1. Copyright © 2011 Pearson Education, Inc. Thermochemistry

2. Copyright © 2011 Pearson Education, Inc. Chapter Outline Overview of Energy in Chemical Reaction First Law of Thermodynamics Heat, Specific heat, Calorimetry Enthalpy of Reaction Standard Enthalpy of Formation Hess’s law 2 Tro: Chemistry: A Molecular Approach, 2/e

3. Copyright © 2011 Pearson Education, Inc. Chemical Hand Warmers and much more Most hand warmers uses the heat from rxn: 4 Fe(s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) The temperature change in your hand depends on the size of the hand warmer the size of your glove, etc. mainly, the amount of heat from the reaction Most chemical rxn involves energy change Chemical energy provides nearly 70% electricity in the US. Nuclear 11%, Hydroelectric 17%. All batteries (car, portable electronic device) 3 Tro: Chemistry: A Molecular Approach, 2/e

4. Copyright © 2011 Pearson Education, Inc. About Energy Even though chemistry is the study of matter, energy affects matter Energy is anything that has the capacity to do work. Example: Kinetic energy in moving water; Potential energy of meteorite near the Earth Chemical energy in gasoline or battery Energy exchange through doing work or heat exchange. 4 Tro: Chemistry: A Molecular Approach, 2/e

5. Copyright © 2011 Pearson Education, Inc. Classification of Energy Kinetic energy is energy of motion or energy that is being transferred Thermal energy is the energy associated with temperature thermal energy is a form of kinetic energy 5 Tro: Chemistry: A Molecular Approach, 2/e

6. Copyright © 2011 Pearson Education, Inc. Some Forms of Energy Electrical kinetic energy associated with the flow of electrical charge Heat or thermal energy kinetic energy associated with molecular motion Light or radiant energy kinetic energy associated with energy transitions in an atom Nuclear potential energy in the nucleus of atoms Chemical potential energy due to the structure/bonding of the atoms/molecules 6 Tro: Chemistry: A Molecular Approach, 2/e

7. Copyright © 2011 Pearson Education, Inc. Exchange of Energy Work = force acting over a distance Energy = Work = Force x Distance Example: Compressor convert electric energy to take away heat from room. Heat is the flow of energy caused by a difference in temperature. Energy can be exchanged between objects through contact How? Fast-moving particles pass the kinetic energy to slow moving particles Atomic level: Heat exchange through molecular collisions 7 Tro: Chemistry: A Molecular Approach, 2/e

8. Copyright © 2011 Pearson Education, Inc. Energy Exchange: System vs. Surroundings System: the material or process within which we are studying the energy changes within. Example: Chemical Rxn. Surroundings: everything else with which the system can exchange energy with. Example: Solvent where rxn occurs. Left: System gives energy to the Surroundings. Right: System takes energy from the Surroundings. Surroundings System Surroundings System 8 Tro: Chemistry: A Molecular Approach, 2/e

9. Copyright © 2011 Pearson Education, Inc. Units of Energy joule (J) is the amount of energy needed to move a 1-kg mass a distance of 1 meter 1 J = 1 N∙m = 1 kg∙m 2 /s 2 calorie (cal) is the amount of energy needed to raise the temperature of one gram of water 1°C kcal = energy needed to raise 1000 g of water 1°C food Calories = kcals Energy Conversion Factors 1 calorie (cal)=4.184 joules (J) (exact) 1 Calorie (Cal)=1000 cal = 1 kcal = 4184 J 1 kilowatt-hour (kWh)=3.60 x 10 6 J 9 Tro: Chemistry: A Molecular Approach, 2/e

10. Copyright © 2011 Pearson Education, Inc. The First Law of Thermodynamics: Law of Conservation of Energy Thermodynamics is the study of energy and its interconversions The First Law of Thermodynamics is the Law of Conservation of Energy This means that the total amount of energy in the universe is constant You can therefore never design a system that will continue to produce energy without some source of energy 10 Tro: Chemistry: A Molecular Approach, 2/e

11. Copyright © 2011 Pearson Education, Inc. Energy Flow and Conservation of Energy energy changes ( Conservation of energy requires that the sum of the energy changes (  E) in the system and the surroundings must be zero  Energy universe = 0 =  Energy system +  Energy surroundings  Is the symbol that is used to mean change  final amount – initial amount 11 Tro: Chemistry: A Molecular Approach, 2/e

12. Copyright © 2011 Pearson Education, Inc. Internal Energy The internal energy is the sum of the kinetic and potential energies of all of the particles that compose the system The change in the internal energy of a system only depends on the amount of energy in the system at the beginning and end  E = E final – E initial Internal energy is a state function, a mathematical function whose result only depends on the initial and final conditions, not on the process used  E reaction = E products − E reactants 12 Tro: Chemistry: A Molecular Approach, 2/e

13. Copyright © 2011 Pearson Education, Inc. State Function From the foothill to the peak, many pathways, the change in potential energy is the same. Potential energy is a state function. It depends only on the difference in elevation between the base and the peak. The actual travelling distance for each pathway is NOT a state function. It depends only many factors, like slope, terrain, etc.12 miles vs. 5 miles. 13 Tro: Chemistry: A Molecular Approach, 2/e

14. Copyright © 2011 Pearson Education, Inc. Energy Diagrams 14 Energy diagrams: “graphical” way of showing the direction of energy flow during a process Internal Energy initial final energy increase Internal Energy initial final energy decrease If the final condition has a larger amount of internal energy than the initial condition, the change in the internal energy will be + If the final condition has a smaller amount of internal energy than the initial condition, the change in the internal energy will be ─ Tro: Chemistry: A Molecular Approach, 2/e

15. Copyright © 2011 Pearson Education, Inc. Energy Flow: System  Surrounding When energy flows out of a system, it must all flow into the surroundings When energy flows out of a system,  E system is ─ Surroundings gains energy,  E surroundings is + So ─  E system =  E surroundings Example: Hot water (system) lose heat to cool air in the room (surroundings) Surroundings  E + System  E ─ 15 Tro: Chemistry: A Molecular Approach, 2/e

16. Copyright © 2011 Pearson Education, Inc. Energy Flow: System  Surroundings When energy flows into a system, it must all come from the surroundings System gains energy,  E system is + Energy flows out of the surroundings,  E surroundings is ─ So  E system = ─  E surroundings Example: Rechargeable battery (system) receives energy from charger (surroundings) Surroundings  E ─ System  E + 16 Tro: Chemistry: A Molecular Approach, 2/e

17. Copyright © 2011 Pearson Education, Inc. Energy Flow in a Chemical Reaction Fact: Total internal energy in 1mol of C(s) and 1 mole of O 2 (g) > Internal energy in 1 mole of CO 2 (g) at the same temperature and pressure Reaction C(s) + O 2 (g) → CO 2 (g) : system loses energy to the surroundings: −  E reaction =  E surroundings Reaction CO 2 (g) → C(s) + O 2 (g) : system gains energy from the surroundings  E reaction = −  E surroundings Internal Energy CO 2 (g) C(s) + O 2 (g) Internal Energy C(s) + O 2 (g) CO 2 (g) 17 Tro: Chemistry: A Molecular Approach, 2/e

18. Copyright © 2011 Pearson Education, Inc. Work and Heat in Energy Exchange Energy is exchanged between the system and surroundings through heat and work q = heat (thermal) energy w = work energy q and w are NOT state functions, their value depends on the process  E = q + w 18 Tro: Chemistry: A Molecular Approach, 2/e

19. Copyright © 2011 Pearson Education, Inc. Energy Exchange in real life Example: Combustion reaction in engine: System (reaction) loses heat (q < 0) and does work (w < 0) to the surroundings. Charging battery: System (battery) loses heat (gets warm, q 0) from the surroundings 19 Tro: Chemistry: A Molecular Approach, 2/e

20. Copyright © 2011 Pearson Education, Inc. Heat Exchange Heat is the exchange of thermal energy between the system and surroundings Heat exchange occurs when system and surroundings have a difference in temperature Temperature is the measure of the amount of thermal energy within a sample of matter Heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature thermal equilibrium 20 Tro: Chemistry: A Molecular Approach, 2/e

21. Copyright © 2011 Pearson Education, Inc. Temperature change and Heat Capacity When a system absorbs heat, its temperature increases The increase in temperature is directly proportional to the amount of heat absorbed The proportionality constant is called the heat capacity, C units of C are J/°C or J/K q = C x  T The larger the heat capacity of the object being studied, the smaller the temperature rise will be for a given amount of heat 21 Tro: Chemistry: A Molecular Approach, 2/e

22. Copyright © 2011 Pearson Education, Inc. Factors Affecting Heat Capacity The heat capacity of an object depends on its amount of matter usually measured by its mass 200 g of water requires twice as much heat to raise its temperature by 1°C as does 100 g of water The heat capacity of an object depends on the type of material 1000 J of heat energy will raise the temperature of 100 g sand by 12°C, but only raise the temperature of 100 g water by 2.4 °C 22 Tro: Chemistry: A Molecular Approach, 2/e

23. Copyright © 2011 Pearson Education, Inc. Specific Heat Capacity The specific heat capacity, or specific heat, is the amount of heat energy required to raise the temperature of one gram of a substance by 1°C C Measure of a substance’s intrinsic ability to absorb heat Unit = J/(g∙°C) 23 Tro: Chemistry: A Molecular Approach, 2/e

24. Copyright © 2011 Pearson Education, Inc. Specific Heat of Water The rather high specific heat of water allows water to absorb a lot of heat energy without a large increase in its temperature Weather pattern: beach vs. inland. The large amount of water absorbing heat from the air keeps beaches cool in the summer Used as a coolant because it can absorb a lot of heat and remove it from the important mechanical parts to keep them from overheating 24 Tro: Chemistry: A Molecular Approach, 2/e

25. Copyright © 2011 Pearson Education, Inc. Quantifying Heat Energy The heat capacity of an object is proportional to its mass and the specific heat of the material Heat calculation from specific heat: Heat absorbed or released can be calculated as if we know the mass, the specific heat, and the temperature change of the object Heat = (mass) x (specific heat) x (temp. change) q = m x c x  T 25 Tro: Chemistry: A Molecular Approach, 2/e

26. Copyright © 2011 Pearson Education, Inc. Find Heat: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from −8.0 °C to 37.0 °C? Specific heat of copper is 0.385 J/g °C  T = 29.0°C q = 53.7 J

27. Copyright © 2011 Pearson Education, Inc. Heat Transfer and Temperature Change When two objects are placed in contact, heat flows from the material at the higher temperature to the material at the lower temperature Heat flows until both materials reach the same final temperature Heat lost by the hot material equals the heat gained by the cold material q hot = −q cold m hot  C s,hot   T hot = −(m cold  C s,cold   T cold ) 27 Tro: Chemistry: A Molecular Approach, 2/e

28. Copyright © 2011 Pearson Education, Inc. 1. Heat lost by the metal > heat gained by water 2. Heat gained by water > heat lost by the metal 3. Heat lost by the metal > heat lost by the water 4. Heat lost by the metal = heat gained by water 5. More information is required 1. Heat lost by the metal > heat gained by water 2. Heat gained by water > heat lost by the metal 3. Heat lost by the metal > heat lost by the water 4. Heat lost by the metal = heat gained by water 5. More information is required A piece of metal at 85 °C is added to water at 25 °C, the final temperature of the both metal and water is 30 °C. Which of the following is true? 28 Tro: Chemistry: A Molecular Approach, 2/e

29. Copyright © 2011 Pearson Education, Inc. Find Specific heat: A 100.-g of unknown metal alloy is heated to 99.8°C and submerged into 200. g water at 25.9°C. The final temperature of the water becomes 32.9°C. What is the specific heat of the metal alloy? 29 q m = −q H2O Tro: Chemistry: A Molecular Approach, 2/e

30. Copyright © 2011 Pearson Education, Inc. Predict  T: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? 30 q Al = −q H2O Tro: Chemistry: A Molecular Approach, 2/e

31. Copyright © 2011 Pearson Education, Inc. Exchanging Energy Between System and Surroundings Exchange of heat energy q = mass x specific heat x  Temperature Exchange of work w = −Pressure x  Volume 31 Tro: Chemistry: A Molecular Approach, 2/e

32. Copyright © 2011 Pearson Education, Inc. Measure Heat from Chem. Rxn: Bomb Calorimeter Calorimeter Calorimeter for Heat measurement Used to measure  E because it is a constant volume system The heat capacity of the calorimeter = heat absorbed by the calorimeter per degree rise in temperature. Also called the calorimeter constant C cal, kJ/ºC 32 Tro: Chemistry: A Molecular Approach, 2/e

33. Copyright © 2011 Pearson Education, Inc. When 1.010 g of sugar (C 12 H 22 O 11 ) is burned in a bomb calorimeter, the temperature rises from 24.92 °C to 28.33 °C. If heat capacity of calorimeter C cal = 4.90 kJ/°C, find  E for burning 1 mole of sugar. 33 Tro: Chemistry: A Molecular Approach, 2/e Mol sugar = 0.0025906 molq cal = 16.7 kJ q rxn = -16.7 kJ  E = -5.66 x 10 -3 kJ

34. Copyright © 2011 Pearson Education, Inc. Enthalpy The enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume H is a state function H = E + PV The enthalpy change,  H, of a reaction is the heat evolved in a reaction at constant pressure  H reaction = q reaction at constant pressure Usually  H and  E are similar in value, the difference is largest for reactions that produce or use large quantities of gas 34 Tro: Chemistry: A Molecular Approach, 2/e

35. Copyright © 2011 Pearson Education, Inc. Endothermic and Exothermic Reactions   H < 0 (or H final < H initial ): heat is released by the system Heat-releasing reactions: exothermic reactions Examples: Chemical heat packs Combustion reaction, Explosion reaction   H > 0 (or H final > H initial ): heat is absorbed by the system Heat-absorbing reactions: endothermic reactions Chemical cold packs contain NH 4 NO 3 that dissolves in water in an endothermic process ─ your hands get cold because the pack is absorbing your heat 35 Tro: Chemistry: A Molecular Approach, 2/e

36. Copyright © 2011 Pearson Education, Inc. Enthalpy of Reaction The enthalpy change in a chemical reaction is an extensive property (aka, depending on the amount) the more reactants you use, the larger the enthalpy change. More C 3 H 8 ? More enthalpy change! By convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written C 3 H 8 (g) + 5 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(g)  H = −2044 kJ 36 Tro: Chemistry: A Molecular Approach, 2/e

37. Copyright © 2011 Pearson Education, Inc. Example: How much heat is evolved in the complete combustion of 13.2 kg of C 3 H 8 (g)? molkJkgg 37 Tro: Chemistry: A Molecular Approach, 2/e C 3 H 8 (g) + 5O 2 (g)→3CO 2 (g) + 4H 2 O(g)  H = −2044 kJ

38. Copyright © 2011 Pearson Education, Inc. Measuring  H Calorimetry at Constant Pressure Reactions done in aqueous solution are at constant pressure open to the atmosphere Instrument: Foam cup calorimeter contains the solution (foams provides great insulation! ) q reaction = ─ q solution = ─(mass solution x C s, solution x  T)   H reaction = q constant pressure = q reaction to get  H reaction per mol, divide by the number of moles of limiting reagent 38 Tro: Chemistry: A Molecular Approach, 2/e

39. Copyright © 2011 Pearson Education, Inc. Determining the Enthalpy Change of an Aqueous Reaction 50.0 mL of 0.500 M NaOH is placed in a coffee-cup calorimeter at 25.00ºC and 25.0 mL of 0.500 M HCl is carefully added, also at 25.00ºC. After stirring, the final temperature is 27.21ºC. Calculate q soln (in J) and the change in enthalpy,  H, (in kJ/mol of H 2 O formed). Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/mL, and that c = 4.184 J/g∙K PLAN: Heat flows from the reaction (the system) to its surroundings (the solution): –q rxn = q soln. Start with heat absorbed by the solution. q soln = 693 Jq rxn = -693 J Find Limiting reactant: HCl; the amount of product H 2 O = 0.0125 mol.  H rxn = q rxn /mol L.R. = 55.4 kJ/mol H 2 O

40. Copyright © 2011 Pearson Education, Inc. More Example: What is  H rxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl 2 (aq) + H 2 (g) if 0.158 g Mg reacts in 100.0 mL of solution and the solution changes the temperature from 25.6°C to 32.8 °C? Assume d soln = 1.00 g/mL, C soln = 4.18 J/g∙°C 40 Q soln = 3.0E+3 J Mol Mg = 6.499E-3 mol  H rxn/mol Mg = -4.6E+5 J/mol Mg

41. Copyright © 2011 Pearson Education, Inc. Relationships Involving  H rxn When reaction is multiplied by a factor,  H rxn is multiplied by that factor because  H rxn is extensive C(s) + O 2 (g) → CO 2 (g)  H = −393.5 kJ 2 C(s) + 2 O 2 (g) → 2 CO 2 (g)  H = 2(−393.5 kJ) = −787.0 kJ If a reaction is reversed, then the sign of  H is changed CO 2 (g) → C(s) + O 2 (g)  H = +393.5 kJ 41 Tro: Chemistry: A Molecular Approach, 2/e

42. Copyright © 2011 Pearson Education, Inc. Relationships Involving  H rxn : Hess’s Law If a reaction can be expressed as a series of steps, then the  H rxn for the overall reaction is the sum of the heats of reaction for each step 42 Tro: Chemistry: A Molecular Approach, 2/e A + 2B → C  H 1 C + E → F  H 2 F → G  H 3 A + 2B + C + E + F → C + F + G Total reaction: A + 2B + E → G  H 4 =  H 1  H 2  H 3

43. Copyright © 2011 Pearson Education, Inc. How to use Hess’s Law from given  H’s Compare the target reaction and given reactions Two basic manipulations: Reverse and/or Multiply If one formula exists in more than one of the given rxns, do not attempt to manipulate eqn using this formula. A + 2B → 3F Example: Find enthalpy change for A + 2B → 3F 43 Tro: Chemistry: A Molecular Approach, 2/e Given: A + 2B → 2C  H 1 3G → 2C  H 2 F → G  H 3 A + 2B → 3F  H =  H 1  H 2 )  H 3  2C → 3G -  H 2 3G → 3F -3 x  H 3

44. Copyright © 2011 Pearson Education, Inc. Example: Calculating ΔHº rxn from given ΔHº Values Find  H rxn for the rxn: 4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(g) Given: N 2 (g) + 3H 2 (g) → 2NH 3 (g)  H 1 = -91.8 kJ N 2 (g) + O 2 (g) → 2NO(g)  H 2 = 180.6 kJ 2H 2 (g) + O 2 (g) → 2H 2 O(g)  H 3 = -483.6 kJ  H rxn = -906.0 kJ

45. Copyright © 2011 Pearson Education, Inc. Using Hess’s Law to Calculate an Unknown  H The catalytic converter in automobiles ultilizes the following reaction to convert pollutants CO and NO into harmless gases: CO (g) + NO (g) → CO 2 (g) + ½N 2 (g) Given the following information, calculate the unknown  H: Equation A: CO (g) + ½O 2 (g) → CO 2 (g)  H A = –283.0 kJ Equation B: N 2 (g) + O 2 (g) → 2NO (g)  H B = 180.6 kJ  H rxn = –373.3 kJ

46. Copyright © 2011 Pearson Education, Inc. Standard Conditions The standard state is the state of a material at a defined set of conditions pure gas at exactly 1 atm pressure pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest  usually 25 °C substance in a solution with concentration 1 M The standard enthalpy change,  H°, is the enthalpy change when all reactants and products are in their standard states 46 Tro: Chemistry: A Molecular Approach, 2/e

47. Copyright © 2011 Pearson Education, Inc. Standard enthalpy of formation,  H f °, Formation Reaction: Reactions of elements in their standard state to form 1 mole of a pure compound The standard states of all metals are solid except Hg; Nonmetals: Only Br 2 is liquid, low atomic mass nonmetals are gas whilst heavy nonmetals are solid The standard enthalpy of formation,  H f °: the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements the elements must be in their standard states the  H f ° for a pure element in its standard state = 0 kJ/mol 47 Tro: Chemistry: A Molecular Approach, 2/e

48. Copyright © 2011 Pearson Education, Inc. Standard Enthalpies of Formation 48 Tro: Chemistry: A Molecular Approach, 2/e

49. Copyright © 2011 Pearson Education, Inc. Writing Formation Reactions Write the Formation Reaction for CO(g) The formation reaction is the reaction between the elements in the compound C + O → CO(g) The elements must be in their standard state C(s, graphite) + O 2 (g) → CO(g) The equation must be balanced, but the coefficient of the product compound must be 1 C(s, graphite) + ½ O 2 (g) → CO(g) 49 Tro: Chemistry: A Molecular Approach, 2/e

50. Copyright © 2011 Pearson Education, Inc. Write the formation reactions for the following: CO 2 (g) Al 2 (SO 4 ) 3 (s) C(s, graphite) + O 2 (g)  CO 2 (g) 2 Al(s) + 3 S(s) + 6 O 2 (g)  Al 2 (SO 4 ) 3 (s) 50 Tro: Chemistry: A Molecular Approach, 2/e

51. Copyright © 2011 Pearson Education, Inc. More example: Writing Formation Equations Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include  Hº f. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. Ag (s) + ½Cl 2 (g) → AgCl (s)  Hº f = –127.0 kJ Ca (s) + C(graphite) + O 2 (g) → CaCO 3 (s)  H° f = –1206.9 kJ 3 2 ½H 2 (g) + C(graphite) + ½N 2 (g) → HCN (g)  H f = 135 kJ

52. Copyright © 2011 Pearson Education, Inc. Calculating Standard Enthalpy Change for a Reaction Any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products The  H° for the reaction is then the sum of the  H f ° for the component reactions  H° reaction =  n  H f °(products) −  n  H f °(reactants)  means sum n is the coefficient of the reaction 52 Tro: Chemistry: A Molecular Approach, 2/e

53. Copyright © 2011 Pearson Education, Inc. Calculate the enthalpy change in the reaction 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(l) 1. Collect the  H f ° for each compound in the equation 2 C(s, gr) + H 2 (g)  C 2 H 2 (g)  H f ° = +227.4 kJ/mol C(s, gr) + O 2 (g)  CO 2 (g)  H f ° = −393.5 kJ/mol H 2 (g) + ½ O 2 (g)  H 2 O(l)  H f ° = −285.8 kJ/mol 53 Tro: Chemistry: A Molecular Approach, 2/e 2.  H° reaction =  n  H f °(products) −  n  H f °(reactants)  H rxn = −2600.4 kJ

54. Copyright © 2011 Pearson Education, Inc. Calculating ΔHº rxn from ΔHº f Values For the following reaction: 4NH 3 (g) + 5O 2 (g) → 4NO (g) + 6H 2 O (g) Calculate  H rxn from  H f values.  H rxn =  m  Hº f (products) –  n  Hº f (reactants) m and n are the coefficients for products and reactants, respectively.  H f (kJ/mol): NH3(g) -45.9, NO(g) 90.3, H2O(g) -241.8  H rxn = –906 kJ

55. Copyright © 2011 Pearson Education, Inc. Example: Determine the standard enthalpy of formation for nitromethane from  H° rxn 2CH 3 NO 2 (l) + 3/2 O 2 (g)  2CO 2 (g) + N 2 (g) + 3H 2 O(l)  H rxn = -709.2 kJ/mol 55 Tro: Chemistry: A Molecular Approach, 2/e 1. Collect the  H f ° for each compound in the equation C(s, gr) + O 2 (g)  CO 2 (g)  H f ° = −393.5 kJ/mol H 2 (g) + ½ O 2 (g)  H 2 O(l)  H f ° = −285.8 kJ/mol 55 N 2 (g) and O 2 (g) are the element at the standard state,  H f ° = 0  Set up equation based on  H° rxn =  n  H f °(products) −  n  H f °(reactants)  H CH3NO2 = −401.6 kJ

56. Copyright © 2011 Pearson Education, Inc. Practice: Determine the standard enthalpy of formation for TNT (C 7 H 5 N 3 O 6 ). The heat of combustion for TNT is -14.5 MJ/kg. C 7 H 5 N 3 O 6 (s) + ?O 2 (g)  7CO 2 (g) + 3/2 N 2 (g) + 5/2 H 2 O(l)  H rxn = -3400 kJ/mol find  H f °(C 7 H 5 N 3 O 6 ) 56 Tro: Chemistry: A Molecular Approach, 2/e  H f °[CO 2 (g)] = −393.5 kJ/mol  H f °(H 2 O(l)) = −285.8 kJ/mol 56 N 2 (g) and O 2 (g) as element at the standard state,  H f ° = 0  H° rxn =  n  H f °(products) −  n  H f °(reactants)

57. Copyright © 2011 Pearson Education, Inc. 57 Practice: How many kg of octane must be combusted to supply 1.0 x 10 11 kJ of energy? MM octane = 114.2 g/mol  H°rxn = −5074.1 kJ Tro: Chemistry: A Molecular Approach, 2/e

58. Copyright © 2011 Pearson Education, Inc. Table Selected Standard Enthalpies of Formation at 25°C (298K) Formula  H° f (kJ/mol) Calcium Ca(s) CaO(s) CaCO 3 (s) Carbon C(graphite) C(diamond) CO(g) CO 2 (g) CH 4 (g) CH 3 OH(l) HCN(g) CS s (l) Chlorine Cl(g) 0 –635.1 –1206.9 0 1.9 –110.5 –393.5 –74.9 –238.6 135 87.9 121.0 Hydrogen Nitrogen Oxygen Formula  H° f (kJ/mol) H(g) H2(g)H2(g) N2(g)N2(g) NH 3 (g) NO(g) O2(g)O2(g) O3(g)O3(g) H 2 O(g) H 2 O(l) Cl 2 (g) HCl(g) 0 0 0 –92.3 0 218.0 –45.9 90.3 143 –241.8 –285.8 107.8 Formula  H° f (kJ/mol) Silver Ag(s) AgCl(s) Sodium Na(s) Na(g) NaCl(s) Sulfur S 8 (rhombic) S 8 (monoclinic) SO 2 (g) SO 3 (g) 0 0 0 –127.0 –411.1 0.3 –296.8 –396.0

59. Copyright © 2011 Pearson Education, Inc. The two-step process for determining  Hº rxn from  Hº f values.

60. Copyright © 2011 Pearson Education, Inc. Find Heat: How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from −8.0 °C to 37.0 °C? q = m ∙ C s ∙  T C s = 0.385 J/gºC (Table 6.4) the unit is correct, the sign is reasonable as the penny must absorb heat to make its temperature rise T 1 = −8.0 °C, T 2 = 37.0 °C, m = 3.10 g q, Jq, J Check: Check Solution: Follow the conceptual plan to solve the problem Conceptual Plan: Relationships: Strategize Given: Find: Sort Information C s m,  T q 60 Tro: Chemistry: A Molecular Approach, 2/e

61. Copyright © 2011 Pearson Education, Inc. C H2O = 4.18 J/gºC Find Specific heat: A 100.-g of unknown metal alloy is heated to 99.8°C and submerged into 200. g water at 25.9°C. The final temperature of the water becomes 32.9°C. What is the specific heat of the metal alloy? m m =100.g, T mi = 99.8°C, m H20 =200.g, T H2Oi =25.9°C, T final = 32.9°C find C m Solution: Plan: Given: Find: 61 q m = −q H2O Tro: Chemistry: A Molecular Approach, 2/e

62. Copyright © 2011 Pearson Education, Inc. Find Specific heat (contd.): A 100.-g of unknown metal alloy is heated to 99.8°C and submerged into 200. g water at 25.9°C. The final temperature of the water becomes 32.9°C. What is the specific heat of the metal alloy? m m = 100. g, T mi = 99.8°C, m H20 = 200. g, T H2Oi = 25.9°C, T final = 32.9°Cfind C m Solution: Given: Find: 62 Tro: Chemistry: A Molecular Approach, 2/e

63. Copyright © 2011 Pearson Education, Inc. q = m ∙ C s ∙  T C s, Al = 0.903 J/gºC, C s, H2O = 4.18 J/gºC(Table 6.4) Predict  T: A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures m Al = 32.5 g, T al = 45.8 °C, m H20 = 105.3 g, T H2O = 15.4 °C T final, °C Check: Solution: Conceptual Plan: Relationships: Given: Find: C s, Al m Al, C s, H2O m H2O  T Al = k   T H2O 63 q Al = −q H2O T final Tro: Chemistry: A Molecular Approach, 2/e

64. Copyright © 2011 Pearson Education, Inc. q = m ∙ C s ∙  T C s, Al = 0.903 J/gºC, C s, H2O = 4.18 J/gºC(Table 6.4) (Continued): A 32.5-g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? the unit is correct, the number is reasonable as the final temperature should be between the two initial temperatures m Al = 32.5 g, T al = 45.8 °C, m H20 = 105.3 g, T H2O = 15.4 °C T final, °C Check: Solution: Conceptual Plan: Relationships: Given: Find: C s, Al m Al, C s, H2O m H2O  T Al = k   T H2O 64 q Al = −q H2O T final Tro: Chemistry: A Molecular Approach, 2/e

65. Copyright © 2011 Pearson Education, Inc. Molar Energy Change: When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92 °C to 28.33 °C. If C cal = 4.90 kJ/°C, find  E for burning 1 mole of sugar. q cal = C cal x  T = −q rxn MM C 12 H 22 O 11 = 342.3 g/mol the units are correct, the sign is as expected for combustion 1.010 g C 12 H 22 O 11, T 1 = 24.92 °C, T 2 = 28.33 °C, C cal = 4.90 kJ/°C  E rxn, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: q rxn, mol EE C cal,  T q cal q rxn 65 Tro: Chemistry: A Molecular Approach, 2/e

66. Copyright © 2011 Pearson Education, Inc. Example: How much heat is evolved in the complete combustion of 13.2 kg of C 3 H 8 (g)? combustion of 1 mole C 3 H 8 gives 2044 kJ heat. 1 kg = 1000 g, Molar Mass = 44.09 g/mol 13.2kg C 3 H 8, C 3 H 8 (g) + 5O 2 (g)→3CO 2 (g) + 4H 2 O(g)  H = −2044 kJ q, kJ/mol Solution: Conceptual Plan: Relationships: Given: Find: molkJkgg 66 Tro: Chemistry: A Molecular Approach, 2/e

67. Copyright © 2011 Pearson Education, Inc. Example: Determining the Enthalpy Change of an Aqueous Reaction PROBLEM:50.0 mL of 0.500 M NaOH is placed in a coffee-cup calorimeter at 25.00ºC and 25.0 mL of 0.500 M HCl is carefully added, also at 25.00ºC. After stirring, the final temperature is 27.21ºC. Calculate q soln (in J) and the change in enthalpy,  H, (in kJ/mol of H 2 O formed). Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/mL, and that c = 4.184 J/g∙K PLAN:Heat flows from the reaction (the system) to its surroundings (the solution). Since –q rxn = q soln, we can find the heat of the reaction by calculating the heat absorbed by the solution.

68. Copyright © 2011 Pearson Education, Inc. SOLUTION: Total mass (g) of the solution = (25.0 mL + 50.0 mL) x 1.00 g/mL = 75.0 g  T soln = 27.21ºC – 25.00ºC = 2.21ºC = 2.21 K q soln = c soln x mass soln x  T soln = (4.184 J/g∙K)(75.0 g)(2.21 K) = 693 J (a) To find q soln : (b) To find  H rxn we first need a balanced equation: HCl (aq) + NaOH (aq) → NaCl (aq) + H 2 O (l)

69. Copyright © 2011 Pearson Education, Inc. Example (contd): For HCl: 25.0 mL HCl x 1 L 10 3 mL 0.500 mol 1 L xx 1 mol H 2 O 1 mol HCl = 0.0125 mol H 2 O For NaOH: 50.0 mL NaOH x 1 L 10 3 mL xx 1 mol H 2 O 1 mol NaOH = 0.0250 mol H 2 O 0.500 mol 1 L HCl is limiting, and the amount of H 2 O formed is 0.0125 mol.  H rxn = q rxn mol H 2 O 0.0125 mol –693 J = 1 kJ 10 3 J x = –55.4 kJ/mol H 2 O

70. Copyright © 2011 Pearson Education, Inc. Example 6.8: What is  H rxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl 2 (aq) + H 2 (g) if 0.158 g Mg reacts in 100.0 mL of solution and changes the temperature from 25.6 °C to 32.8 °C? the heat released from reaction (-) is transferred to the solution, so solution gains heat (+) 0.158 g Mg, 100.0 mL, T 1 = 25.6 °C, T 2 = 32.8 °C, assume d = 1.00 g/mL, C s = 4.18 J/g∙°C  H, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: q rxn, mol HH C s,  T, m q sol’n q rxn q sol’n = m x C s, sol’n x  T = −q rxn, MM Mg= 24.31 g/mol 70 Tro: Chemistry: A Molecular Approach, 2/e

71. Copyright © 2011 Pearson Education, Inc. Example 6.8: What is  H rxn/mol Mg for the reaction Mg(s) + 2 HCl(aq) → MgCl 2 (aq) + H 2 (g) if 0.158 g Mg reacts in 100.0 mL of solution and changes the temperature from 25.6 °C to 32.8 °C? the sign is correct and the value is reasonable because the reaction is exothermic 0.158 g Mg, 100.0 mL, T 1 = 25.6 °C, T 2 = 32.8 °C, assume d = 1.00 g/mL, C s = 4.18 J/g∙°C  H, kJ/mol Check: Solution: Conceptual Plan: Relationships: Given: Find: q rxn, mol HH C s,  T, m q sol’n q rxn q sol’n = m x C s, sol’n x  T = −q rxn, MM Mg= 24.31 g/mol 71 Tro: Chemistry: A Molecular Approach, 2/e

72. Copyright © 2011 Pearson Education, Inc. More example: Calculating ΔHº rxn from given ΔHº Values Find  H rxn for the rxn: 4NH 3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(g) Given: N 2 (g) + 3H 2 (g) → 2NH 3 (g)  H 1 = -91.8 kJ N 2 (g) + O 2 (g) → 2NO(g)  H 2 = 180.6 kJ 2H 2 (g) + O 2 (g) → 2H 2 O(g)  H 3 = -483.6 kJ NH 3 : Rxn# 1 reverse and x 2 4NH 3 (g) → 2N 2 (g) + 6H 2 (g)  H = 183.6 kJ O 2 : in more than one rxn, skip NO: Rxn# 2 times 2 2N 2 (g) + 2O 2 (g) → 4NO(g)  H = 361.2 kJ H 2 O: Rxn# 3 times 3 6H 2 (g) + 3O 2 (g) → 6H 2 O(g)  H = -1450.8 kJ Sum of all reactions:  H rxn = 183.6 kJ+361.2 kJ + (-1450.8 kJ) = 906.0 kJ 4NH 3 (g) + 2N 2 (g) + 2O 2 (g) + 6H 2 (g) + 3O 2 (g) → 2N 2 (g) + 6H 2 (g) + 4NO(g) + 6H 2 O(g)

73. Copyright © 2011 Pearson Education, Inc. 73 Example 6.12: How many kg of octane must be combusted to supply 1.0 x 10 11 kJ of energy? MM octane = 114.2 g/mol, 1 kg = 1000 g the units and sign are correct, the large value is expected 1.0 x 10 11 kJ mass octane, kg Check: Solution: Conceptual Plan: Relationships: Given: Find:  H f °’s  H rxn ° Write the balanced equation per mole of octane kJ mol C 8 H 18 g C 8 H 18 kg C 8 H 18 from above C 8 H 18 (l) + 25/2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g) Look up the  H f ° for each material in Appendix IIB  H°rxn = −5074.1 kJ Tro: Chemistry: A Molecular Approach, 2/e

74. Copyright © 2011 Pearson Education, Inc. 74 Example 6.12: How many kg of octane must be combusted to supply 1.0 x 10 11 kJ of energy? MM octane = 114.2 g/mol, 1 kg = 1000 g the units and sign are correct, the large value is expected 1.0 x 10 11 kJ mass octane, kg Check: Solution: Conceptual Plan: Relationships: Given: Find:  H f °’s  H rxn ° Write the balanced equation per mole of octane kJ mol C 8 H 18 g C 8 H 18 kg C 8 H 18 from above C 8 H 18 (l) + 25/2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g)  H°rxn = −5074.1 kJ Tro: Chemistry: A Molecular Approach, 2/e

75. Copyright © 2011 Pearson Education, Inc. 75 Example 6.12: How many kg of octane must be combusted to supply 1.0 x 10 11 kJ of energy? MM octane = 114.2 g/mol, 1 kg = 1000 g the units and sign are correct, the large value is expected 1.0 x 10 11 kJ mass octane, kg Check: Solution: Conceptual Plan: Relationships: Given: Find:  H f °’s  H rxn ° Write the balanced equation per mole of octane kJ mol C 8 H 18 g C 8 H 18 kg C 8 H 18 from above C 8 H 18 (l) + 25/2 O 2 (g) → 8 CO 2 (g) + 9 H 2 O(g)  H°rxn = −5074.1 kJ Tro: Chemistry: A Molecular Approach, 2/e

76. Copyright © 2011 Pearson Education, Inc. More examples: Using Hess’s Law to Calculate an Unknown  H PROBLEM:Two gaseous pollutants that form in auto exhausts are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following reaction: CO (g) + NO (g) → CO 2 (g) + ½N 2 (g)  H = ? Given the following information, calculate the unknown  H: Equation A: CO (g) + ½O 2 (g) → CO 2 (g)  H A = –283.0 kJ Equation B: N 2 (g) + O 2 (g) → 2NO (g)  H B = 180.6 kJ PLAN: Manipulate Equations A and/or B and their  H values to get to the target equation and its  H. All substances except those in the target equation must cancel.

77. Copyright © 2011 Pearson Education, Inc. More example: Writing Formation Equations PROBLEM:Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include  Hº f. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. PLAN:Write the elements as reactants and 1 mol of the compound as the product formed. Make sure all substances are in their standard states. Balance the equations and find the value of  Hº f in Table 6.3 or Appendix B. (c) Hydrogen cyanide, HCN, a gas at standard conditions.

78. Copyright © 2011 Pearson Education, Inc. SOLUTION: (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO 3, a solid at standard conditions. Ag (s) + ½Cl 2 (g) → AgCl (s)  Hº f = –127.0 kJ Ca (s) + C(graphite) + O 2 (g) → CaCO 3 (s)  H° f = –1206.9 kJ 3 2 (c) Hydrogen cyanide, HCN, a gas at standard conditions. ½H 2 (g) + C(graphite) + ½N 2 (g) → HCN (g)  H° f = 135 kJ