1 / 96

Energy, Potential Energy, Kinetic Energy, Energy Conversions, Work, 1st Law of Thermodynamics, Thermochemistry, Enthalpy

Learn about energy, potential energy, kinetic energy, energy conversions, work, the 1st law of thermodynamics, thermochemistry, enthalpy, and specific heat. Understand the formulas and examples related to these concepts.

burk
Télécharger la présentation

Energy, Potential Energy, Kinetic Energy, Energy Conversions, Work, 1st Law of Thermodynamics, Thermochemistry, Enthalpy

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. Energy

  2. Energy Energy – the ability to do work or produce heat Potential energy – energy due to composition or position of an object Kinetic energy – the energy of motion

  3. Energy The SI unit for energy is the joule (J) 1 J = 1 Kgm2 / s2 Another unit of energy that you may be more familiar with is the calorie calorie – amount of energy required to raise 1 g of water 1°C 1 cal = 4.18 J 1000 calories = 1 Kilocalorie = 1 Calorie

  4. Energy Conversions Convert 15,500 joules into Calories 15500 J x 1 cal x 1 Cal = 4.18 J 1000 cal 3.71 Cal

  5. Formulas – Kinetic Energy Kinetic energy KE = ½ mv2 KE = kinetic energy (joules) m = mass (must be in Kg) V = velocity (must be in m/s)

  6. Formulas – Potential Energy Potential Energy PE = mgh PE = Potential Energy (J) m = mass (Kg) g = gravitational constant = 9.8 m/s2 h = height (m)

  7. Formulas - Work Work (w) – force acting over a distance Force (F) – a push or pull on an object W = mgd = Fd = PE Work and potential energy can be looked at in the same light No distance means no work

  8. Examples A bowler lifts a 5.4 kg bowling ball 1.6m and then drops it to the ground. How much work was required to raise the ball? W = mgd W = (5.4 kg)(9.8 m/s2)(1.6m) 85 Kgm2/s2 = 85 J

  9. Examples How much potential energy does that ball have at this height? 85 J

  10. Examples If the ball is dropped and we assume that all of the potential energy is turned into kinetic energy, at what velocity will the bowling ball hit the ground? KE = PE = 85J m = 5.4 Kg V = ?

  11. Examples KE = ½ mv2 85 J = ½ (5.4 Kg) v2 v2 = 31.5 v = 5.6 m/s

  12. More examples What is the kinetic energy of 1 atom of Ar moving at 650 m/s? KE = ½ mv2 1atom Ar x 1mol Ar x 39.95 g Ar x 1 Kg Ar = 6.02 x 1023 atoms 1 mol Ar 1 x 103 g Ar 6.64 x 10-26 kg Ar

  13. More Examples KE = ½ mv2 KE = ½ (6.64 x 10-26)(6502) KE = 1.4 x 10 -20 J

  14. 1st Law of Thermodynamics 1st Law of Thermodynamics – energy is conserved The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.

  15. 1st Law of Thermodynamics Since energy can neither be gained nor lost, the change in E can be calculated using: E = Ef – Ei In a chemical reaction i indicates reactants and f indicated products

  16. E E has 3 parts: A # indicating the magnitude A sign (+/-) indicating the direction A unit

  17. Thermochemistry Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes. In thermochemistry, the system is the specific part of the universe that contains the reaction or process you wish to study.

  18. Thermochemistry Everything in the universe other than the system is considered the surroundings. Therefore, the universe is defined as the system plus the surroundings. universe = system + surroundings

  19. Relating E to heat & work The system can exchange energy with its surroundings in 2 ways: as heat or work E = q + w E = change in energy q = heat w = work

  20. q & w Don’t forget q & w must have signs In order to get the sign you must look at the system as a box and the surroundings as everything else System Surroundings

  21. q & w Anything going INTO the box will be + Anything going OUT of the box will be – + -

  22. Examples A system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings. Calculate E. E = q + w E = (+ 140J) + (-85J) E = + 55 J

  23. Endothermic & Exothermic Endothermic system absorbs heat Heat flows into the system Temperature goes down Exothermic Heat flows out of the system and into the surroundings Temperature goes up Only look at heat (q) to determine if the system is endo or exo

  24. State Function State function-in going from one state to another, the pathway isn’t important. Energy is a state function, work and heat aren't. SURROUNDINGS SYSTEM HEAT LEAVES SYSTEM EXOTHERMIC HEAT ENTERS SYSTEM ENDOTHERMIC

  25. Enthalpy EnthaIpy (H) is a state function equal to E +pV H = E +  (pV), with constant presure, H = E + p V is equal to the expression for q at constant pressure. Therefore, at constant pressure H = qp. This means that the change in enthalpy is equal to the heat flow. H = H final - H initial = qp Enthalpy of reaction: H = H products- H reactants

  26. Enthalpy When ΔH is negative, heat is leaving the system and the reaction is exothermic. When ΔH is positive, heat is added to the system and the reaction is endothermic.

  27. Enthalpy When 1 mole of methane is burned at constant pressure, 890 kJ of energy is released as heat. Calculate the ΔH for a process in which a 5.8 g sample of methane is burned at constant pressure.

  28. Enthalpy qp = ΔH = -890 kJ/mol CH4 5.8 g CH4 x 1 mol CH4 = 0.36 mol CH4 16.0 g CH4 0.36 mol CH4 x -890 kJ = -320 kJ mol CH4

  29. Specific Heat Specific Heat (c) – the amount of heat required to raise the temperature of 1 g of a substance 1 °C The units for specific heat are J/g°C The specific heat of water (in a liquid form is 4.18 J/g°C) All substances have a particular specific heat

  30. Specific heat equation q = mcT q = heat gained or lost (J) m = mass (grams) note that this is different than the energy calculations c = specific heat (J/g°C) T = change in temperature (°C) = Tf-Ti

  31. Specific heat calculations How much heat is required to raise 250 g of water from 22°C to 98°C? q = mcT q = (250g)(4.18J/g°C)(90-22°C) q = 7.9 x 104 J

  32. A piece of metal with a mass of 4.68 g absorbs 256 J of heat when its temperature is increased by 182°C. What is its specific heat? q = mcT 256 J = (4.68g)(c)(182°C) c = 0.301J/g°C Specific heat calculations

  33. 60.0 J of heat are applied to a 5.00 g sample of calcium (c = 0.647J/g°C). If the final temperature is 51.1°C, calculate the original temperature. q = mcT 60.0 J = (5.00 g)(0.647J/g°C)(T) T = 18.55 °C T = Tf – Ti 18.55 = 51.1 – Ti Ti = 32.6 °C Specific heat calculations

  34. Calorimetry A calorimeter is an insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.

  35. Calorimetry When using calorimetry, you are usually trying to determine the identity of an unknown metal by finding its specific heat The heat lost from the metal will be gained by the water q metal = - q water

  36. Calorimetry q metal = - q water (m metal)(c metal)(T metal) = - (m water)(c water)(T water)

  37. Calorimetry Examples A 58.0 g sample of a metal at 100.0 °C is placed in a calorimeter containing 60.0 g of water at 18.0 °C. The temperature of that water increases to 22.0 °C. Calculate the specific heat of the metal.

  38. q metal = - q water (m metal)(c metal)(T metal) = - (m water)(c water)(T water) (58.0)(c metal)(22.0-100.0) = - (60.0)(4.18)(22.0-18.0) -4524 (c metal) = -1004 (c metal) = 0.222 J/g°C Calorimetry Examples

  39. A piece of metal with a mass of 4.68 g at 135°C is placed in a calorimeter with 25.0 g of water at 20.0 °C. The temperature rises to 35.0 °C. What is the specific heat of the metal? Calorimetry Examples

  40. q metal = - q water (m metal)(c metal)(T metal) = - (m water)(c water)(T water) (4.68)(c metal)(35.0-135.0) = - (25.0)(4.18)(35.0-20.0) -468 (c metal) = -1567.5 (c metal) = 3.35 J/g°C Calorimetry Examples

  41. More calorimetry 3.25 g Mg is placed into 125 mL of HCl. The initial temperature of the calorimeter is 18.5°C & the final temperature is 26.6°C. If the heat capacity of the calorimeter is 4.86 J/g°C, calculate the enthalpy of the reaction.

  42. More calorimetry Enthalpy will be in KJ/mol Mg + 2HCl  MgCl2 + H2 q = mcT q = (128.3)(4.86)(8.1) q = 5050 J  5.050 KJ 3.25 g Mg  0.134 mol Mg H = KJ/mol = 5.050KJ/0.134 mol 37.7 KJ/mol (need sign) -37.7 KJ/mol (because temp went up)

  43. Bomb Calorimetry For constant volume, (bomb calorimeter) -P  V =0, so  E = q + w, but w = 0, So E = qv qv = c  T c= heat capacity, the energy required to change the temperature of the bomb 1C

  44. Hess’s Law Heat of Formation Heat of Combustion Bond Enthalpy Thermochemistry 2

  45. Hess’s Law Hess's Law -the Heat of Reaction, Enthalpy (ΔH) of a reaction is the sum of the individual ΔHfº for each step of the reaction mechanism .

  46. Hess’s Law (cont’d) ΔHfº for Step #1 + ΔHfº for Step #2 + ΔHfº for Step #3 ΔH for the reaction

  47. Look at the example below.

  48. Try this one.

  49. Answer: ΔHo=? = (-790 kJ) - (-196 kJ) = -594 kJ for 1 mole of S02, ΔHo= 1/2 ( -594 kJ) = -297 kJ

  50. Heats of Formation Heat of Formation (Hf) -amount of energy needed to form 1 mole of product. Standard Heat of FormationHf- Heat of Formation at 298K and 1 atm. To determine the amount of energy needed to decompose a reactant, change the sign of ΔHfº, The more negative the ΔHfº, the more stable the compound.

More Related