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Unit 10 (Chapter 17): Thermodynamics

Unit 10 (Chapter 17): Thermodynamics. Cartoon courtesy of NearingZero.net. 90 0 C. I. Introduction. Thermochemistry : Study of heat flow in chemical reactions. Heat (Q) – The energy that transfers from one object to another, because of a temperature difference between them.

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Unit 10 (Chapter 17): Thermodynamics

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  1. Unit 10 (Chapter 17): Thermodynamics Cartoon courtesy of NearingZero.net

  2. 900C I. Introduction Thermochemistry: Study of heat flowin chemical reactions. Heat (Q)– The energy that transfers from one object to another, because of a temperature difference between them. • flows from warmer  cooler object Enthalpy (Δ H) – The heat released or absorbed. • Used interchangeably with “Q”! Energy units: Joules (J), Kilojoules (kJ), calories (cal), kilocalories (kcal) Food: “C”

  3. Exothermic and Endothermic Processes • In studying heat changes, think of defining these two parts: • the system- the part of the universe on which you focus your attention. • the surroundings- includes everything else in the universe.

  4. Heat (Enthalpy) Change Two types of heat change exist. Endothermic: Processes in which energy is absorbed into the system as it proceeds, and surroundings become colder. + qsystem - qsurroundings Exothermic: Processes in which energy is releasedas it proceeds, and surroundings become warmer. - qsystem + qsurroundings

  5. Activation Energy (EA) The minimum energy required to initiate a chemical reaction. Examples: Flame, spark, high temperature, radiation are all sources of activation energy

  6. Endothermic Reactions *** Where does the released energy come from and the absorbed energy go? Breaking and forming bonds

  7. Exothermic Reactions

  8. Exothermic and Endothermic • Fig. 17.2, page 506 - on the left, the system (the people) gain heat from it’s surroundings (the fire) • this is endothermic • On the right, the system (the body) cools as perspiration evaporates, and heat flows to the surroundings • this is exothermic

  9. Exothermic and Endothermic • Every reaction has an energy change associated with it. • Exothermic reactions release energy, usually in the form of heat. • Endothermic reactions absorb energy. • Energy is stored in bondsbetween atoms.

  10. The Effect of a Catalyst • Catalyst: A substance that speeds up a reaction without being consumed. It lowers the activation energy! • Enzyme: A large molecule (usually a protein) that catalyzes biological reactions.

  11. Endothermic Reaction witha Catalyst

  12. Exothermic Reaction with a Catalyst

  13. Change in Enthalpy (____) = _______ content = Heat of _______ • 1. ____ = Heat of __________ - Heat of __________ • 2. Exothermic reactions: _____ = ____ • 3. Endothermic reactions:_____ = ____ • Units of _____: • 5. Example: Combustion of methane (_____) ____= -802 kJ/mol • Example: Decomposition of limestone (_______) • _____= +177.8 kJ/mol Heat  H reaction  H product reactants  H –  H +  H or CH4  H CH4 + 2O2 CO2 + 2 H2O + 802 kJ CaCO3  H CaCO3 + 177.8 kJ + CO2 CaO

  14. H 3 Techniques Determination of ______? 1. Experimentally 2. Using Heats of __________ 3. Using ______ Law Formation Hess’s

  15. ^ Calculating _____ Experimentally H Heat exchange: Heat released in a chemical reaction = Heat absorbed by water in calorimeter Q released = Q absorbed Simple Calorimeter: Heat of solution Heat of combustion Thermometer can Styrofoam cup Solution

  16. Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured… …usually by the change in temperature of a known quantity of water • 1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C • 1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

  17. The Joule The unit of heat used in modern thermochemistry is the Joule. 4.184 joules = 1 calorie

  18. A Bomb Calorimeter

  19. A Cheaper Calorimeter

  20. Heat Capacity/Specific Heat (c) • Heat Capacity/ Specific Heat- the amount of heat needed to increase the temperature of an object exactly 1 oC. • Depends on both the object’s mass and its chemical composition. • Units are either J/(g oC) or cal/(g oC)

  21. Specific Heat cont. 4.184 J/goC 1 cal/goC 0.444 J/goC * Specific heat of water: ________ = ___________ * Specific heat of iron: __________  Example 1: compare heating an empty stainless steel pan to a pan full of water… Example 2: Which has a higher heat capacity – water or sand? Example 3: Which as a higher heat capacity – cheese or pizza crust? • For a constant amount of heat (Q), as specific heat decreases, the change in temperature ___________. 5. Conversion between J and cal: ______________________ It takes less heat to raise the temp. of iron than water. Water: It takes a lot of heat to warm water. The crust increases 4.184 J = 1 cal

  22. Calculations Involving Specific Heat OR c = Specific Heat Capacity q = Heat lost or gained T = Temperature change

  23. - Page 510

  24. Sample Problem #1 Q = 200 kcal = 200,000 cal What is the temperature change of 2.0 L of water heated with a 200 Calorie (kcal) candy bar? m = 2.0 L 2.0 L 1 mL 1 g = 2000 g L 1 mL 10-3 c = 1 cal/goC T = ? Q = mcT T = 100 oC

  25. Q = ? m = 150.0 g 2. How much heat energy (kJ) must be added to change the temperature of 150.0 mL of water from 20.0C to 85.0C? T = final (Tfinal) - initial (Tinitial) T = 85.0 oC – 20.0 oC = 65.0 oC c = 4.184 J/goC Q = mcT Q = (150.0 g)(4.184 J/goC)(65.0 oC) Q = 40,794 J Q = 40.8 kJ

  26. m = 100.0 mL = 100.0 g T = Tfinal - Tinitial 3. What is the final temperature of 100.0 mL of 10.0C water if 1500 calories of heat is absorbed? Tf = T + Ti Tf = T + 10.0 oC Q = 1500 cal c = 1 cal/goC Q = mcT T = 15 oC Tf = 15 oC + 10.0 oC Tf = 25 oC

  27. mcT = mcT mAlcAl(Thot – Tfinal) = mwcw(Tfinal – Tcold) 0.834 0.900 Experimental Setup #1: Place a 150 g block of aluminum into a pot of boiling water (100.0 oC). Once the block is hot, place it into 200.0 mL of 10.0C water. As the block cools, the water warms up. The final temperature is 21.7C for the mixture. What is the specific heat of aluminum (cAl)?   Heat lost by aluminum = Heat gained by water QH (Hot object) = QC (Cold object) TH = 100.0 oC TC = 10.0 oC TF = 21.7 oC mAl = 150 g mw = 200.0 g cw = 4.184 J/g oC = (200.0g H2O)(4.184 J/g oC)(21.7 oC – 10.0 oC) (150 g Al)(cAl)(100.0 oC – 21.7 oC) = (200.0 g H2O)(4.184 J/g oC)(11.7 oC) (150 g Al)(cAl)(78.3 oC) = 9790.56 J (11745 g oC)(cAl) cAl = 0.834 J/g oC % error = 0.900 = 7.3 %

  28. mcT = mcT m?c?(Thot – Tfinal) = mwcw(Tfinal – Tcold) Sample problem #2: A student places an 85.5 g piece of metal at 100.0C into 122 mL of 16.0C water. If the final temperature is 20.2C, what is the specific heat of the metal?   Heat lost by the metal = Heat gained by water QH (Hot object) = QC (Cold object) TH = 100.0 oC TC = 16.0 oC TF = 20.2 oC m? = 85.5 g mw = 122 g cw = 4.184 J/g oC = (122 g H2O)(4.184 J/g oC)(20.2 oC – 16.0 oC) (85.5 g)(c?)(100.0 oC – 20.2 oC) = (122 g H2O)(4.184 J/g oC)(4.2 oC) (85.5 g)(c?)(79.8 oC) = 2143.88 J (6822.9 g oC)(c?) c? = 0.31 J/g oC

  29. IQ #1 • Calorimetry is based on what concepts? • How can enthalpy changes be shown in a chemical reaction? • How much heat is required is to raise a 150g piece of aluminum 35°C?

  30. Specific Heat Activity Procedure: • Record the mass of your cup (place in volume of water row) and keep the same cup throughout the experiment. • Fill the your cup with enough water to cover the metal. Record mass of water (mc). (Do not use too much water!!) • Record the temperature of the water (Tc). (Record temp to tenths) • Specific heat of water (cc) = 4.184 J/goC. • Record the temperature of the metal (Th). (Iron, Marble, Aluminum, & Brass). 6. Transfer hot metal to your Styrofoam cup (Splash/Thermometer). 7. Let temperature balance out and record the final temperature (Tf). Remember, this is the final temperature for water and for the metal as well! 8. Dry of the metal and record the mass of the metal (mh). • Determine the specific heat of each metal in J/goC & cal/goC • Determine % Error for Known metal (#1)

  31. Book Work #2 Ans. 12) 1.46 KJ 13) 146 J

  32. 1. A student mixes 165 mL of water at 14.5C with a piece of metal with a mass of 112 g and a temperature of 100.0C. The final temperature of the mixture is 32.2C. Calculate the specific heat of the unknown metal.   Heat lost by the metal = Heat gained by water QH (Hot object) = QC (Cold object) mcT = mcT m?c?(Thot – Tfinal) = mwcw(Tfinal – Tcold) TH = 100.0 oC TC = 14.5 oC TF = 32.2 oC m? = 112 g mw = 165 g cw = 4.184 J/g oC = (165 g H2O)(4.184 J/g oC)(32.2 oC – 14.5 oC) (112 g)(c?)(100.0 oC – 32.2 oC) = (165 g H2O)(4.184 J/g oC)(17.7 oC) (112 g)(c?)(67.8 oC) = 12219.37 J (7593.6 g oC)(c?) c? = 1.61 J/g oC

  33. 2. If a student mixes 125 mL of 85.0C water and 225 mL of 22.0C water, what will be the final temperature just after mixing? Heat lost by the metal = Heat gained by water QH (Hot object) = QC (Cold object) mcT = mcT mwcw(Thot – Tfinal) = mwcw(Tfinal – Tcold) TH = 85.0 oC TC = 22.0 oC TF = ? mw = 125 g mw = 225 g cw = 4.184 J/g oC cw of hot water = cw of cold water So cancel = (225 g H2O)(TF – 22.0 oC) (125 g H2O)(85.0 oC – TF) = 225 TF 4950 10625 125 TF = 350 TF 15575 = 44.5 oC TF

  34. 3. If a student puts a piece of 100.0C aluminum into 500.0 mL of water at 20.0C in an insulated calorimeter. The final temperature of the mixture is 23.7C. What was the mass of aluminum? (cAl = 0.900 J/gC).  Heat lost by the Aluminum = Heat gained by water QH (Hot object) = QC (Cold object) mcT = mcT malcal(Thot – Tfinal) = mwcw(Tfinal – Tcold) TH = 100.0 oC TC = 20.0 oC TF = 23.7 oC mAl = ? g mw = 500.0 g cw = 4.184 J/g oC (mAl)(0.900 J/g oC)(100.0 oC – 23.7 oC) = (500.0 g H2O)(4.184 J/g oC)(23.7 oC – 20.0 oC) (mAl)(0.900 J/g oC)(76.3 oC) = (500.0 g H2O)(4.184 J/g oC)(3.7 oC) = (7740.4 J) (mAl)(68.67 J/g) (mAl) = 110 g

  35. 4. If a student puts a 228 g piece of aluminum metal (at 100.0C) into 125 mL of 10.0C water, what will the final temperature of the mixture be? (specific heat of Al = 0.900 J/goC) Heat lost by the aluminum = Heat gained by water QH (Hot object) = QC (Cold object) mcT = mcT mwcw(Thot – Tfinal) = mwcw(Tfinal – Tcold) TH = 100.0 oC TC = 10.0 oC TF = ? mAl = 228 g mw = 125 g cw = 4.184 J/g oC = (125 g H2O)(4.184 J/g oC)(TF – 10.0 oC) (228 g Al))(0.900 J/goC)(100.0 oC – TF) = 523 TF 5230 20520 205.2 TF = 728.2 TF 25750 = 35.4 oC TF

  36. II. Change in enthalpy of a reaction (Using Q to determine H) A. B. Units of ____: C. Exothermic reactions: ___ is ________. D. Endothermic reactions: ___ is _______. E. Heat transfer between reaction and water: Q released by reaction = Q absorbed by water in calorimeter Q = mcT H J/g; J/mol; kJ/mol; kJ/g; cal/mol H negative H20 ↑ H positive H20 ↓

  37. Sample Problems: 1.When 5.00 g of liquid toluene, C7H8 is burned, it causes 500.0 mL of water at 20.0C to rise to 40.3C. What is the heat of combustion of liquid toluene in kJ/mol? Q = mcT Q = (500.0 g)(4.184J/goC)(40.3 oC – 20.0 oC) Q = 42467.6 J = 42500 J = 42.5 kJ 5.00 g C7H8 1 mol C7H8 = 0.0543 mol C7H8 92.15 g C7H8 = - 783 kJ/mol of C7H8 exothermic

  38. 2. What is the H of a reaction that lowers the temperature of 1.00 L of water by 7.50C when 5.00 g of the substance (molar mass = 120.0 g/mol) is reacted. Give your answer in (a) kJ/mol, and (b) cal/g. Q = mcT Q = (1000.0 g)(4.184J/goC)(7.50 oC) Q = 31380 J = 31.4 kJ 5.00 g 1 mol = 0.0417 mol 120.0 g + endothermic = 753 kJ/mol 752.52 kJ 103 J 1 cal 1 mol = 1498.80 cal/g mol 1 KJ 4.184 J 120.0 g = 1.50 x 103 cal/g

  39. Experimental Determination of ΔH • When a 0.764 g almond is burned, it heats 225 mL of water from 14.8oC to 18.4oC. Calculate the heat of combustion of the almond in kcal/g. • What is the ∆H of a reaction that lowers the temperature of a liter of water by 7.50oC when 5.00 g of the substance is dissolved in the water? (Assume the molar mass = 120 g/mol). Give your answer in (a) cal/g; (b) kcal/mol; (c) kJ/mol. • A 4.00 g substance (with molar mass = 80.0 g/mol) heats 750.0 mL from 15.0oC to 22.0oC. Calculate the ∆H in (a) kJ/mol; (b) kcal/mol. • When gasoline burns in air, heat is released. If 11.0 mL of gasoline burns and the temperature of 10.0 liters of water increase by 10.0oC, what is the heat of combustion (in kJ/mol)? Assume that gasoline is all octane (C8H18) and has a density of 0.785 g/mL.

  40. IQ #3 m= 9.1g • What mass of water is heated 2.3oC when 87.4 J of heat is added to it? • Calculate its specific heat if a 35.0 g sample of a metal at 58.0oC is immersed in 52.1 g of water at 16.3oC, warming the water to 20.7oC. 3. Calculate the final temperature after 30.0 g of a metal (c = 0.950 J/goC) at 71.3oC is immersed in 155 g of water at 21.3oC. c= 0.735 J/g°C Tf= 23.4°C

  41. Section 17.3Heat in Changes of State • OBJECTIVES: • Classify the enthalpy change that occurs when a substance melts, freezes, boils, condenses, or dissolves.

  42. Section 17.3Heat in Changes of State • OBJECTIVES: • Solve for the enthalpy change that occurs when a substance melts, freezes, boils, condenses, or dissolves.

  43. Heat in Changes of State • Molar Heat of Fusion (Hfus) : the heat absorbed by one mole of a substance in melting from a solid to a liquid • q= mass x Hfus(there is no temperature change) • Molar Heat of Solidification (Hsolid):heat lost when one mole of liquid solidifies (or freezes) to a solid • q = mass x Hsolid(no temperature change)

  44. Heat in Changes of State • Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies • Thus, Hfus = -Hsolid • Note Table 17.3, page 522

  45. - Page 521

  46. Heats of Vaporization and Condensation • When liquids absorb heat at their boiling points, they become vapors. • Molar Heat of Vaporization (Hvap) =the amount of heat necessary to vaporize one mole of a given liquid. • q = mass x Hvap(no temperature change) • Table 17.3, page 522

  47. Heats of Vaporization and Condensation • Condensation is the opposite of vaporization. • Molar Heat of Condensation (Hcond) = amount of heat released when one mole of vapor condenses to a liquid Hvap = - Hcond

  48. Heats of Vaporization and Condensation • Note Figure 17.10, page 523 • The large values for water Hvapand Hcondare the reason hot vapors such as steam are very dangerous • You can receive a scalding burn from steam when the heat of condensation is released! H20(g) H20(l) Hcond = - 40.7kJ/mol

  49. -Page 524

  50. Heat and Phase Changes endothermic A. Solid Liquid 1. Melting = Solid  Liquid; an ___________ process. 2. Freezing = Liquid  Solid; an __________ process. 3. When heat is added during a phase change, __________remains constant. (All energy is being used for the change of _____). 4. Enthalpy of _______ (____)= Energy needed to ____ one gram of solid. 5. Hfus of water = _______ 6. Calculation of heat (Q) of fusion: exothermic temperature phase fusion Hfus melt 334 J/g Qfus =mHfus

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