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Introduction to Thermochemistry

Introduction to Thermochemistry. That’s Me!. Okay…not really. …Or is it?. Why study thermochemistry?. We depend on energy for nearly every aspect of our lives Think about how you use energy every day… Electricity Cooking Automobiles Heating/Cooling Batteries

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Introduction to Thermochemistry

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  1. Introduction to Thermochemistry That’s Me! Okay…not really …Or is it?

  2. Why study thermochemistry? • We depend on energy for nearly every aspect of our lives • Think about how you use energy every day… Electricity Cooking Automobiles Heating/Cooling Batteries • Cars, cell phones, iPods, watches Human Body • Temperature regulation • Conversion of food into usable cellular energy

  3. Seriously, why are we studying this? • Energy crisis in the world! • Over-use/dependence on fossil fuels (petroleum, gasoline, diesel, methane, propane) • Research/Development of alternative energy resources • Nuclear • Fuel Cells (H2, Methanol, Solid oxide) • Hydro • Wind • Solar NEWS FLASH: This is going to be OUR problem!

  4. Major Sources of Energy over the last 150 years

  5. An interesting quote… “So called “global warming” is just a secret ploy by wacko tree-huggers to make America energy independent, clean our air and water, improve the fuel efficiency of our vehicles, kick-start 21st century industries, and make our cities safer and more livable. Don’t let them get away with it!” - Chip Giller Founder of Grist.com, where environmentally minded people gather online.

  6. Organizing the Universe Any process can be defined by the following… • Universe = everything, the big enchilada, the whole kit and caboodle, all that and a bag of chips, the whole shabang, all that jazz, tout, todos, sarra, alles, everything-ay, Chuck Norris, etc. • System = the part of the universe we are studying • i.e. a particular chemical reaction, etc. • Surroundings = everything else in the universe that is NOT being studied • i.e. the environment, the lab, the students, your mom, etc. Universe System Surroundings

  7. Internal Energy • Total Energy = Kinetic Energy + Potential Energy • Endothermic process: energy enters the system (from the surroundings) • Exothermic process: energy leaves the system (to the surroundings) due to motion stored energy System Energy Surroundings System Energy Surroundings

  8. Kinetic Energy in Atoms & Molecules • Temperature  average KE of particles in a sample • Components of Kinetic Energy: • KEtot = all components added together is proportional to  Translation Rotation Vibration

  9. Potential Energy (PE) • Chemical Change:intramolecular potential energy • Breaking and forming chemical bonds (PEnucleus-electron) Example: 2H2 + O2 2H2O • Physical Change:intermolecular potential energy • Disrupting or forming intermolecular attractions (PEIMF) Example: H2O(ℓ) H2O(g)

  10. The Law of Conservation of Energyis also known asThe First Law of Thermodynamics • Energy cannot be created or destroyed ΔEuniverse = 0 (reminder: Δ = “change in”) ΔEuniverse = ΔEsystem + ΔEsurroundings = 0 ΔEsystem = -ΔEsurroundings Energy can be transferred as: heat (q) work (w) ΔEsys = q + w

  11. The First Law of Thermodynamics System Energy Surroundings System Energy Surroundings Endothermic process: Exothermic process: Universe Universe In either case, the energy of the universe remains constant

  12. Conservation of Energy in a Chemical Reaction Endothermic Reaction Reactant + Energy Product Surroundings System In this example, the energy of the reactants and products increases, while the energy of the surroundings decreases. In every case, however, the total energy does not change. Surroundings Energy System Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

  13. Conservation of Energy in a Chemical Reaction Exothermic Reaction Reactant Product + Energy Surroundings System In this example, the energy of the reactants and products decreases, while the energy of the surroundings increases. In every case, however, the total energy does not change. Surroundings System Energy Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

  14. Heat (q) • Heat: the transfer of energy between objects due to a temperature difference • Flows from higher-temperature object to lower-temperature object If T1 > T2 q system = - q surroundings = + exothermic System (T1) Heat Surroundings (T2) If T1 < T2 q system = + q surroundings = - endothermic System (T1) Heat Surroundings (T2)

  15. Pressure/Volume Work (PV work) work = forcedistance force= pressure  area = P  m2 distance = m work = P  m2m = P  V Assuming external pressure (Pext) is constant, the volume can change (ΔV) wsystem = -PextΔV (in L atm) Where ΔV = Vf – Vi (in L)

  16. Sign Conventions In chemistry, always take the system’s perspective (physics takes the surr’s view) • Heat (q) q > 0 heat is added to the sys by surr (q is +) q < 0 heat is leaving sys to surr (q is -) • Work (w) w > 0 work done on sys by surr (work added to sys by surr, so w is +) w < 0 work done by sys on surr (work is leaving surr, so w is -)

  17. Sign Conventions: Heat and Work Surroundings System - w Heat + q (endo) Work - q (exo) + w Work Heat

  18. Practice with Work and Heat Pext = 2.0 atm Pext = 2.0 atm V2 = 12.0 L V1 = 5.0 L An ideal gas expands from 5.0 L to 12.0 L against a constant external pressure of 2.0 atm. How much work is done? w = -PextΔV w = -(Pext)(Vfinal – Vinitial) w = -(2.0 atm)(12.0 L – 5.0 L) w = -14 L atm 1 L atm = 101.325 J w = -14 L atm (101.325 J/1 L atm) w = -1400 J If ΔE = 0 for this process, what is q? ΔE = q + w = q + (-1.42  103 J) = 0 q = 1400 J

  19. Practice with Work and Heat Calculate the change in energy for a system that loses 15 kJ of heat and expands from a volume of 10. L to 200. L under a constant external pressure of 2.0 atm. ΔE = q + w w = -PextΔV w = -(2.0 atm)(200L -10L) = w = -380 L atm(101.325 J/1 L atm) (1 kJ/1000 J) = -38.5035 kJ ΔE = 15 kJ + -38.5035 kJ = -53.5035 kJ Remember: 101.325 J = 1 L atm 1000 J = 1 kJ • -380 L atm - -54 kJ

  20. Calorimetry(Calorie Measuring) • Calorimetry: the science of measuring heat • C = heat capacity • the heat required to raise temp by 1°C (or 1 K) • Units → Joules/oC (J/oC or J/K) Because the temperature increase depends on the amount of “stuff”… • Specific heat capacity (Cp) • heat capacity per gram = J/goC or J/gK • Molar heat capacity (Cm) • heat capacity per mole = J/moloC or J/mol K

  21. Various Heat Capacities Specific heat capacity (J/K g) Molar heat capacity (J/K mol) Molar mass (g/mol) Substance Gold Silver Copper Iron Aluminum H2O(l) H2O(s) H2O(g) 0.129 0.235 0.385 0.449 0.897 4.184 2.03 1.998 197.0 107.9 63.55 55.85 26.98 18.02 18.02 18.02 25.4 25.4 24.5 25.1 24.2 75.3 36.6 36.0

  22. Using heat capacities… q = m  C ΔT q (J) = mass (g) C (J/goC)  ΔT (oC) q = joules (J) q (J) = moles (mol) C (J/K mol)  ΔT (K) q = joules (J) Mnemonic device: q = m “CAT”

  23. Heating Curve of water Gas Boiling Point Liquid Melting Point Solid 140 120 l ↔ g 100 80 60 40 Temperature (oC) 20 s ↔ l 0 -20 -40 -60 -80 -100 Energy Added

  24. Heating Curves • Temperature Change within phase • change in KE (molecular motion) • depends on heat capacity of phase C H2O (l) = 4.184 J/goC C H2O (s) = 2.03 J/goC C H2O (g) = 1.998 J/goC • Phase Changes (s ↔ l ↔ g) • change in PE (molecular arrangement) • temperature remains constant • overcoming intermolecular forces • (requires the most heat) • (requires the least heat)

  25. Calculating Energy Changes Same Formulas Different C Values 140 q = ? q = ? 120 100 80 q = mass x Cgas xDT 60 40 Temperature (oC) 20 q = mass x Cliquid xDT 0 -20 -40 -60 q = mass x Csolid xDT -80 -100 Energy Added

  26. Phase Transitions • During phase transitions, added heat is used to overcome intermolecular forces rather than to increase temperature ΔHfusion = energy needed to convert solid to liquid • For water, ΔHfusion = 6.02 kJ/mol • For liquid to solid, ΔH = - 6.02 kJ/mol ΔHvaporization = energy needed to convert liquid to gas • For water, ΔHvap = 40.7 kJ/mol • For gas to liquid, ΔH = - 40.7 kJ/mol

  27. Depending on the process occurring, all q’s are calculated individually and then added together Note: steps 2 and 4 → kJ steps 1, 3, 5 → J Unit Conversions! 140 4. q = mol xDHvap 2. q = mol xDHfus 120 100 80 5. q = mass x Cgas xDT 60 40 Temperature (oC) 20 3. q = mass x Cliquid xDT 0 -20 -40 -60 1. q = mass x Csolid xDT -80 -100 Energy Added

  28. Heating Curve of Water From Ice to Steam in Five Easy Steps 4 5 q1: Heat the ice to 0°C q2: Melt the ice into a liquid at 0°C q3: Heat the water from 0°C to 100°C q4: Boil the liquid into a gas at 100°C q5: Heat the gas above 100°C q = m CiceΔTice q = molΔHfus q = m CwaterΔTwater q = molΔHvap q = m CsteamΔTsteam 3 2 1 Heat Heat

  29. Heating Curve Practice How much energy (J) is required to heat 12.5 g of ice at –10 oC to water at 0.0 oC? 4 5 3 2 1 q1: Heat the ice from -10 to 0°C q2: Melt the ice at 0°C to liquid at 0 oC 253.75 J q = 12.5 g (2.03 J/g oC)(0.0 - -10.0 oC) = q = 12.5 g ice 6.02 kJ 1 mol 4.177 kJ = 18.016 g 1 mol qtot= q1 + q2 4,430 J = 253.75 J + 4,177 J =

  30. Heating Curve Practice How much energy (J) is required to heat 25.0 g of ice at –25.0 oC to water at 95.0 oC? 4 5 Notice that your q values are positive because heat is added… 3 2 q1: Heat the ice from -25 to 0°C q2: Melt the ice at 0°C to liquid at 0 oC q3: Heat the water from 0°C to 95 °C 1 1268.75 J q = 25.0 g (2.03 J/g oC)(0.0 - -25.0 oC) = q = 25.0 g ice 6.02 kJ 1 mol 8.352 kJ = 18.016 g 1 mol 9937 J q = 25.0 g (4.184 J/g oC)(95.0 – 0.0oC) = qtot= q1 + q2 + q3 19,560 J = 1268.75 J + 8,352 J + 9937 J =

  31. Heating Curve Practice How much energy (J) is removed to cool 50.0 g of steam at 115.0 oC to ice at -5.0 oC? Notice that your q values are negative because heat is removed… 4 5 q5: Cool the steam from 115.0 to 100°C q4: Condense the steam into liquid at 100°C q3: Cool the water from 100°C to 0 °C q2: Freeze the water into ice at 0 °C q = 50.0 g H2O q1: Cool the ice from 0°C to – 5.0 °C 3 -1498.5 J q = 50.0 g (1.998 J/g oC)(100.0 - 115.0 oC) = q = 50.0 g H2O 2 1 - 40.7 kJ 1 mol -112.96 kJ = 18.016 g 1 mol -20920 J q = 50.0 g (4.184 J/g oC)(0.0 – 100.0oC) = 1 mol - 6.02 kJ -16.71 kJ = 18.016 g 1 mol -507.5 J q = 50.0 g (2.03 J/g oC)(- 5.0 – 0.0oC) = qtot= q1 + q2 + q3+ q4 + q5 = -1498.5 J + -112,960 J + -20920 J + -16,710 J + -507.5 J = -153,000 J

  32. Heating Curve Challenge Problems 140 DH = mol xDHvap DH = mol xDHfus 120 100 80 60 Heat = mass xDt x Cp, gas 40 20 Heat = mass xDt x Cp, liquid Temperature (oC) 0 -20 -40 -60 Heat = mass xDt x Cp, solid -80 -100 Time 1. A sample of ice at -25oC is placed into 75 g of water initally at 85oC. If the final temperature of the mixture is 15oC, what was the mass of the ice? 52.8 g ice • A 38 g sample of ice at -5oC is placed into 250 g of water at 65oC. Find the final temperature of the mixture assuming that the ice sample completely melts. • A 35 g sample of steam at 116oC are bubbled into 300 g water at 10oC. Find the final temperature of the system, assuming that the steam condenses into liquid water. 45.6 oC 76.6 oC

  33. What will happen over time? Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 291

  34. Let’s take a closer look… Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 291

  35. Eventually, the temperatures will equalize Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 291

  36. Calorimetry • Allows us to measure the flow of heat • Think back to the ΔHfus expt.. q system = - q surroundings • In other words, whatever energy is lost by one is gained by the other (direct E transfer) • Major Assumption: No E is lost to other parts of the surroundings

  37. Typical ApparatusA Coffee Cup Calorimeter Thermometer Styrofoam cover Styrofoam cups Stirrer Used for calorimetry under constant pressure i.e. normal lab conditions Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 302

  38. Constant-Pressure Calorimetry Example: 5.8 g of NaCl dissolves in 50.0 mL H2O NaCl(s) Na+(aq) + Cl-(aq) Ti = 22.2 °C Tf = 20.5 °C What is qrxn? qrxn = qsystem qsurr= m C  ΔT qsurr = (55.8 g)(4.184 J/°C g)(20.5 °C – 22.2°C) qsurr = -397 J qsystem = -qsurr = 397 J qrxn? Coffee cup calorimeter In this case, of the surroundings… We can’t directly measure q sys = -qsurroundings

  39. Bomb Calorimeter • Constant V calorimetry (as opposed to P) • “Bomb” has own heat capacity (kJ/0C) • Multiply heat capacity by the ΔT to find the total heat transferred • Factor heat into # of g or mol burned (kJ/g or kJ/mol)

  40. Bomb Calorimeter Example A 1.800 g sample of sugar, C12H22O11, was burned in a bomb calorimeter whose total heat capacity is 15.34 kJ/oC. Once burned, the temperature of the calorimeter plus contents increased from 21.36oC to 28.78oC. What is the heat of combustion per gram of sugar? per mole of sugar? More sugar = more energy! q = Cbomb x ΔT q = 15.34 kJ/oC (28.78 oC – 21.36 oC) q = 113.8 kJ 113.8 kJ released per 1.800 g sugar 113.8 kJ / 1.800 g = 63.23 kJ/g 63.23 kJ 342.3 g 21640 kJ/mol = g mol

  41. Food and Energy Caloric Values Food joules/grams calories/gram “Calories”/gram Protein 17,000 4,000 4 Fat 38,000 9,000 9 Carbohydrates 17,000 4,000 4 1000 calories = 1 “Calorie” 1 calorie = 4.184 joules "science" "food" or… 1 Kcal = 1 “Calorie” Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51

  42. Does water have negative calories…? How many Calories (nutritional) will you burn by drinking 1.0 L of water, initially at 36.5 oF (standard refrigeration temperature)? Assume that the body must expend energy to heat the water to body temperature at 98.6 oF. 37 oC 1 L = 1000 mL 2.5 oC 1 mL = 1 g 1 calorie = 4.184 joules 1000 calories = 1 “Calorie” q = 1.0 x 103 g (4.184 J/g oC)(37oC- 2.5oC) = 144,348 J 144348 J 1 cal 1 “Cal” 35 Cal = 4.184 J 1000 cal

  43. Heat Transfer Experiments Cu - qCu = qwater q = m x C xΔT for both cases, although specific values differ Plug in known information for each side Density of water = 1 g/mL(150 mL H2O) = 150 g H2O -mCuCCuΔT = mwaterCwater ΔT -20 g (0.385 J/goC)(Tf – 250 oC) = 150 g (4.184 J/goC)(Tf – 20oC) Solve for Tf ... Tf =22.8 °C What is the final temperature, Tf, of the mixture? 150. mL 20.0 °C 20.0 g 250.0 °C C = 0.385 J/°C g

  44. T = 20oC mass = 240 g 240. g of water (initially at 20.0oC) are mixed with an unknown mass of iron initially at 500.0oC (CFe = 0.4495 J/goC). When thermal equilibrium is reached, the mixture has a temperature of 42.0oC. Find the mass of the iron. T = 500oC Fe mass = ? grams - LOSE heat = GAIN heat -q1= q2 - [ (mass) (CFe ) (DT)] = (mass) (CH2O) (DT) - [ (X g) (0.4495 J/goC) (42oC - 500oC)] = (240 g) (4.184 J/goC) (42oC - 20oC)] - [ (X) (0.4495) (-458)] = (240 g) (4.184) (22) 205.9 X = 22091 X = 107 g Fe

  45. T = 15.0 oC mass = 323 g - LOSE heat = GAIN heat A 97.0 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15.0oC. If gold has a specific heat of 0.129 J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. T = 785oC Au mass = 97.0 g - [(C Au) (mass) (DT)] = (CH2O) (mass) (DT) - [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] - [(12.5) (Tf - 785oC)] = (1.35) (Tf - 15oC)] -12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 104 3 x 104 = 1.36 x 103 Tf Tf = 22.1oC

  46. T = 72.0 oC T = 13.0 oC mass = 87.0 g mass = 59.0 g - LOSE heat = GAIN heat If 59.0 g of water at 13.0 oC are mixed with 87.0 g of water at 72.0 oC, find the final temperature of the system. - [ (mass) (C H2O) (DT)] = (mass) (C H2O) (DT) - [ (59 g) (4.184 J/goC) (Tf - 13oC)] = (87 g) (4.184 J/goC) (Tf - 72oC)] - [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)] -246.8 Tf + 3208 = 364 Tf - 26208 29416 = 610.8 Tf Tf = 48.2oC

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