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Phase Changes PowerPoint Presentation
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Phase Changes

Phase Changes

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Phase Changes

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  1. The thermal or internal energyof a sample is the sum of all the kinetic and potential energies of all the atoms and molecules in a sample

  2. Vaporization Melting Freezing Condensation Phase Changes Solid Gas Liquid

  3. Heating Curves Heating Curves Animation A plot of temperature vs. time that represents the process in which energy is added at a constant rate Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  4. Gas - KE  Boiling - PE  Liquid - KE  Melting - PE  Solid - KE  Heating Curves 140 120 100 80 60 40 Temperature (oC) 20 0 -20 -40 -60 -80 -100 Time

  5. A plot of temperature vs. time that represents the process in which energy is added at a constant rate

  6. Temperature and Phase Change • The temperature doesn’t change during a phase change. • If you have a mixture of ice and water, the temperature is 0ºC • At 1 atm, boiling water is 100ºC • You can’t get the temperature higher until it boils

  7. Chemical Energy 2 parts of the universe as it relates to a chemical reaction: • System • the reactants and the products   • Surroundings • everything else in the universe (such as container, the room, etc.) Law of Conservation of Energy: the total energy of the universe is constant and can neither be created nor destroyed; it can only be transformed.

  8. surroundings system Chemical energy lost by combustion = Energy gained by the surroundings The First Law of Thermodynamics:The total energy content of the universe is constant Signs (+/-) will tell you if energy is entering or leaving a system + indicates energy enters a system - indicates energy leaves a system

  9. Chemical Energy Two types of processes based on energy flow: • Exothermic • produces energy (heat flows out of the system) • Endothermic • absorbs energy (heat flows into the system)

  10. Exothermic Reaction Reactant Product + Energy Surroundings System Conservation of Energy in a Chemical Reaction In this example, the energy of the reactants and products decreases, while the energy of the surroundings increases. In every case, however, the total energy does not change. Surroundings System Energy Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

  11. Surroundings Endothermic Reaction Reactant + Energy Product System Conservation of Energy in a Chemical Reaction In this example, the energy of the reactants and products increases, while the energy of the surroundings decreases. In every case, however, the total energy does not change. Surroundings Energy System Before reaction After reaction Myers, Oldham, Tocci, Chemistry, 2004, page 41

  12. Thermochemistry • Every reaction has an energy change associated with it • Energy is stored in bonds between atoms • Making bonds gives energy • Breaking bonds takes energy

  13. Enthalpy and enthalpy changes • To more easily measure and study the energy changes that accompany chemical reactions, chemists have defined a property called enthalpy. • Enthalpy(H) is the heat content of a system at constant pressure.

  14. Enthalpy and enthalpy changes • Although you cannot measure the actual energy or enthalpy of a substance, you can measure the change in enthalpy, which is the heat absorbed or released in a chemical reaction. • The change in enthalpy for a reaction is called theenthalpy (heat) of reaction(∆Hrxn). • You have already learned that a symbol preceded by the Greek letter ∆ means a change in the property.

  15. Enthalpy and enthalpy changes • Thus, ∆Hrxn is the difference between the enthalpy of the substances that exist at the end of the reaction and the enthalpy of the substances present at the start. • Because the reactants are present at the beginning of the reaction and the products are present at the end, ∆Hrxn is defined by this equation.

  16. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.4

  17. reaction reaction Energy Change in Chemical Processes Exothermic, heat given off & temperature of water rises Exothermic process: H < 0 (at constant pressure) Endothermic, heat taken in & temperature of water drops Endothermic process: H > 0 (at constant pressure)

  18. 2H2(g) + O2(g) 2H2O (l) + energy energy + H2O (s) H2O (l) Exothermic processis any process that gives off heat – transfers thermal energy from the system to the surroundings. H2O (g) H2O (l) + energy Endothermic processis any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (s) 2Hg (l) + O2(g) 6.2

  19. Enthalpy and enthalpy changes

  20. Endothermic Reactions

  21. Exothermic Reactions

  22. Endothermic or Exothermic? exothermic endothermic exothermic endothermic endothermic

  23. No catalyst Effect of Catalyst on Reaction Rate What is a catalyst? What does it do during a chemical reaction? Catalyst lowers the activation energy for the reaction. activation energy for catalyzed reaction reactants Energy products ReactionProgress

  24. Calculating Energy Changes - Heating Curve for Water 140 DH = mol xDHvap DH = mol xDHfus 120 100 80 Heat = mass xDt x Cp, gas 60 40 Temperature (oC) 20 Heat = mass xDt x Cp, liquid 0 -20 -40 -60 Heat = mass xDt x Cp, solid -80 -100 Time

  25. Heat of Fusion and Solidification • The heat that is absorbed by one mole of a substance in melting at a constant temperature is the molar heat of fusion DHfus • The heat lost when one mole of a liquid solidifies at a constant temperature is the molar heat of solidification DHsol H2O(s) H2O(l) DHfus= 6.01 kJ/mol H2O(l) H2O(g) DHsol = - 6.01 kJ/mol

  26. Heat of Vaporization&Condensation • The molar heat of vaporization: • Heat needed to change one mol of a liquid to gas DHvap • The molar heat of condensation: • Heat needed to change one mol of a gas to liquid DHcon H2O(l) H2O(g) DHvap= 40.7 kJ/mol H2O(g) H2O(l) DHcon = - 40.7 kJ/mol

  27. Heat of Reaction • The heat that is released or absorbed in a chemical reactionis equivalent to DH C + O2(g) CO2(g) +394 kJ C + O2(g) CO2(g) DH = -394 kJ • In thermochemical equation it is important to say what state H2O(g) H2(g) + ½O2 (g) DH = 241.8 kJ H2O(l) H2(g) + ½ O2 (g) DH = 285.8 kJ Difference = 44.0 kJ

  28. Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

  29. Hess’s Law

  30. Standard Heat of Formation • The change in heat that accompanies the formation of a mole of a compound from its elements at standard conditions • Standard conditions 25°C and 1 atm. • Symbol is H◦f • The standard heat of formation of an element at its most stable form is 0 • This includes the diatomics

  31. Hess’s Law • There are tables of heats of formations (pg. 316) • For most compounds it is negative • Because you are making bonds • Making bonds is exothermic • The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products H = H◦f (products) - H◦f (reactants)

  32. Rules • If a reaction is reversed, the sign of ∆H must be reversed as well. • because the sign tells us the direction of heat flow as constant P • The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction. If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way. • because ∆H is an extensive property

  33. Hess’s Law If H2(g)+ 1/2 O2(g) H2O(l)  H=-285.5 kJ/mol then H2O(l) H2(g)+ 1/2 O2(g)  H =+285.5 kJ/mol If you turn an equation around, you change the sign 2 H2O(l) 2 H2(g)+ O2(g)  H =+571.0 kJ/mol If you multiply the equation by a number, you multiply the heat by that number. • Twice the moles, twice the heat

  34. Hess’s Law • You make the products, so you need their heats of formation • You “unmake” the reactants so you have to subtract their heats. https://www.youtube.com/watch?v=_NLAgSnqNOE&noredirect=1

  35. Hess’s Law Example Problem Calculate the heat of combustion of methane, CH4 CH4(g) +2O2(g) CO2(g) + 2 H2O(g) H◦fCH4 (g) = -74.86 kJ/mol H◦fO2(g) = 0 kJ/mol H◦fCO2(g) = -393.5 kJ/mol H◦ fH2O(g) = - 241.8 kJ/mol pg. 316 H2 (g) + ½ O2 (g) H2O(g) 2x(- 241.8)= - 483.6kJ/mol Step #1: since 2 moles of water are produced by each mole of methane, we multiply the H◦f. of water by 2.

  36. Hess’s Law Example Problem Calculate the heat of combustion of methane, CH4 CH4(g) +2O2(g) CO2(g) + 2 H2O(g) H◦fCH4 (g) = +74.86 kJ/mol H◦fO2(g) = 0 kJ/mol H◦fCO2(g) = -393.5 kJ/mol H◦ fH2O(g) = -483.6 kJ/mol pg. 316 H◦f=[-393.5 kJ/mol + (-483.6 kJ/mol)]- [-74.86 kJ/mol + (0 kJ/mol )] H◦f= [-393.5 -483.6] + 74.86 = -877.1 + 74.86 = -802.2 kJ/mol Hrxn = Hf(products) - Hf(reactants) Step #2: sum up all the H◦f. :

  37. How do we relate change in temperature to the energy transferred? Specific Heat capacity (J/oC) = heat supplied (J) temperature (oC) Specific Heat Capacity = heat required to raise the temperature of 1 gram of a substance object by 1oC • Affected by • What the substance is • Mass of the object

  38. Calorimetry The amount of heat absorbed or released during a physical or chemical change can be measured… …usually by the change in temperature of a known quantity of water 1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C 1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

  39. Calorimeter • A device used to experimentally determine the amount of heat released or absorbed during a physical or chemical change heat gained = heat lost

  40. Units of energy Most common units of energy 1. S unit of energy is the joule (J), defined as 1 (kilogram•meter2)/second2, energy is also expressed in kilojoules (1 kJ = 103J). 2. Non-S unit of energy is the calorie. One cal = 4.184 J or 1J = 0.2390 cal. Units of energy are the same, regardless of the form of energy

  41. Specific Heat Capacity aka Specific Heat The amount of heat required to raise the temperature of one gram of substance by one degree Celsius. Page 296

  42. Specific Heat Capacity • The higher the specific heat the more energy it takes to change its temperature. • Pizza burning the roof of your mouth • The same amount of heat is released when an object cools down

  43. Calculations Involving Specific Heat Q = m. C. T Q= Heat lost or gained ( J) C= Specific Heat (J/ ºC.g) T = Temperature change = Tf – Ti (ºC)