 (g)  H = kJ ½N 2 (g) + ½O 2 (g) → NO(g)  H = kJ NO(g) + ½O 2 (g) → NO 2 (g)  H = kJ

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 43 of 58 Hess’s Law Schematically

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 44 of 58  The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states.  The standard enthalpy of formation of a pure element in its reference state is 0. Hf°Hf° 7-8 Standard Enthalpies of Formation

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 45 of 58 Standard States

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 46 of 58 Standard Enthalpy of Formation

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 47 of 58 Standard Enthalpies of Formation

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 48 of 58 Standard Enthalpies of Reaction  H overall = -2  H f ° NaHCO 3 +  H f ° Na 2 CO 3 +  H f ° CO 2 +  H f ° H 2 O

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 49 of 58 Enthalpy of Reaction  H rxn =   H f ° products -   H f ° reactants

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 50 of 58 Table 7.3 Enthalpies of Formation of Ions in Aqueous Solutions

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 51 of Fuels as Sources of Energy  Fossil fuels.  Combustion is exothermic.  Non-renewable resource.  Environmental impact.

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 52 of 58 World Primary Energy Consumption petroleum coal natural gas wind nuclear

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 53 of 58 Greenhouse Effect

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 54 of 58 CO 2 in the Atmosphere Global Average

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 55 of 58 Actual and Predicted Emission of CO 2 petroleum coal natural gas total

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 56 of 58 Focus on: Fats Carbohydrates and Energy Storage

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 57 of 58 Energy Values of Fats and Carbohydrates 1 cal = J

Prentice-Hall © 2007 General Chemistry: Chapter 7 Slide 58 of 58  Seek help with problems  But only after struggling with them.  The initial effort pays off  Understanding occurs when solutions are explained by peers, tutors or professors.  Do not go straight to the solution manual  You only think that you understand the solution if you have not worked hard on it first. The struggle to solve the problem is a key part of the memory and understanding process. End of Chapter Questions