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Chemistry-140 Lecture 13

Chemistry-140 Lecture 13. Chapter Highlights energy transfer, specific heat and heat transfer heat of fusion & heat vapourization exothermic & endothermic reactions first law of thermodynamics enthalpy changes calorimetry Hess’s Law.

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Chemistry-140 Lecture 13

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  1. Chemistry-140 Lecture 13 • Chapter Highlights • energy transfer, specific heat and heat transfer • heat of fusion & heat vapourization • exothermic & endothermic reactions • first law of thermodynamics • enthalpy changes • calorimetry • Hess’s Law Chapter 6: Enthalpy Changes for Chemical Reactions

  2. Chemistry-140 Lecture 13 • Thermodynamics: energy and its transformations. • Thermochemistry: energy changes and chemical reactions • Force: "push" or "pull" exerted on an object. • Work: energy required to overcome a force: • Work = force x distance. • Heat: energy transferred from one object to another because of a temperature difference. Thermodynamics

  3. Chemistry-140 Lecture 13 • Kinetic energy (EK): energy of an object due to its motion. • EK = mv2 (m = mass, v = velocity) • Potential energy (EP): energy stored by an object • due to its relative position. Kinetic and Potential Energy

  4. Chemistry-140 Lecture 13 • Energy can be expressed in a wide variety of units. • The joule (J) is the metric unit of energy and is the energy possessed by a 2 kg object moving at a velocity of 1 m/s. • 1 J (joule) =1 kg-m2/s2. • The calorie (cal) is the energy required to raise the temperature of 1 g of water by 1°C. • 1 cal = 4.184 J • 1 Cal = 1 kcal = 1000 cal = 4.184 kJ Energy

  5. Chemistry-140 Lecture 13 • The environment of a chemical reaction is separated into system and surroundings. • System: reactants, products, solvents, etc. in the reaction. • Surroundings: the universe, including the vessel. • 2 H2(g) + O2(g) 2 H2O(l) + energy • Processes that lower a system's internal energy are spontaneous. Processes that increase a system's internal energy are nonspontaneous. Energy Transfer

  6. C = Chemistry-140 Lecture 13 • Heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C, • The specific heat of a substance, C, is the amount of heat required to raise the temperature of a 1 g sample of the substance by 1°C, Energy Transfer & Specific Heat heat transferred = q = C x DT

  7. Chemistry-140 Lecture 13 Example 6.2: A lake has a surface area of 2.6 x 106 m2 and an average depth of 10 m. What quantity of heat (kJ) must be transferred to the lake to raise the temperature by 1 oC? (Assume a density of 1.0 g/cm3 the lakewater) Energy Transfer & Specific Heat

  8. Answer: Determine the mass of water and calculate the energy required using the concept of specific heat. Volume(H2O) = (2.6 x 106 m2)(10 m) = 2.6 x 107 m3 = Mass(H2O) = (2.6 x 1013 cm3)(1.0 g/cm3) = Since: q = m x C x DT = (2.6 x 1013 g)(4.184 J/g-K)(1.0 K) = 2.6 x 1013 cm3 2.6 x 1013 g C = 1.1 x 1011 kJ Chemistry-140 Lecture 13

  9. 500 kJ 1100oC Iron, 1.0 kg, 0oC 0oC 500 kJ Ice, 2.0 kg, 0oC 0oC Chemistry-140 Lecture 13 • Heat transfer: heat is transferred to a substance with no structural change • Phase changes or phase transitions: a substance's structure is altered as in melting, freezing, vaporizing, and condensing. Energy and Changes of State

  10. Chemistry-140 Lecture 13 • Heat of fusion: the enthalpy change associated with melting a substance, DHfus (kJ/mol). • Heat of vapourization; the enthalpy associated with vapourizing a substance, DHvap (kJ/mol). • Heating a substance is endothermic. BUT... during a phase transition the temperature remains constant Energy and Changes of State

  11. Chemistry-140 Lecture 13 • The energy required (q) for the five possible processes is determined by the amount of sample. • For warming: • q = C x mass x DT, • (specific heat, C, is different for each physical state) • For the phase transitions: q = (# of moles) x DH. Energy and Changes of State

  12. Chemistry-140 Lecture 13 Question (similar to example 6.4): Calculate the enthalpy change upon converting 1.00 mol of ice at -25 oC to water vapour (steam) at 125 oC under a constant pressure of 1 atm. The specific heat of ice, water and steam are 2.09 J/g-K, 4.18 J/g-K and 1.84 J/g-K, respectively. Also for water; DHfus = 6.01 kJ/mol and DHvap = 40.67 kJ/mol. Energy and Changes of State

  13. Chemistry-140 Lecture 13 Answer: Divide the total change into calculable segments and calculate the enthalpy change for each segment. Sum to get the total enthalpy change (Hess's Law). Segment 1: Heat ice from -25 oC to 0 oC. Segment 2: Convert ice to water at 0 oC. Segment 3: Heat water from 0 oC to 100 oC. Segment 4: Convert water to steam at 100 oC. Segment 5: Heat steam from 100 oC to 125 oC. Energy and Changes of State

  14. Heat liberated 5: Steam 4: Boiling Heat absorbed 3: Water 2: Melting 1: Ice Chemistry-140 Lecture 13 Energy and Changes of State Temperature oC Heat added (kJ)

  15. Chemistry-140 Lecture 13 Segment 1: Heating the ice from -25 oC to 0 oC. DH1 = (1.00 mol)(18.0 g/mol)(2.09 J/g-K)(25 K) = Segment 2: Convert ice to water at 0 oC. DH2 = (1.00 mol)(6.01 kJ/mol) = Segment 3: Heating the water from 0 oC to 100 oC. DH3 = (1.00 mol)(18.0 g/mol)(4.18 J/g-K)(100 K) = Energy and Changes of State 940 J 6.01 kJ 7520 J

  16. Chemistry-140 Lecture 13 Segment 4: Convert water to steam at 100oC. DH4 = (1.00 mol)(40.67 kJ/mol) = Segment 5: Heating the steam from 100oC to 125oC. DH5 = (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) = Total Enthalpy Change: DH = DH1 + DH2 + DH3 + DH4+ DH5 = (0.94 kJ) + (6.01 kJ) + (7.52 kJ) + (40.67 kJ) + (0.83 kJ) = 40.67 kJ Energy and Changes of State 830 J 55.97 kJ

  17. Chemistry-140 Lecture 13 • Energy is neither created nor destroyed; only exchanged between system and surroundings. • Internal energy is the total energy of a system. Only changes in internal energy, DE, can be measured. • DE = Efinal - Einitial • (+ve = gain by system & -ve = loss to surroundings) • DE = q + w • (heat added is positive & heat withdrawn is negative) • (work done is positive & work done by is negative) The First Law of Thermodynamics work done internal energy heat added

  18. Chemistry-140 Lecture 13 The First Law of Thermodynamics Energy transferred from surroundings to system Energy transferred from system to surroundings

  19. Chemistry-140 Lecture 13 Question: The H2(g) and O2(g) in a cylinder are ignited and the system loses 550 J to its surroundings. The expanding gas does 240 J of work on its surroundings. What is the change in internal energy of the system? Heat Transfer

  20. Answer: Energy flows from the system, so q = -550 J. Work is done by the system, so w = -240 J Therefore: DE = q + w = (-550 J) + (-240 J) = -790 J Chemistry-140 Lecture 13 Heat Transfer

  21. Chemistry-140 Lecture 13 Term Test #1 Friday October 12th, 2001 6:30 P.M.

  22. Chemistry-140 Lecture 13 A to H Education Gym I to Z Rm 104 Odette Bldg

  23. Chemistry-140 Lecture 13 Duration: 75 minutes Contents: SIX “Problems”!! Covers Material From Chapters 4 - 6 A Periodic Table & ALL Required Constants will be Supplied

  24. Chemistry-140 Lecture 14 • Chapter Highlights • energy transfer, specific heat and heat transfer • heat of fusion & heat vapourization • exothermic & endothermic reactions • first law of thermodynamics • enthalpy changes • calorimetry • Hess’s Law Chapter 6: Enthalpy Changes for Chemical Reactions

  25. Chemistry-140 Lecture 14 • Energy is neither created nor destroyed; only exchanged between system and surroundings. • Internal energy is the total energy of a system. Only changes in internal energy, DE, can be measured. • DE = Efinal - Einitial • (+ve = gain by system & -ve = loss to surroundings) • DE = q + w • (heat added is positive & heat withdrawn is negative) • (work done is positive & work done by is negative) The First Law of Thermodynamics work done internal energy heat added

  26. Chemistry-140 Lecture 14 The First Law of Thermodynamics Energy transferred from surroundings to system Energy transferred from system to surroundings

  27. Enthalpy: amount of heat energy possessed by a substance. Enthalpy change corresponds to the heat change of the system at constant pressure: • Endothermic reaction: heat is added (positive value). • Exothermic reaction: heat is withdrawn (negative value). DH = qp DH = Hfinal - Hinitial Chemistry-140 Lecture 14 Heat & Enthalpy Changes

  28. Enthalpy change: the sum of the absolute enthalpies of the products minus the absolute enthalpies of the reactants. • 2 H2(g) + O2(g) 2 H2O(l) + energy • 2 H2(g) + O2(g) 2 H2O(l) DH = -483.6 kJ DH = H(products) - H(reactants) Chemistry-140 Lecture 14 Enthalpies of Reaction

  29. Enthalpy is an extensive property. The magnitude of DH is proportional to the amount of reactant consumed. • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) • DH = -802 kJ • 2 CH4(g) + 4 O2(g) 2 CO2(g) + 4 H2O(g) • DH = -1604 kJ Chemistry-140 Lecture 14 Enthalpies of Reaction

  30. The enthalpy change is equal in magnitude but opposite in sign to DH for the reverse reaction. • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) • DH = -802 kJ • CO2(g) + 2 H2O(g) CH4(g) + 2 O2(g) • DH = +802 kJ Chemistry-140 Lecture 14 Enthalpies of Reaction

  31. The enthalpy change depends on the states of • the reactants and the products. • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) • DH = -802 kJ • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) • DH = -890 kJ • 2 H2O(g) 2 H2O(l) • DH = -88 kJ Chemistry-140 Lecture 14 Enthalpies of Reaction

  32. Chemistry-140 Lecture 14 Question: Ammonium nitrate can decompose by the reaction: NH4NO3(s) N2O(g) + 2 H2O(g) DH = -37.0 kJ Calculate the quantity of heat produced when 2.50 g of NH4NO3 decomposes at constant pressure. Enthalpies of Reaction

  33. Chemistry-140 Lecture 14 Answer: We know the heat produced from 1 mole of NH4NO3. Calculate moles in 2.50 g and determine the heat produced by that amount. heat = (2.50 g) = Enthalpies of Reaction -1.16 kJ

  34. Chemistry-140 Lecture 14 • Calorimetry: The measurement of heat flow. Measurements are made using a calorimeter. • Recall that: The heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C, • In constant-pressure calorimetry, the pressure remains constant because the apparatus is open to the atmosphere. At constant pressure, the heat change of the reaction is the enthalpy change, DH. Calorimetry heat transferred = q = C x DT

  35. Chemistry-140 Lecture 14 Question: When 50 mL of 1.0 M HCl(aq) and 50 mL of 1.0 M NaOH(aq) are mixed in a constant-pressure calorimeter, the temperature increases from 21.0 to 27.5 oC. Calculate DH for this reaction assuming a specific heat of 4.18 J/g-oC and a density of 1.0 g/mL for the final solution. Calorimetry

  36. Chemistry-140 Lecture 14 Answer: Recall, C = q = C x m x DT Since the total solution is 100 mL and the density is 1.0 g/mL, then m = (100 mL) (1.0 g/mL) = 100 g DT = (27.5 - 21.0 oC) = 6.5 oC and C = 4.18 J/g-oC Calorimetry

  37. Chemistry-140 Lecture 14 Answer: then qp = (4.18 J/g-oC) (100 g) (6.5 oC) = -2700 J = Since moles in solution were (0.050 L)(1.0 M ) = 0.050 mol DH = = Calorimetry -2.7 kJ -54 kJ/mol

  38. Chemistry-140 Lecture 14 Bomb Calorimetry • In bomb calorimetry, the apparatus is sealed and the experiment is a constant-volume process. The heat change of the reaction is the internal energy change. DE = qevolved = -Ccalorimeter x DT

  39. Chemistry-140 Lecture 14 Bomb Calorimetry

  40. Chemistry-140 Lecture 14 Thermometer A Student “Bomb” Calorimetry Cardboard or Styrofoam Lid Nested Styrofoam Cups Exothermic Reaction Occurs in Solution

  41. Chemistry-140 Lecture 14 Question: Hydrazine, N2H4, and its derivatives are widely used as rocket fuels. N2H4(l) + O2(g) N2(g) + 2 H2O(g) When 1.00 g of hydrazine is burned in a bomb calorimeter, the temperature of the calorimeter increases by 3.51 oC. If the calorimeter has a heat capacity of 5.510 kJ/oC, what is the quantity of heat evolved? Bomb Calorimetry

  42. Chemistry-140 Lecture 14 Answer: Recall that: qevolved = (-5.510 kJ/oC) x (3.51 oC) qevolved = -19.3 kJ Calculate this on a per mole basis: qevolved = = DE = qevolved = -Ccalorimeter x DT Bomb Calorimetry -618 kJ/mol

  43. Chemistry-140 Lecture 14 Term Test #1 Friday October 12th, 2001 6:30 P.M.

  44. A to H Education Gym I to Z Rm 104 Odette Bldg Chemistry-140 Lecture 14

  45. Chemistry-140 Lecture 14 Duration: 75 minutes Contents: SIX “Problems”!! Covers Material From Chapters 4 - 6 A Simple Periodic Table & ALL Required Constants will be Supplied

  46. Chemistry-140 Lecture 16 • Chapter Highlights • energy transfer, specific heat and heat transfer • heat of fusion & heat vapourization • exothermic & endothermic reactions • first law of thermodynamics • enthalpy changes • calorimetry • Hess’s Law Chapter 6: Enthalpy Changes for Chemical Reactions

  47. Chemistry-140 Lecture 16 • If a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy changes for each step. • For example: • (1) CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) • DH = -802 kJ • (2) 2 H2O(g) 2 H2O(l) • DH = -88 kJ Hess’s Law

  48. Add: (1)+(2) (1)+(2) CH4(g) + 2 O2(g) + 2 H2O(g) CO2(g) + 2 H2O(l) + 2 H2O(g) DH = (-802 - 88) kJ = -890 kJ Net Equation: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) DH = -890 kJ Chemistry-140 Lecture 16 Hess’s Law

  49. Question: Calculate DH for the reaction: 2 C(s) + H2(g) C2H2(g) Chemistry-140 Lecture 16 Hess’s Law

  50. Question: Given the following: (1) C2H2(g) + O2(g) 2 CO2(g) + H2O(l) DH = -1299.6 kJ (2) C(s) + O2(g) CO2(g) DH = -393.5 kJ (3) H2(g) + O2(g) H2O(l) DH = -285.9 kJ Chemistry-140 Lecture 16 Hess’s Law

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