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Unit 3: Thermochemistry

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  1. May 11 Unit 3: Thermochemistry Chemistry 3202

  2. May 11 Unit Outline • Temperature and Kinetic Energy • Heat/Enthalpy Calculation • Temperature changes (q = mc∆T) • Phase changes (q = n∆H) • Heating and Cooling Curves • Calorimetry (q = C∆T & above formulas)

  3. May 11 Unit Outline • Chemical Reactions • PE Diagrams • Thermochemical Equations • Hess’s Law • Bond Energy • STSE: What Fuels You?

  4. May 11 Temperature and Kinetic Energy Thermochemistryis the study of energy changes in chemical and physical changes eg. dissolving burning phase changes

  5. May 11 Temperature, T, measures the average kinetic energy of particles in a substance - a change in temperature means particles are moving at different speeds - measured in either Celsius degrees or degrees Kelvin Kelvin = Celsius + 273.15

  6. May 11 The Celsius scale is based on the freezing and boiling point of water The Kelvin scale is based on absolute zero- the temperature at which particles in a substance have zero kinetic energy.

  7. May 11 p. 628

  8. May 11

  9. 300 K 500 K # of particles May 11 Kinetic Energy

  10. May 11 Heat/Enthalpy Calculations system - the part of the universe being studied and observed surroundings - everything else in the universe open system - a system that can exchange matter and energy with the surroundings eg. an open beaker of water a candle burning closed system - allows energy transfer but is closed to the flow of matter.

  11. May 11 isolated system – a system completely closed to the flow of matter and energy heat - refers to the transfer of kinetic energy from a system of higher temperature to a system of lower temperature. - the symbol for heat is q WorkSheet: Thermochemistry #1

  12. May 12 Part A: Thought Lab (p. 631)

  13. May 12 Part B: Thought Lab (p. 631)

  14. May 12 Heat/Enthalpy Calculations specific heat capacity – the energy , in Joules (J), needed to change the temperature of one gram (g) of a substance by one degree Celsius (°C). • The symbol for specific heat capacity is a lowercase c

  15. May 12 • A substance with a large value of c can absorb or release more energy than a substance with a small value of c. ie. For two substances, the substance with the larger c will undergo a smaller temperature change with the same loss or gain of heat.

  16. q = mc∆T q = heat (J) m = mass (g) c = specific heat capacity ∆T = temperature change = T2 – T1 = Tf – Ti May 12 FORMULA

  17. May 12 eg. How much heat is needed to raise the temperature of 500.0 g of water from 20.0 °C to 45.0 °C? Solve q = m c ∆T for c, m, ∆T, T2 & T1 • p. 634 #’s 1 – 4 • p. 636 #’s 5 – 8 WorkSheet: Thermochemistry #2

  18. May 13 heat capacity- the quantity of energy , in Joules (J), needed to change the temperature of a substance by one degree Celsius (°C) • The symbol for heat capacity is uppercase C • The unit is J/ °C or kJ/ °C

  19. C = mc q = C ∆T C = heat capacity c = specific heat capacity m = mass ∆T = T2 – T1 May 13 FORMULA Your Turn p.637 #’s 11-14 WorkSheet: Thermochemistry #3

  20. May 18 Enthalpy Changes enthalpy change - the difference between the potential energy of the reactants and the products during a physical or chemical change AKA: Heat of Reaction or ∆H

  21. Products ∆H Reactants May 18 PE Endothermic Reaction Reaction Progress

  22. Products ∆H Reactants May 18 PE Endothermic Reaction Enthalpy ∆H Reaction Progress

  23. Products Reactants May 18 Enthalpy ∆H is + Endothermic

  24. products ∆H is - reactants May 18 Enthalpy Exothermic

  25. May 18 Enthalpy Changes in Reactions • All chemical reactions require bond breaking in reactants followed by bond making to form products • Bond breaking requires energy (endothermic) while bond formation releases energy (exothermic) see p. 639

  26. May 18

  27. May 18 Enthalpy Changes in Reactions endothermic reaction - the energy required to break bonds is greater than the energy released when bonds form. ie. energy is absorbed exothermic reaction - the energy required to break bonds is less than the energy released when bonds form. ie. energy is produced

  28. May 18 Enthalpy Changes in Reactions ∆H can represent the enthalpy change for a number of processes • Chemical reactions ∆Hrxn – enthalpy of reaction ∆Hcomb – enthalpy of combustion (see p. 643)

  29. May 18 • Formation of compounds from elements ∆Hof– standard enthalpy of formation The standard molar enthalpy of formation is the energy released or absorbed when one mole of a compound is formed directly from the elements in their standard states. (see p. 642) eg. C(s) + ½ O2(g) → CO(g) ΔHfo = -110.5 kJ/mol

  30. May 18 Use the equation below to determine the ΔHfo for CH3OH(l) 2 C(s) + 4 H2(g) + O2(g) → 2 CH3OH(l) + 477.2 kJ 1 C(s) + 2 H2(g) + ½ O2(g) → 1 CH3OH(l) + 238.6 kJ ∆H = -238.6 kJ/mol

  31. May 18 Use the equation below to determine the ΔHfo for CaCO3(s)  2 CaCO3(s) + 2413.8kJ → 2 Ca(s) + 2 C(s) + 3 O2(g) 2 Ca(s) + 2 C(s) + 3 O2(g) → 2 CaCO3(s) + 2413.8kJ 1 Ca(s) + 1 C(s) + 1.5 O2(g) → 1 CaCO3(s) + 1206.9 kJ ∆H = -1206.9 kJ/mol

  32. May 18 Use the equation below to determine the ΔHfo for PH3(g) 4 PH3(g) → P4(s) + 6 H2(g) + 21.6 kJ a) +21.6 kJ/mol b) -21.6 kJ/mol c) +5.4 kJ/mol d) -5.4 kJ/mol

  33. May 18 • Phase Changes (p.647) ∆Hvap – enthalpy of vaporization (l → g) ∆Hfus – enthalpy of melting(fusion: s → l) ∆Hcond – enthalpy of condensation (g → l) ∆Hfre – enthalpy of freezing (l → s) eg. H2O(l)  H2O(g) ΔHvap = Hg(l)  Hg(s) ΔHfre = +40.7 kJ/mol -23.4 kJ/mol

  34. May 18 • Solution Formation(p.647, 648) ∆Hsoln– enthalpy of solution • eg. • ΔHsoln, of ammonium nitrate is +25.7 kJ/mol. • NH4NO3(s) + 25.7 kJ → NH4NO3(aq) • ΔHsoln, of calcium chloride is −82.8 kJ/mol. • CaCl2(s) → CaCl2(aq) + 82.8 kJ

  35. May 18 Three ways to represent an enthalpy change: 1. thermochemical equation - the energy term written into the equation. 2. enthalpy term is written as a separate expression beside the equation. 3. enthalpy diagram.

  36. May 18 eg. the formation of water from the elements produces 285.8 kJ of energy. 1. H2(g) + ½ O2(g) → H2O(l) + 285.8 kJ 2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = -285.8 kJ/mol thermochemical equation

  37. H2(g) + ½ O2(g) ∆Hf = -285.8 kJ/mol H2O(l) enthalpy diagram 3. Enthalpy (H) May 18 examples: pp. 641-643 questions p. 643 #’s 15-18 WorkSheet: Thermochemistry #4

  38. FORMULA: q = n∆H q = heat (kJ) n = # of moles ∆H = molar enthalpy (kJ/mol) May 24 Calculating Enthalpy Changes

  39. May 24 eg. How much heat is released when 50.0 g of CH4forms from C and H ? (p. 642) q = nΔH = (3.115 mol)(-74.6 kJ/mol) = -232 kJ

  40. May 24 eg. How much heat is released when 50.00 g of CH4 undergoes complete combustion? (p. 643) q = nΔH = (3.115 mol)(-965.1 kJ/mol) = -3006 kJ

  41. May 24 eg. How much energy is needed to change 20.0 g of H2O(l) at 100 °C to steam at 100 °C ? Mwater = 18.02 g/mol ΔHvap = +40.7 kJ/mol q = nΔH = (1.110 mol)(+40.7 kJ/mol) = +45.2 kJ

  42. May 24 ∆Hfre and ∆Hcond have the opposite sign of the above values.

  43. May 24 eg. The molar enthalpy of solution for ammonium nitrate is +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves? q = nΔH = (0.4996 mol)(+25.7 kJ/mol) = +12.8 kJ

  44. May 24 What mass of ethane, C2H6, must be burned to produce 405 kJ of heat? ΔH = -1250.9 kJ q = - 405 kJ m = ? q = nΔH n = 0.3238 mol m = n x M = (0.3238 mol)(30.08 g/mol) = 9.74 g

  45. May 24 Complete: p. 645; #’s 19 – 23 pp. 648 – 649; #’s 24 – 29 p. 638 #’ 4 – 8 pp. 649, 650 #’s 3 – 8 p. 657, 658 #’s 9 - 18 WorkSheet: Thermochemistry #5

  46. 19. (a) -8.468 kJ (b) -7.165 kJ 20. -1.37 x103 kJ 21. (a) -2.896 x 103 kJ (b) -6.81 x104 kJ 21. (c) -1.186 x 106 kJ 22. -0.230 kJ 23. 3.14 x103 g 24. 2.74 kJ 25.(a) 33.4 kJ (b) 33.4 kJ 26.(a) absorbed (b) 0.096 kJ 27.(a) NaCl(s) + 3.9 kJ/mol → NaCl(aq) (b) 1.69 kJ (c) cool; heat absorbed from water 28. 819.2 g 29. 3.10 x 104 kJ

  47. May 25 Heating and Cooling Curves Demo: Cooling of p-dichlorobenzene

  48. May 25 KE PE KE Cooling curve for p-dichlorobenzene 80 Temp. (°C ) liquid 50 freezing solid 20 Time

  49. May 25 KE KE PE Heating curve for p-dichlorobenzene 80 Temp. (°C ) 50 20 Time

  50. May 25 What did we learn from this demo?? • During a phase change temperature remains constant and PE changes • Changes in temperature during heating or cooling means the KE of particles is changing