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Thermodynamics

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Published in: Chemistry
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This presentation provides an in-depth overview of the fundamental principles of thermodynamics, the branch of physics that deals with heat, energy, and work. It begins with the basic definitions and laws of thermodynamics, including the First, Second, and Third Laws. Key concepts such as systems, surroundings, and state functions are introduced. The PPT then explores critical topics like enthalpy, entropy, internal energy, and Gibbs free energy, demonstrating their real-world applications in physical and chemical processes.

Mudasir A / Jammu

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  1. Unit 6 Thermodynamics
  2. VIDEO Endo and Exothermic Processes
  3. Endothermic Reactions C] Endothermic reactions gain energy during the reaction C] They absorb heat and energy Examples of Endothermic reactions • Photosynthesis Water Evaporating Ill ENDOTHERMIC HEAT
  4. Exothermic Reactions C] Exothermic reactions lose energy during the reaction C] They release heat and energy Examples of Exothermic reactions • Combustion Ice melting Ill EXOTHERMIC HEAT
  5. Examples of Endo and Exothermic Graphs Endothermic Reaction c net energy absorbed Reaction Coordinate Exothermic Reaction net energy released c Reaction Coordinate
  6. Endo and Exothermic Processes in Word problems You can determine whether a reaction is endo or exothermic by the change in enthalpy, Or AH. If the change in enthalpy is positive then the reaction is endothermic. If the change in enthalpy is negative the reaction is exothermic.
  7. Practice 1. N02+F2—2N02F AH=-284 kJ/mol According to the information given, is the proposed mechanism an endothermic or exothermic reaction? Why? 1. According to the graph is the reaction endothermic or exothermic? Why? Reaction Coordinate
  8. emß7øøEND0 & PLAON o e u; EXOTHERMIC —PROCESSES & ENERGY DIAGRAMS
  9. reacti0h repreceht tAe ehergy cÅah9ec tAat occur dur;h9 a poteht;a/ eher9 product to cohduct CA e react;oh. VA ey A/s'o CA ow tke reactioh ehd0tAermic or 6 2 - ENERGY DIAGRAMS
  10. ExotAeFht;c React;oh". a decrea?e potent;a/ ehergy as' peactahte are converted ;hto products'. React;ohç a increaee ih potentiA/ ehergy ae producte are converted ihto reactAhtç. ee,ergy
  11. is' a h-.eaCure OF ee,ergy ;hvo/ved a reacti0h If tAe ehtAa/py is' (ecc tkah O, tAe reacti0h is' exotkermc. it ic greater tAah O, tAe rat;oh e h e rht;c. axis' repreceettc . x axic reprecehtc tAe ecc tAe react;oh dup;h9 tine pÅace cAAh9e. Ah upward C/ope wi// ihd;cate tÅe is' "orbed durih9 reacti0h, or Ah ehd0tAerhGC react;oh. Ah reacti0h w;// ave a dowhward s'/ope ihdicated 4 _ I _ _
  12. Determhe tAe poteht;a/ eherqy of tke products' Ahd tAe reactahtc, Ahd wketker tAe FO//0wih9 react;0hC are ehd0tÅermc or For TAPA: Etivated Reaction pathway Progress Of the reaction
  13. https://create.kahoot.it/details/583a0f23-75cb-4961-ba88-6e2478a00c 1 b
  14. TRANSFER AND THERMAL EQUILIBRIUh i 6.3 Heat Transfer and Thermal Equilibrium
  15. TOPIC: 6.3 HEAT TRANSFER AND THERMAL EQUILIBRIUM ENDURING UNDERSTANDING: ENE•2 Changes in a substance's properties or change into a different substance requires an exchange of enero•. LEARNING OBJECTIVE: ENE•2.C Explain the relationship between the transfer of thermal enerv and molecular collisions, ESSENTIAL KNOWLEDGE: ENE.2.C1 The particles in a warmer body have a greater average kinetic energy than those in a cooler body. ENE•2.C2 Collisions between particles in thermal contactcan result in the transfer ofenergy. Thisprocessiscalled •heat transfer. • •heat excha e. • or -transfer of ene as heat.- ENE-2.C3 Eventually, thermal equilibrium is reached as the particles continue to collide. At thermal equilibrium. the aver kinetic of both bodies is the same and hence their tem ratures are the same. EQUATION(S): N/A
  16. Intro to Heat Transfer/ Thermal Eq. Heat transfer refers to the movement of thermal energy between objects due to a temperature difference. Thermal equilibrium is reached when two objects in contact with each other reach the same temperature and there is no net heat transfer between them. Three main methods of heat transfer: Conduction: Transfer of heat through direct contact between particles in a material. Convection: Transfer of heat through the movement of fluids (liquids or gases). Radiation: Transfer of heat through electromagnetic waves, without the need for a medium
  17. Specific Heat and Heat Capacity Specific heat (c) is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. Specific Heat Heat capacity (C) is the amount of heat required to raise the temperature of an entire sample of a substance by one degree Celsius. The formula for calculating heat (q) is q = mcAT, where: q is the heat transferred, m is the mass of the substance, c is the specific heat, and AT is the change in temperature.
  18. Application of Heat Transfer Principles Understanding heat transfer and thermal equilibrium is essential in various chemical processes, including: Calorimetry: Measuring heat changes in chemical reactions or physical processes using calorimeters. Phase changes: Heat transfer is involved in processes such as melting, freezing, vaporization, and condensation. WHEN EXOTHERMIC EXPERIMENT, Thermodynamics: Heat transfer principles are ITS TO FEEL THE fundamental to the study of energy changes in chemical reactions and systems.
  19. Practice A 50.0 g piece of copper at 100.00C is placed into 200.0 g of water at 20.00C. Assuming no heat is lost to the surroundings, what will be the final temperature of the mixture? (Specific heat of copper = 0.385 J/g0C, specific heat of water = 4.18 J/g0C)
  20. Practice 1 Answer 836xfi nal temperature-836x20.00C= 19.25x100.00C—19.25xfinaI temperature 836xfinal temperature+19.25xfinal temperature= 19.25x100.OOC+836x20.OOC (836+19.25)xfinal temperature-(1925+16720) temperature=(1925+16720)J 855.25xfinal temperature= 18645 J855.25xfinal temperature=18645J final temperature=18645/855.25 final temperature= final temperature= 21.8 oc final temperature
  21. Practice A 100.0 g block of aluminum at 80.00C is placed into a container with 300.0 g of water at 25.00C. After reaching thermal equilibrium, the final temperature of the mixture is found to be 30.00C. Calculate the specific heat of the container material assuming no heat is lost to the surroundings. (Specific heat of aluminum = 0.897 J/g0C, specific heat of water = 4.18 J/g0C)
  22. Practice 2 Answer c=4485 J/(IOO.O cz4485/J500.O gx5.OOC cz 500.Ogx5.OOC 4485 J cz4485/J2500 J cz1.79J/gOC So, the specific heat of the container material is approximately 1.79 J/g0C
  23. Extra Study sources https://quizlet.com/846987972/ap-daily-63-ap-chemistry-flash-cards/ https://quizlet.com/665705823/ap-chem-unit-63-heat-transfer-and- thermal-equilibrium-flash-cards/ https://quizlet.com/54078203/ap-chemistry-61-63-flash-cards/
  24. 8l3Wld01V0Å ??? ??? ?? ?? pue ("Alp eöH*
  25. 00 Heat Capacity and Calorimetry o o The first law of thermodynamics states that energy is conserved in physical and chemical processes. The heating of a cooler sample by a warmer sample is an important form of energy transfer between the two systems. a The study of this heat transfer is called calorimetry. The amount of energy can be calculated using this equation: a m: mass a temperature change (Tf-Ti) a heat capacity WHEN EVEN THE DOG STARTS
  26. Specific Heat and Molar Heat o o O molar Specific Heat Capacity (c) Energy required to raise lg of a substance IOC or 1 K Molar Heat Capacity Energy required to raise 1 mol of a substance 1 oc or 1 K mol•K o o O
  27. Chemical Systems Change Chemical systems change their energy through 3 main processes: a Heating or cooling a Phase change a Chemical reaction Calorimetry: A measurement of changing the heat in a system a 1st Law: I thing gains heat= 1 thing loses heat If we know how much heat is being transferred, we can calculate how much heat is being lost.
  28. O o O Practice Problem 2: O o Problem 1: How much heat is required to heat IOOg of water If 55gof a metal (antimony) at 1000C heated the from 250C to 270C ? water, what is the heat capacity of that metal?
  29. N/A
  30. AP Chem Topic 6 5 . - Energy of IF ENERGLCAUSES Phase Changes CHÄNGE„. AP Classroom 6.5 Video THEN,WHAtlCAUSES
  31. Endothermic Phase Changes Point Q = me-AT Progress of Reaction (TIME) Liquid end This graph shows the temperature (which directly correlates with the energy present) v.s. the progression of time as we change a solid, into a liquid, and into a gas. These processes of breaking bonds in order to allow the molecules to be more spread (as shown in the diagram at the bottom) consumes heat/energy, making the reactions endothermic. Take note of the plateaus in the upper diagram. While the substance is in the process of melting and/or vaporizing, there is no change in temperature, and there is both solid and liquid material (for melting), or liquid and gas material (for vaporizing) present.
  32. 100 freezing Solid Exothermic Phase Changes I OOOC pen t Condensing Progress of Reaction (TIME) Liquid Now take a look at the same processes happening in reverse. We changed a gas into a liquid, and a liquid into a solid. These processes form bonds between the molecules of the substance, which causes the molecules to stay closer together depending on which state of matter they are in (as seen in the diagram below). This process of forming bonds between molecules releases energy, making the processes exothermic. Take note of the plateaus of the upper diagram. They represent condensation, and freezing. Note that they, just like in the endothermic processes, have no change in temperature. These stages also, like the endothermic reactions, contain some of both of states of matter that
  33. Molar Enthalpy Molar Enthalpy is the energy (in joules) required to conduct an endothermic process (so either melting or vaporization) per mole of substance. It is the energy consumed during the reaction. The energy released during exothermic reactions is the same magnitude, but the opposite direction, as the molar enthalpy for the opposite process. This means that (AH of vaporization) = the molar enthalpy of vaporization, and that The energy released when that same substance is condensed = (-AH of vaporization) The same is true for (AH of melting) and the energy released when the same substance freezes
  34. Remember... Gaseous substances tend to have more kinetic energy, and solid substances tend to have less kinetic energy If gaseous substances have more kinetic energy, then it makes sense that the process of condensation releases energy, and is exothermic If solid substances have less kinetic energy, the it makes sense that the process of melting consumes energy, and is endothermic The same amount of energy is consumed/released when traveling between two states of matter; (AH of vaporization) = the molar enthalpy of vaporization, and the energy released of condensation = (-AH of vaporization)
  35. Practice 1: Question Answer the following questions about gallium, a metal Often used to make computer chips. Molar heat capacity of solid Heat of fusion Heat of vaporization Melting point Gallium can be purified by melting. originally at 293 K? 25.861 mol-I ICI 5.59 mot-I 256 k/ mol-I 303 K How much energy in kJ is required to melt 2 moles of gallium
  36. Practice 1: Answer Answer the following questions about gallium, a metal often used to make computer chips. 25.86] moriK-1 Molar heat capacity of solid 5.59 mori Heat of fusion 256 k/ mori Heat of vaporization Melting point 303 K Gallium can be purified by melting. How much energy in kJ is required to melt 2 moles of gallium originally at 293 K? Step one: heat to melting point q = (2 - 293) = 520/ q (2 KJ mot-I ) 11.2 Step two: melting Step three: convert and add "q" values 11.2k] + = 11.7k]
  37. Practice 2: Question Answer the following questions about gallium, a metal Often used to make computer chips. Molar heat capacity of solid Heat of fusion Heat of vaporization Melting point 25.86] moriK-1 5.59 kJ mot-I 256 k] mol-I 303 K use principles of bonding to explain why the heat of vaporization is much greater than the heat of
  38. Practice 2: Answer Answer the following questions about gallium, a metal Often used to make computer chips. Molar heat capacity of solid Heat of fusion Heat of vaporization Melting point 25.86 J mot-I VI 5.59 k] mori 256 kJmor1 303 K Use principles of bonding to explain why the heat of vaporization is much greater than the heat of fusion. Gallium has metallic bonding with metal ions arranged in a lattice with delocalized "sea of electrons" moving freely through the structure. During melting, small separations form within the lattice, allowing the metal to flow. During vaporization, the metal atoms must be completely separated, and this requires far more energy.
  39. Practice Here it is
  40. Unit 6.6 Enthalpy of Reaction James Moroch m U«OINTRODUCTION ONS,.. ENTHALPY OF REACTION
  41. Overview Enthalpy is a measure of the total internal energy of a system. This is including the energy required to change the temperature and the pressure of the system. It can be used to to describe heat energy changes in chemical reactions and phase transitions AH refers to the heat transferred at constant pressure under standard conditions and states FINO THE
  42. Calculating AW = the standard enthalpy change AHO is found by using the equation: AW = m x s x AT M is the mass of the reactants, S is the specific heat of the product, and AT is the change in temperature from the reaction Energy Specific heat mass x capacity Temperature change m x s
  43. Standard Enthalpies Standard Enthalpy Change of. Reacti0Q Formation Combustion Neutralisation Definition The change When the reactants in the Exothermic/ Symbol Endothermic stoichiometric equation react to give the products AHOr under standard conditions The enthalpy change when one mole of a compound is formed from its elements under standard conditions The enthalpy change when one mole of a substance is burnt in excess oxygen under standard conditions The enthalpy change when one mole Of water is formed by reacting an acid and alkali under Standard conditions Both Both Exothermic AH neut Exothermic
  44. Practice Problems One mole of water is formed from hydrogen and oxygen releasing 286 kJ H2 (g) + 1/202 (g) H20 (l) AHOr= -286 k] mol-I Calculate AHr for the reaction below: 2H2 (g) +02 (g) 2H20 (l) Calculate AHr for the reaction below 4Fe (s) +02 (g) 2Fe203 (s) given that AHOf [Fe203 824 k] mol-I
  45. Review https://quizizz.com/admin/quiz/5e78f0a()373c7f()()ldbb3e8d/ent halpy-change-of-reaction
  46. SAME ENERGY Bond Enthalpies UNIT By Sophia Brochetti Link to video 1 Link to video 2
  47. Breaking and orming bonds • When bonds are formed, it requires a RELEASE of energy (ENDOthermic) • When bonds are broken, it requires an INPUT of energy (EXOthermic) BOND ENERGY - The energy stored in a bond. If the bond energy is negative, it represents the energy that is released when the bond forms If the bond energy is positive, it represents the energy that is required to break the bond For example: The bond energy Of H-H bonds is 436 kJ/moI, Which means that: • 436 kJ would be released when I mole Of H-H bonds is formed, and • 436 kJ would be required to break 1 mole of H-H bonds.
  48. Average Bond Energies Energy. c-c Notce that bond energy increases as bond order This is because it takes more energy to break double Il or triple Ill bonds. Bond energies or bond enthalpies are useful because we can use the average changes to estimate the overall enthalpy/energy change in a reaction. This is represented by AH. AH = E (B.E. of bonds broken [positive + E (B.E. of bonds formed [negative
  49. Practice problem 1 CI-CI H-CI Based on the bond energies shown in the table below. which Of the following diagrams best represents the change in energy as the reaction represented below proceeds? A) 0) Bond 1-1 Bond Energy ( kJ/niol) 240 430
  50. Practice problem 1 SOLVED Reactants For I H_H: 430 kJ For 1 240 kJ Total Reactants: 430 + 240 = 670 kJ These bonds will be broken, so we will make 670 positive Products From 2 HO: kJ These bonds will be formed, so we will make 8}) negative + 670 -860 -190 kJ We are looking for an exothermic reaction. So since we know that the reaction is EXOTHERMIC and the enthalpy change is 190... the answer is A! A)
  51. Practice problem 2 2 H202(aq)— 2 H20(t) + 02(g) = -196 kJ/molr„ The decoagx»sition of H707(aq) represented by the equation abm•e. AS.urne that tmd the in are Significantly in HA, the o' 01 the of the be the (in and formed in the C) D)
  52. Practice problem 2: SOLVED 600- +100 300 - 500 = -200 1000 - 300 = +702 500 - 300 +200 Reactants H-O-O-H H-O-O-H Products H-O-H H-O-H 2 0-0 2 0-0-1 = around -200 So since we calculated -200 as the bond enthalpy, the correct answer would be B!
  53. Cooper Hall Video! (watch to 4:10 for this topic) Enthalpy of Formation
  54. The Basics AHOf = Standard Enthalpy of Formation A reference values that exists for compounds and elements Represents the change in enthalpy when a substance forms Can be used to estimate the overall enthalpy change of a reaction (AHOrxn) Gives information about exothermic or endothermic reactions The degree symbol indicates that everything is happening under standard conditions Gas standard state = 1 atm of pressure Solid and Liquid standard state = most common form of element at 1 atm and 250C Solutions standard state: concentratitn of 1 M
  55. + •Where is AHOf coming from?? • + Compounds = enthalpy change when 1 mol of a compound forms from its elements in their standard states Pure element in standard state will always be have a standard enthalpy of formation of O Not true for pure elements not in their standard state F2 has a AHOf of O F has a AHOf of 83.88
  56. FLOURINE Being alone Being with another F Remember diatomie elements are only in their standard state if they are bonded together Diatomic Elements: H, N, O, F, Cl, l, and Br 0'
  57. Calculating Enthalpy of Formation - AHOrxn— f(products)— f(reactants) Multiply stoichiometry coefficients by standard enthalpy of formation for each value Remember that pure elements in their standard states have a standard enthalpy of formation of O
  58. Practice 1 Using the table of standard enthalpies of formation below, determine the standard enthalpy change of the reaction when 2.00 moles of propene combust. The balanced equation for the combustion of 2.00 moles of propene is 2 C3H6(g) + 9 02(g) —+ 6 C02(g) + 6 H20(g). Substance Standard Enthalpy Of Formation (kJ/mol) C3H6(g) 21 C02(g) -394 02(g) H20(g) -242
  59. Practice 2 3 C2H2(g) -i C6H6(g) What is the standard enthalpy change AHO, for the reaction represented above? AHO, of C2H2(g) is 230 kJ mol-I; of C6H6(g) is 83 kJ mol-I. a) - 607 kJ b) — 147kJ c) - 19kJ d) + 19kJ + 773 kJ
  60. Kahoot! https://create.kahoot.it/share/6-8-enthalpy-of- formation/e22d9565-8d8e-4f92-b21f- 4f2571aOb9fb Games
  61. Topic 6.9: Hess's Law 5 LAONSo.o AND HESSS LAW
  62. Hess's Law Describes how the enthalpy of a reaction changes if reactions are reversed, combined, and/or multiplied. The overall enthalpy change of a reaction is the sum of the enthalpy change in the products minus the sum of the enthalpy changes of the reactants. AHOrxn = EnpAHtP (products) — EnrAHtP (reactants) O
  63. Example of Hess's Law in Action The reaction: mol Can take place as a series Of two steps: step 1: + 02(g) C02(g) step 2: C02(g) -o + 02(g) AH, = -394 AHF 396— mol (Exotherma:) (Ent%thermic) AHI + = -394 -EL + 396 Y = 2 mol mol mol
  64. Factors Impacting Hess's Law If a chemical reaction is reversed, the magnitude of enthalpy stays the same but takes the opposite signs. - 396 kJ/mol turns to 396 kJ/mol If a chemical reaction is multiplied by a factor, the change in O enthalpy is also multiplied by that factor. —396 — m 01 ICJ AHrxn = -792 — mol For example: + 02(g) -5 C02(g) 2Cdiamond(S) + 202(g) 2C02(g) A Hrxn
  65. Factors Impacting Hess's Law • If reactions are combined, their enthalpies can be combined in the same manner. AH - -787.0 kJ AH- +566.0 kJ O 2C(s) +02(g) 2CO(g) AH - -787.0 +566.0kJ --2210 kJ
  66. Practice Problem #1 N2(g) + 02(9) -0 2NO(9) 2N02(9) 2NO(9) + 02(g) 180 —112 Given the information above, what is the overall enthalpy change for the reaction given below? N2(9) + 202(9) 2N02(g) O
  67. Practice Problem #2 HI(g) 26 kJ/molrxn % H2(g) % 12(g) — HI(g) -5.0 Based on the information above, What is the enthalpy change for the sublimation Of iodine, represented belON9 D E 15 kJ/molrxn 21 kJ/mOlrxn 31 kJ/molrxn 42 kJ/molrxn 62 kJ/moI 12(s) '2(g)
  68. Practice Problem #3 The enthalpy change for the reaction A — X is This reaction can be tyoken down into a senes 01 steps as shown in ttæ diagram: All A relationship that exist the varimjs enth*.y chanws is
  69. Quizizz Review Ck12 Review Another quizizz review

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