1 / 87

Chapter 6 Notes

Chapter 6 Notes. Thermochemistry. Part 1: Energy, Heat and Work. Thermochemistry. Energy is the capacity to do work or to produce heat. The law of conservation of energy states that energy can be converted from one form to another, but can be neither created nor destroyed.

layne
Download Presentation

Chapter 6 Notes

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6 Notes Thermochemistry

  2. Part 1: Energy, Heat and Work Thermochemistry

  3. Energy is the capacity to do work or to produce heat. The law of conservation of energy states that energy can be converted from one form to another, but can be neither created nor destroyed.

  4. Potential energy is energy due to position or composition. Kinetic energy is energy due to the motion of the object and depends on the mass and velocity of the object. KE = ½mv2 **Mass must be in kilograms and velocity must be in meters/second!!!!! The unit kgm2 = J s2

  5. Heat and temperature are different. Temperature is a measure of the kinetic energy of the molecules. Heat refers to the transfer of energy between two objects due to a temperature difference. Heat is not a substance contained by an object, although we often talk of heat as if this were true.

  6. The pathway is likened to a “path” or “route.” For instance, I can get to the stadium by walking out the front door or the back door. A state function or state property depends only on the characteristics of the present state – not on the pathway.

  7. The universe is divided into two parts: a. The system is the part of the universe on which we focus.b. The surroundings include everything else in the universe. For a reaction, the system includes the reactants and products. The surroundings includes the reaction container, the room, etc. (i.e. anything else other than reactants and products.)

  8. The study of energy and its interconversions is called thermodynamics. The law of conservation of energy is often called the first law of thermodynamics. It states: The energy of the universe is constant.

  9. Thermodynamic quantities always consist of two parts: a number, giving the magnitude of the change, and a sign, indicating the direction of the flow. The sign reflects the system’s point of view.

  10. q = heat*If energy flows INTO the system via heat (endothermic), then q = +.*If energy flows OUT OF the system via heat (exothermic), then q = .

  11. w = work*If the surroundings do work on the system (energy flows into the system), then w = +.* If the system does work on the surroundings (energy flows out of the system), then w = .

  12. Work is defined as force acting over a distance. The formula used to solve for work is: W = PΔV Pressure is measured in atmospheres and volume is measured in liters. Convert to joules using 101.3 J = 1 Latm The sign changes depending on: compressed gas = +PΔV expanding gas = PΔV

  13. Memorize!!! gas endothermic liquid exothermic solid sublimation deposition vaporization condensation melting freezing

  14. The internal energy, E, of a system is defined as the sum of the kinetic and potential energies. The formula is ΔE = q + w where q is heat and w is work. The sign convention is that anything that leaves the system is negative. *q is negative: system releases heat *q is positive: system absorbs heat *w is negative: system does work*w is positive: surroundings do work

  15. Part 2: Properties of Enthalpy Thermochemistry

  16. Enthalpy, H, is defined as: H = E + PV where E = internal energyP = pressureV = volume

  17. Internal energy, pressure and volume are all state functions (independent of the pathway) therefore enthalpy is also a state function.

  18. At constant pressure, the change in enthalpy (ΔH) of the system is equal to the energy flow as heat. Therefore, ΔH = q at constant P. At constant pressure, exothermic means that ΔH is negative and endothermic means that ΔH is positive.

  19. Stoichiometric Calculations When a mole of methane (CH4) is burned at constant pressure, 890 kJ of energy are released as heat. Calculate H when 5.8 grams of methane are burned at constant pressure. 16.0 g/mol CH4 + 2 O2 CO2 + 2 H2O H = –890 kJ 5.8 g x kJ – 890. kJ 1 mol CH4 5.8 g CH4 = – 320 kJ 1 mol CH4 16.0 g CH4

  20. Part 3: Calorimetry and Heat Capacity Thermochemistry

  21. Calorimetry Calorimetry is the study of heat flow and heat measurement.

  22. Calorimetry experiments determine the heats (enthalpy changes) of reactions by making accurate measurements of temperature changes produced in a calorimeter.

  23. The formula used incalorimetric calculations is: q = mcTwhereq = heat (J)m = mass (g)c = specific heat capacity (J/g C)T = change in temperature (C)

  24. The heat capacity of an object is the amount of heat needed to raise the temperature of the object by 1C.The heat capacity of one gram of a substance is called its specific heat.

  25. The specific heat is a physical property of the substance, like its color and melting point. Substances have different specific heat capacities.

  26. The specific heat capacity of water = 4.18 J/gC

  27. When calculating T, always subtract the smaller temperature FROM the larger temperature: T = Tlarger  Tsmaller**If the temperature rises,(ex: from 25C to 30C) then q will be negative and the reaction is exothermic.**If the temperature drops,(ex: from 40C to 30C) then q will be positive and the reaction is endothermic.

  28. Example 1: What is the specific heat capacity of iron if the temperature of a 12.3-g sample of iron is increased by 10.2C when 56.7 J of heat is added?

  29. q = mcT Example 1: What is the specific heat capacity of iron if the temperature of a 12.3-g sample of iron is increased by 10.2C when 56.7 J of heat is added? 56.7 J = (12.3 g) (c) (10.2 C) 56.7 J = (12.3 g)( c)(10.2C) (12.3 g)(10.2C) (12.3 g)(10.2C) 0.452 J/gC = c

  30. Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4C to 19.7C. Calculate H for the solution process.Pb(NO3)2 (s)  Pb+2 (aq) + 2 NO31 (aq)  H = ?

  31. q = mcT Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4C to 19.7C. Calculate H for the solution process.Pb(NO3)2 (s)  Pb+2 (aq) + 2 NO31 (aq)  H = ? q = mc(Tlarger Tsmaller) q = (85.0 g) (4.18 J/gC ) (23.4 C19.7C ) q = 1310 J **Since the temperature dropped, q will be positive and the reaction is endothermic.

  32. Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4C to 19.7C. Calculate H for the solution process.Pb(NO3)2 (s)  Pb+2 (aq) + 2 NO31 (aq)  H = ? Calculate the molar mass of Pb(NO3)2: MM = 331 g/mol Remember that from the first part, q = +1310 J

  33. Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4C to 19.7C. Calculate H for the solution process.Pb(NO3)2 (s)  Pb+2 (aq) + 2 NO31 (aq)  H = ? 331 g Pb(NO3)2 1 mol Pb(NO3)2 +1310 J 13.7 g Pb(NO3)2 1 mol Pb(NO3)2 = +31,700 J = 31.7 kJ

  34. Energy Calculations

  35. Important Information:q = mcΔTq = ΔHfusion molesq = ΔHvaporization molesspecific heat capacity of water = 4.18 J/gC specific heat capacity of ice = 2.1 J/gC specific heat capacity of steam = 1.8 J/gCΔHfusion of water = 6.0 kJ/moleΔHvaporization of water = 40.7 kJ/mole

  36. How much energy does it take to convert 130. grams of ice at 40.0C to steam at 160.C?

  37. Convert grams to moles of water: 1 mol H2O 130. g H2O 18.0 g H2O = 7.22 mol H2O

  38. Plan: a. Heat ice from 40.0C to 0.00C.

  39. q = mcT q = (130. g)( 2.1 J/gC)(40.0C)q = 10,920 Jq  10,900 Jq = 10.9 kJ

  40. b. Add heat to convert ice to liquid water at 0C.

  41. q = Hfusion  molesq = (6.0 kJ/mol)(7.22 mol)q = 43.32 kJ q  43.3 kJ

  42. c. Heat liquid water from 0.00C to 100.C.

  43. q = mcT q = (130. g)( 4.18 J/gC)(100.0C)q = 54,340 Jq  54,300 Jq = 54.3 kJ

  44. d. Add heat to convert liquid water to steam at 100C.

  45. q = Hvaporization  molesq = (40.7 kJ/mol)(7.22 mol)q = 293.854 kJq  294 kJ

  46. e. Heat steam from 100.C to 160.C.

  47. q = mcT q = (130. g)( 1.8 J/gC)(60.0C)q = 14,040 J q  14.0 kJ

  48. Add the energy values:Total energy = a + b + c + d + e

  49. Total energy = 10.9 kJ43.3 kJ54.3 kJ294 kJ14.0 kJ 416.5 kJ  417 kJ

  50. Part 4: Hess’ Law Thermochemistry

More Related