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Unit 4: Thermochemistry & Thermodynamics

Unit 4: Thermochemistry & Thermodynamics. 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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Unit 4: Thermochemistry & Thermodynamics

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  1. Unit 4: Thermochemistry & Thermodynamics 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. 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

  6. 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

  7. 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

  8. 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)

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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)

  14. 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)

  15. 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 -)

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

  17. 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 (C) • heat capacity per gram = J/g oC or J/g K • Molar heat capacity (Cm) • heat capacity per mole = J/moloC or J/mol K

  18. 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

  19. 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”

  20. 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

  21. Heating Curves • Temperature Change within phase • change in KE (molecular motion) • depends on heat capacity of phase C (H2O liq) = 4.184 J/goC C (H2O sol) = 2.03 J/goC C (H2O gas) = 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)

  22. 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

  23. 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 = 6020 J/mol • For liquid to solid, ΔH = - 6020 J/mol ΔHvaporization = energy needed to convert liquid to gas • For water, ΔHvap = 40,700 J/mol • For gas to liquid, ΔH = - 40,700 J/mol

  24. Depending on the process occurring, all q’s are calculated individually and then added together 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

  25. 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 = mass (g) Cice (J/g°C)  ΔTice(°C) q = moles  ΔHfus(J/mol) q = mass (g)  Cwater (J/g°C)  ΔTwater(°C) q = moles  ΔHvap(J/mol) q = mass (g)  Csteam (J/g°C)  ΔTsteam(°C) 3 2 1 Heat Heat

  26. 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 6020 J 1 mol 4177 J = 18.016 g 1 mol qtot= q1 + q2 4,430 J = 253.75 J + 4,177 J =

  27. 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 6020 J 1 mol 8352 J = 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 =

  28. 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 - 40,700 J 1 mol -112,960 J = 18.016 g 1 mol -20920 J q = 50.0 g (4.184 J/g oC)(0.0 – 100.0oC) = 1 mol - 6020 J -16,710 J = 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

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

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

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

  32. Calorimetry • Allows us to measure the flow of heat • Think back to conservation of energy… 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

  33. 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

  34. 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 = (50.0 g)(4.184 J/°C g)(20.5 °C – 22.2°C) qsurr = -356 J qsystem = -qsurr = 356 J qrxn? Coffee cup calorimeter In this case, of the surroundings… We can’t directly measure q sys = -qsurroundings

  35. 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)

  36. 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

  37. 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

  38. 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

  39. 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

  40. Enthalpy (H) • The heat transferred sys ↔ surr during a chemical rxn @ constant P • Can’t measure H, only ΔH • At constant P, ΔH = q = mCΔT, etc. • Literally,ΔH = Hproducts - Hreactants • ΔH = +(endothermic) • Heat goes from surr into sys • ΔH = - (exothermic) • Heat leaves sys and goes into surr

  41. Reactant + Energy Product Surroundings System In this example, the energy of the system (reactants and products) ↑, while the energy of the surroundings ↓ Notice that the total energy does not change Endothermic Reaction Surroundings Energy System Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

  42. Reactant Product + Energy Surroundings System In this example, the energy of the system (reactants and products) ↓, while the energy of the surroundings ↑ Notice again that the total energy does not change Exothermic Reaction Surroundings System Energy Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

  43. Reaction Coordinate Diagrams: Endothermic Reaction Activation Energy Products Ea D(PE) ΔHrxn= + Reactants PE Progress of the Reaction

  44. Reaction Coordinate Diagrams: Exothermic Reaction Activation Energy ΔHrxn= - Ea D(PE) Reactants Products PE Progress of the Reaction

  45. Reaction Coordinate Diagrams Activation Energy ΔHrxn= -458.1 kJ Ea D(PE) C + O2 CO2 Draw the reaction coordinate diagram for the following rxn: C(s) + O2(g)  CO2 + 458.1kJ EXOTHERMIC PE Progress of the Reaction

  46. Enthalpies of Reaction - 483.6 kJ - 483.6 kJ - 483.6 kJ 1 mol H2O mol O2 1 mol H2 • All reactions have some ΔH associated with it H2(g) + O2(g) → H2O(ℓ)ΔH = - 483.6 kJ • How can we interpret this ΔH? • Amount of energy released or absorbed per specific reaction species • Use balanced equation to find several definitions or or Able to use like conversion factors in stoichiometry

  47. Enthalpies of Reaction • Formation of water H2(g) + O2(g) → H2O(l)ΔH = - 483.6 kJ • ΔH is proportional to amount used and will change as amount changes 2H2(g) + O2(g) → 2H2O(l) • For reverse reactions, sign of ΔH changes 2H2O(l)→ 2H2(g) + O2(g) • Treat ΔH like reactant or product H2(g) + O2(g) → H2O(l) ΔH = - 483.6 kJ ΔH = - 967.2 kJ ΔH = + 967.2 kJ H2(g) + O2(g) → H2O(l) + 483.6 kJ (exo)

  48. Enthalpies of Reaction Practice Consider the following rxn: C(s) + O2(g)  CO+ 458.1kJ Is the ΔH for this reaction positive or negative? NEGATIVE (E released as a product) What is the ΔH for 2.00 moles of carbon, if all the carbon is used? 2.00 mol C - 458.1 kJ = - 916 kJ 1 mol C What is the ΔH if 50.0g of oxygen is used? 50.0 g O2 1 mol O2 - 458.1 kJ = -1430 kJ 32.0 g O2 0.5 mol O2 What is the ΔH if 50.0 g of carbon monoxide decompose, in the reverse reaction? 50.0 g CO 1 mol CO 458.1 kJ = 818 kJ 1 mol CO 28.0 g CO

  49. Hess’s Law Reactants  Products The change in enthalpy is the samewhether the reaction takes place in one step or a series of steps Why? Because enthalpy is a state function Victor Hess To review: 1. If a reaction is reversed, ΔH is also reversed 2 CH4 + O2  2 CH3OH ΔHrxn = -328 kJ 2 CH3OH  2 CH4 + O2ΔHrxn = +328 kJ 2. If the coefficients of a reaction are multiplied by an integer, ΔH is multiplied by that same integer CH4 + 2 O2  CO2 + 2 H2O ΔHrxn = -802.5 kJ 2(CH4 + 2 O2  CO2 + 2 H2O) ΔHrxn = -1605 kJ

  50. Example: Methanol-Powered Cars 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)ΔHrxn = ? - ( ) - ( ) 2 CH4(g) + O2(g)  2 CH3OH(l)ΔHrxn = -328 kJ 2 ( ) 2 ( ) CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)ΔHrxn = -802.5 kJ 2 CH3OH(l) 2 CH4(g) + O2(g)ΔHrxn = +328 kJ 3 2 CH4(g) + 4 O2(g)  2 CO2(g) + 4 H2O(g)ΔHrxn = -1605 kJ 2 CH3OH + 3 O2 2 CO2 + 4 H2O ΔHrxn = -1277 kJ

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