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Energy Relationships in Chemical Reactions

Energy Relationships in Chemical Reactions. Energy is the capacity to do work Thermal/kinetic energy is the energy associated with the random motion of atoms and molecules Chemical Potential energy is the energy stored within the bonds of chemical substances.

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Energy Relationships in Chemical Reactions

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  1. Energy Relationships in Chemical Reactions

  2. Energyis the capacity to do work • Thermal/kinetic energy is the energy associated with the random motion of atoms and molecules • Chemical Potential energy is the energy stored within the bonds of chemical substances

  3. Energyis the capacity to do work • Nuclear energy is the energy stored within the collection of neutrons and protons in the atom • Electrical energy is the energy associated with the flow of electrons

  4. Kinetic Energy Ek = ½ mv2 kinetic energy depends on BOTH mass and speed. Compare two chemical systems. O2(g) at same speed as H2(g) More Ek because of greater mass of molecules Less Ek because mass of molecules is less. Example 2 : compare O2(g) with H2(g) at same EK Average speed of molecules is faster Average speed of molecules is slower

  5. JOCKO JAKE Kinetic Energy Ek = ½ mv2 kinetic energy depends on BOTH mass and speed. equal velocity – unequal mass

  6. Contributions to the kinetic energy: • The molecule moving through space Ek(translation) • The molecule rotating Ek(rotation) • The bound atoms vibrating Ek(vibration) • The electrons moving within each atom Ek(electron)

  7. Contributions to the potential energy: • Forces between the bound atoms vibrating, Ep(vibration) • Forces between nucleus and electrons and between electrons in each atom, Ep(atom) • Forces between the protons and neutrons in each nucleus, Ep(nuclei) • Forces between nuclei and shared electron pair in each bond, Ep(bond)

  8. Intermolecular Forces forces of attraction that exist between molecules. That is: the force between one molecule of water and another molecule of water. These are the forces that must be overcome when a phase change occurs. We must break these forces when we melt a solid or evaporate a liquid.

  9. Intramolecular Forces forces within a molecule that hold it together. That is: covalent bonds. These must be overcome if we take a molecule apart.

  10. SURROUNDINGS SYSTEM Thermochemistry is the study of heat change in chemical reactions. closed isolated open energy nothing Exchange: mass & energy

  11. Thermochemical Definitions System: That part of the Universe whose change we are going to measure. Surroundings:Every thing else that is relevant to the change is defined as the “surroundings”. Heat (Q):Is the energy transferred between a system and it’s surroundings as result in the differences in their temperatures only!

  12. Temperature = Thermal Energy 900C 400C Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. greater thermal energy

  13. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Thermite reaction: 2 Al + Fe2O3→ Al2O3 + 2 Fe

  14. energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

  15. Thermodynamics State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

  16. First Law of Thermodynamics ( Law of Conservation of Energy ) “The Total Energy of the Universe is Constant” ΔE of Universe = ΔESystem + ΔESurroundings = 0

  17. The specific heat capacity (c) of a substance is the amount of heat (Q) required to raise the temperature of one gram of the substance by one degree Celsius. Q = m c Dt Q = quantity of heat (J) m = mass (g) c = specific heat capacity (J/g oC or kJ/kg oC or J/g K) ∆t = temperature change (oC or K)

  18. How much heat is given off when an 869 g iron bar cools from 940C to 50C? Q = m c Δt Q = (869 g) x (0.449 J/g • 0C) x (890C) = 3.5 x 104 J IRON BAR COOLS => LOSES HEAT

  19. If 36.5 kJ of heat is required to raise the temperature of 1.50 kg of a metal from 30.0 oC to 55.0 oC , find the specific heat capacity of the metal. Q c = mΔt 36.5 kJ = (1.50 kg)(25.0oC) = 0.973 kJ/kgoC

  20. Calculate the final temperature for a 150 mL sample of water at 15.0 0 C if it absorbs 5.40 kJ of heat energy. tf =15.0 oC + 8.59 oC Q ∆t = mc = 23.6 oC 5.40 kJ = (0.150 kg)(4.19 kJ/kgoC) = 8.59 oC

  21. 100.5 J of energy is added to four 50.0 g samples of the following metals at room temperature: aluminium, gold, silver and sodium. Arrange the metals from lowest to highest final temperature. Q ∆t = mc 100.5 J = (50 g)(0.897 J/goC) = 2.24 oC

  22. Thermal capacity The thermal capacity (C, unit → J/oC) of a substance is the amount of heat (Q) required to raise the temperature of a given quantity of the substance by one degree Celsius. Q = C Δt A calorimeter is known to absorb 5.48 KJ of energy for every degree increase in temperature. If the temperature of the calorimeter increases from 10.0oC to 78.0oC, how much energy is absorbed? Q = C Δt Q = (5.48 KJ/oC)(78oC – 10oC) Q = 373 KJ

  23. Phase changes

  24. Heat of Fusion The heat of fusion of a substance is the amount of energy required to change the state of the substance from solid to liquid. Calculate the amount of energy needed to melt 50.0 g of H2O(s) at 0.00 oC. ΔH = nΔHfus 50.0 g x 6.01 kJ/mol 18.02 g/mol = 16.7 kJ

  25. Heat of vaporization • The heat of vaporization of a substance is the amount of energy required to change the state of the substance from liquid to gas. ΔH = n ΔHvap Calculate the amount of energy needed to evaporate 200 g of H2O(l) at 100.0 oC. Answer: 451 kJ

  26. H2O(s) H2O(l) H2O(g) boiling point gas Temperature liquid melting point solid Energy in J PHASE TRANSITIONS ΔH = (nliquid)(ΔHvap) Q = (mgas)(cgas)(DT) ΔH = (nsolid)(ΔHfusion) Q = (mliquid)(cliquid)(DT) Q = (msolid)(csolid)(DT)

  27. H2O(s) H2O(l) H2O(g) ΔEk Temperature ΔEk ΔEk Energy in J PHASE TRANSITIONS ΔEp ΔEp Energy is increasing

  28. H2O(s) H2O(l) H2O(g) ΔEk ΔEp ΔEk Temperature ΔEp ΔEk Energy in J PHASE TRANSITIONS Energy is decreasing

  29. Quiz A. The flat lines on a heating curve represent 1) a temperature change 2) a constant temperature 3) a change of state B. The sloped lines on a heating curve represent 1) a temperature change 2) a constant temperature 3) a change of state

  30. Quiz A. Water condenses at a temperature of 1) 0°C 2) 50°C 3) 100°C B. At a temperature of 0°C, water 1) freezes 2) melts 3) changes to a gas C. When a gas condenses, heat is 1) released 2) absorbed D. Freezing is 1) endothermic 2) exothermic

  31. Quiz Is energy absorbed (1) or released (2) in each of the following: ____A. Ice to liquid water ____B. Water vapor to rain ____C. Water to ice When it rains, the air becomes 1) warmer 2) cooler 3) does not change 1 2 2

  32. Quiz Complete using the terms gains or loses • In the cooling coils of a refrigerator, liquid Freon ___________ heat from the food and changes to a gas • Food ___________heat and becomes colder • In the back of the refrigerator, Freon _________ heat and condenses back to a liquid gains loses loses

  33. Combining Heat Calculations To reduce a fever, an infant is packed in 1250 g of ice. If the ice at 0.00 °C melts and warms to body temperature (37.0°C) how much heat in joules is absorbed? • Step 1: Diagram the change of state 100oC 0oC

  34. Step 2: Calculate the heat to melt ice (fusion) ΔH = n Δ Hfus = (1250 g /18.02 g/mol) x 6.01 KJ/mol = 416.897 KJ Step 3: Calculate the heat from 0°C to 37°C Q = m c ΔT = 1250 g x 37.0°C x 4.19 J / g °C = 193 325 J Qtotal = 416 897 J + 193,325 J = 6.10 x 105 J

  35. Calorimetry The measurement of heat in a reaction. For processes occurring at constant pressure the change in energy equals the heat gained or lost. Qlost = Qgained m c ΔT = m c ΔT

  36. Mass(g) c(J/g*ºC) Tinitial Tfinal DT solid 25.64 ? 100.00 28.49 71.51 H2O 50.00 4.184 25.10 28.49 3.39 50.00g x 4.19J/g oC x 3.390C 50.00g x 4.19J/g ºC x 3.39ºC = 0.387J/g oC 25.64g x 71.51ºC A 25.64g sample of a solid was heated to 100.000C and added to a calorimeter containing 50.00g of water. The water temperature increased from 25.100C to 28.490C. What is the specific heat capacity of the solid? Qlost = Qgained m c ΔT = m c ΔT 25.64g x csolid x 71.510C = csolid =

  37. How much cold water at 15.0oC is needed to cool a 300 kg bath from 70.0 oC to a final temperature of 45.0oC? Qlost = Qgained mc∆t = mcΔt (300 kg)(4.19 kJ/kgoC)(25.0oC) = (m)(4.19 kJ/kgoC)(30.0oC) m = 250 kg

  38. A calorimeter system is known to absorb 15.2 kJ for each degree rise in temperature. 250.0 g of a metal is heated to 450.0 oC and placed in the calorimeter. It causes the temperature to rise from 22.0 oC to 28.0 oC. Calculate the specific heat capacity of the metal. ( C = 15.2kJ ) Thermal Capacity ( C ) of a calorimeter is defined as the Calorimeter value for each change in degree celcius which Includes the water and the calorimeter ( absorbing heat energy) Qlost = Qgained mc∆t = C∆t (.250 kg)(c)(422 oC) = (15.2 kJ/oC)(6.0 oC) c = 0.864 kJ/kg oC

  39. 90.0 g of a substance at 70.0 oC is placed into 150 mL of water at 20.0oC in an 80.0 g aluminum calorimeter. If the final temperature is 32.0oC, calculate the specific heat capacity of this substance. Qlost = Qgained H2O(l) + CALORIMETER SUBSTANCE ?? mc∆t = mc∆t + mc∆t (90.0 g)( c )(38oC) = (150g)(4.19 J/g oC)(12 oC) + (80.0g)(0.897 J/goC)(12oC) c = 2.46 J/g oC

  40. 60.0 g of copper at 90.0 oC is placed into 120.0 g of H2O(l) at 20.0 oC in a 75.0 g aluminum calorimeter. Calculate the final temperature. tf cool hot Δt = (90 – tf) Δt = (tf – 20) 90.0 oC tf 20.0 oC Qlost = Qgained (copper) (H2O + cal.) mcΔt = mcΔt + mcΔt (.06kg)(.385kJ/kgoC)(90-tf) = (.12kg)(4.19kJ/kgoC)(tf-20) + (.075kg)(.897kJ/kgoC)(tf-20) (.0231)(90-tf) = (.5028)(tf-20) + (.067275)(tf-20)

  41. 2.079 .0231tf = .5028tf –10.056 +.067275tf –1.3455 .593175tf = 13.4805 That was tough! But not as tough as me. tf = 22.7oC

  42. 80.0g of steam at 120 oC is bubbled through 800.0g of H2O(l) at 40.0 oC. Calculate the final temperature of the water. 120 Temp. (oC) tf 40 time (s) Qlost = Qgained (steam) (water) mc∆t + nΔHvap + mcΔt = mc∆t

  43. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. Cannot measure enthalpy. DH = change in enthalpy; heat given off or absorbed during a reaction at constant pressure

  44. Heat of Reaction If the heat transfer involves a chemical reaction then Q is called the heat of reaction.(ΔH) It is the energy required to return a system to the given temperature at the completion of the reaction. Q = ΔH UNITS: kJ

  45. Molar Heats A molar heat can be defined as the amount of energy released or absorbed when one mole of a substance is put through a defined process. • Melting (fusion) ΔHfus • Evaporating Δ Hvap • Combustion Δ Hcom • Formation Δ Hf • Decomposition Δ Hdec • Solvation Δ Hsol • Neutralization Δ HN • Reaction Δ Hrxn • Condensation Δ Hcon • Freezing Δ Hfr

  46. HEAT OF COMBUSTION If the heat transfer involves a specified amount of substance burning in oxygen then q is called the heat of combustion.(ΔHcom) Molar heat of combustion Q = ΔH = n Δ Hcom moles UNITS: kJ/mol

  47. Example • How much energy is released by the burning of one mole of methane? • How much energy is released by the burning of two moles of methane? • How much energy is released by the burning of 0.12 mol of methane? -890.5 KJ/mol 2 x -890.5 KJ/mol -1781 KJ 0.12 x -890.5 KJ/mol -106.86 KJ

  48. Example • How many moles of octane must there have been if 6987 KJ of energy was absorbed by a calorimeter? • One mole of octane releases 5470.1 KJ. • 6987 KJ x 1 mol = 5470.1 KJ 1.277 mol

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