Thermochemistry and Thermodynamics 1 Chemistry: The central Science AP version (10th edition, chapter 5) Additional References and Resources Powerpoint presentation from Melissa Brophy’s website (modified for our needs) Chemtour Animations and Vocab flashcards from W.W. Norton site: http://www.wwnorton.com/college/chemistry/c hemistry3/ch/05/studyplan.aspx (a) State functions and path functions (b) Internal Energy (c) PV work (d) Calorimetry (e) Heating Curves POGILS: Thermochemistry and Calorimetry Hess’s Law: Enthalpy is a state function Internal Energy and Enthalpy Thermochemistry package: vocab, diagrams and equations, practice problems (M. Brophy) © 2006 Brooks/Cole - Thomson Lab experiments: (1) Heat capacity of a metal (2) Heat of neutralization (HCl + NaOH) Activity (sign conventions for q and w) from chap 5 online resources and movies (thermite; AlBr3; NI3). Online practice quiz Chemguy videos for introduction to energy and thermochemistry or thermodynamics http://www.youtube.com/watch?v=DIpGoOurt8&p=93017BDB7EDFE758&index=1&feature= BF Chem guy videos – thermodynamics 1-5 – one video clip runs into the next one for more advanced aspect of thermodynamics For thermodynamics 1: spontaneous process, enthalpy, entropy (system, surroundings, universe) http://www.youtube.com/watch?v=94p9ZfTjeKc &p=2FDA79D443854B33&index=2&playnext=2 Chemistry and Chemical Reactivity 6th Edition 2 John C. Kotz Paul M. Treichel Gabriela C. Weaver CHAPTER 6 Principles of Reactivity: Energy and Chemical Reactions Lectures written by John Kotz ©2006 2006 Brooks/Cole Thomson © Brooks/Cole - Thomson 3 Thermodynamics and Thermochemistry Thermodynamics • Study of energy and its transformations (or Interconversions) © 2006 Brooks/Cole - Thomson Thermochemistry • That part of thermodynamics that deals with the relationships between chemical reactions and energy changes involving heat Energy & Chemistry 4 ENERGY is the capacity to do work or transfer heat. work is the energy used to cause an object with mass to move (w= F x d). When chemical reactions involve gases, the work done may involve compression or expansion of gases. (w = -P∆V where P = pressure and ∆V = Vfinal-Vinitial) heat is the form of energy that flows between 2 objects because of their difference in temperature. Other forms of energy -Light, electrical, kinetic and potential © 2006 Brooks/Cole - Thomson Internal Energy (E) • PE + KE = Internal energy (E or U) • Internal E of a chemical system depends on • number of particles • type of particles • temperature © 2006 Brooks/Cole - Thomson 5 6 GO TO chemtours animation on Internal Energy • http://www.wwnorton.com/college/chemistry/ chemistry3/ch/05/studyplan.aspx © 2006 Brooks/Cole - Thomson 7 Changes in Internal Energy • When a system undergoes any chemical or physical change, the change in internal energy is • ∆E = q + w [ ∆ E = Efinal- Einitial ] • q = heat added or liberated from the system • w = work done on or by the system © 2006 Brooks/Cole - Thomson 8 heat transfer in (endothermic), +q heat transfer out (exothermic), -q SYSTEM ∆E = q + w w transfer in (+w) © 2006 Brooks/Cole - Thomson w transfer out (-w) 9 Go To TB animations and to Chem Tours • Activity (sign conventions for q and w) from chap 5 online resources and movies (thermite; AlBr3; NI3). • (To Do as Homework) • PV work from Chem tours: • http://www.wwnorton.com/college/chemistry/ chemistry3/ch/05/studyplan.aspx © 2006 Brooks/Cole - Thomson ENTHALPY Most chemical reactions occur at constant P, so Heat transferred at constant P = qp qp = ∆H where H = enthalpy and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = Hfinal – Hinitial More on Enthalpy Later © 2006 Brooks/Cole - Thomson 10 Internal Energy (E) • Recall: PE + KE = Internal energy (E or U) • Internal E of a chemical system depends on • number of particles • type of particles • temperature © 2006 Brooks/Cole - Thomson 11 Potential Energy on the Atomic Scale • Positive and negative particles (ions) attract one another. • Two atoms can bond • As the particles attract they have a lower potential energy © 2006 Brooks/Cole - Thomson NaCl — composed of Na+ and Cl- ions. 12 Potential Energy on the Atomic Scale • Positive and negative particles (ions) attract one another. • Two atoms can bond • As the particles attract they have a lower potential energy © 2006 Brooks/Cole - Thomson + 13 _ 14 Electrostatic potential energy (Eel) • Equation: • The chemical energy associated with compounds is due to the potential energy stored in the arrangement of atoms in a substance (ex: bond energies) © 2006 Brooks/Cole - Thomson Potential & Kinetic Energy Kinetic energy — energy of motion. rotate vibrate translate © 2006 Brooks/Cole - Thomson 15 Internal Energy (E) • PE + KE = Internal energy (E or U) © 2006 Brooks/Cole - Thomson 16 Internal Energy (E) • The higher the T the higher the internal energy • So, use changes in T (∆T) to monitor changes in E (∆E). © 2006 Brooks/Cole - Thomson 17 Thermodynamics • Thermodynamics is the science of heat (energy) transfer. Heat energy is associated with molecular motions. Heat transfers until thermal equilibrium is established. ∆T measures energy transferred. © 2006 Brooks/Cole - Thomson 18 System and Surroundings • SYSTEM – The object under study • SURROUNDINGS – Everything outside the system © 2006 Brooks/Cole - Thomson 19 Heat The flow of thermal energy from one object to another. Cup gets cooler while hand 20 gets warmer Heat always flows from warmer to cooler objects. Ice gets warmer while hand gets cooler © 2006 Brooks/Cole - Thomson Directionality of Heat Transfer 21 • Heat always transfer from hotter object to cooler one. • EXOthermic: heat transfers from SYSTEM to SURROUNDINGS. T(system) goes down T(surr) goes up © 2006 Brooks/Cole - Thomson Directionality of Heat Transfer • Heat always transfers from hotter object to cooler one. • ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM. T(system) goes up T (surr) goes down © 2006 Brooks/Cole - Thomson 22 23 Go To – chapter 5 – online resources for our TB • Go to Activity banner and view sign conventions for q and w • View movies (thermite; AlBr3; NI3) • Write key points in margins of this page • Reminder Online practice quiz for chapter 5 from our TB needs to be completed © 2006 Brooks/Cole - Thomson Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY. • The total energy is unchanged in a chemical reaction. • If PE of products is less than reactants, the difference must be released as KE. © 2006 Brooks/Cole - Thomson 24 Energy Change in Chemical Processes PE Reactants Kinetic Energy Products PE of system dropped. KE increased. Therefore, you often feel a T increase. © 2006 Brooks/Cole - Thomson 25 UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) But we use the unit called the JOULE 1 cal = 4.184 joules © 2006 Brooks/Cole - Thomson James Joule 1818-1889 26 27 HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C. Which has the larger heat capacity? © 2006 Brooks/Cole - Thomson Specific Heat Capacity How much energy is transferred due to T difference? The heat (q) “lost” or “gained” is related to a) b) sample mass change in T and c) specific heat capacity Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K) © 2006 Brooks/Cole - Thomson 28 Specific Heat Capacity Substance Spec. Heat (J/g•K) H2O 4.184 Ethylene glycol 2.39 Al 0.897 glass 0.84 Aluminum © 2006 Brooks/Cole - Thomson 29 30 Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? Specific heat capacity = heat lost or gained by substance (J) (mass, g)(T change, K) © 2006 Brooks/Cole - Thomson Specific Heat Capacity If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? heat gain/lose = q = (sp. ht.)(mass)(∆T) where ∆T = Tfinal - Tinitial q = (0.897 J/g•K)(25.0 g)(37 - 310)K q = - 6120 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al. © 2006 Brooks/Cole - Thomson 31 Heat/Energy Transfer No Change in State q transferred = (sp. ht.)(mass)(∆T) © 2006 Brooks/Cole - Thomson 32 Heat Transfer • Use heat transfer as a way to find specific heat capacity, Cp • 55.0 g Fe at 99.8 ˚C • Drop into 225 g water at 21.0 ˚C • Water and metal come to 23.1 ˚C • What is the specific heat capacity of the metal? (Lab Activity) © 2006 Brooks/Cole - Thomson 33 Heat Transfer Because of conservation of energy, q(Fe) = –q(H2O) (heat out of Fe = heat into H2O) or q(Fe) + q(H2O) = 0 q(Fe) = (55.0 g)(Cp)(23.1 ˚C – 99.8 ˚C) q(Fe) = –4219 • Cp q(H2O) = (225 g)(4.184 J/K•g)(23.1 ˚C – 21.0 ˚C) q(H2O) = 1977 J q(Fe) + q(H2O) = –4219 Cp + 1977 = 0 Cp = 0.469 J/K•g © 2006 Brooks/Cole - Thomson 34 Heat Transfer with Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid water q = (heat of fusion)(mass) © 2006 Brooks/Cole - Thomson 35 36 Heat Transfer and Changes of State Liquid ---> Vapor Requires energy (heat). This is the reason a) you cool down after swimming b) you use water to put out a fire. © 2006 Brooks/Cole - Thomson + energy 37 GO TO chemtours animation on Heating Curves • http://www.wwnorton.com/college/chemistry/ chemistry3/ch/05/studyplan.aspx © 2006 Brooks/Cole - Thomson Heating/Cooling Curve for Water Note that T is constant as ice melts © 2006 Brooks/Cole - Thomson 38 Heat & Changes of State What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 oC? Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 J/g•K Heat of vaporization = 2260 J/g +333 J/g © 2006 Brooks/Cole - Thomson +2260 J/g 39 Heat & Changes of State What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 oC? 1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J 2. To raise water from 0 oC to 100 oC q = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J 3. To evaporate water at 100 oC q = (500. g)(2260 J/g) = 1.13 x 106 J 4. Total heat energy = 1.51 x 106 J = 1510 kJ © 2006 Brooks/Cole - Thomson 40 Chemical Reactivity What drives chemical reactions? How do they occur? The first is answered by THERMODYNAMICS and the second by KINETICS. Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED. • formation of a precipitate • gas formation • H2O formation (acid-base reaction) • electron transfer in a battery © 2006 Brooks/Cole - Thomson 41 Chemical Reactivity But energy transfer also allows us to predict reactivity. In general, reactions that transfer energy to their surroundings are productfavored. So, let us consider heat transfer in chemical processes. © 2006 Brooks/Cole - Thomson 42 Heat Energy Transfer in a Physical Process CO2 (s, -78 oC) ---> CO2 (g, -78 oC) Heat transfers from surroundings to system in endothermic process. © 2006 Brooks/Cole - Thomson 43 Heat Energy Transfer in a Physical Process • CO2 (s, -78 oC) ---> CO2 (g, -78 oC) • A regular array of molecules in a solid -----> gas phase molecules. • Gas molecules have higher kinetic energy. © 2006 Brooks/Cole - Thomson 44 Energy Level Diagram for Heat Energy Transfer CO2 gas ∆E = E(final) - E(initial) = E(gas) - E(solid) CO2 solid © 2006 Brooks/Cole - Thomson 45 Heat Energy Transfer in Physical Change CO2 (s, -78 oC) ---> CO2 (g, -78 oC) Two things have happened! • Gas molecules have higher kinetic energy. • Also, WORK is done by the system in pushing aside the atmosphere. © 2006 Brooks/Cole - Thomson 46 FIRST LAW OF THERMODYNAMICS heat energy transferred ∆E = q + w energy change work done by the system Energy is conserved! © 2006 Brooks/Cole - Thomson 47 48 heat transfer in (endothermic), +q heat transfer out (exothermic), -q SYSTEM ∆E = q + w w transfer in (+w) © 2006 Brooks/Cole - Thomson w transfer out (-w) ENTHALPY Most chemical reactions occur at constant P, so Heat transferred at constant P = qp qp = ∆H where H = enthalpy and so ∆E = ∆H + w (and w is usually small) ∆H = heat transferred at constant P ≈ ∆E ∆H = change in heat content of the system ∆H = Hfinal - Hinitial © 2006 Brooks/Cole - Thomson 49 ENTHALPY ∆H = Hfinal - Hinitial If Hfinal > Hinitial then ∆H is positive Process is ENDOTHERMIC If Hfinal < Hinitial then ∆H is negative Process is EXOTHERMIC © 2006 Brooks/Cole - Thomson 50 USING ENTHALPY Consider the formation of water H2(g) + 1/2 O2(g) --> H2O(g) + 241.8 kJ Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ © 2006 Brooks/Cole - Thomson 51 USING ENTHALPY 52 Making liquid H2O from H2 + O2 involves two exothermic steps. H2 + O2 gas H2O vapor © 2006 Brooks/Cole - Thomson Liquid H2O USING ENTHALPY Making H2O from H2 involves two steps. H2(g) + 1/2 O2(g) ---> H2O(g) + 242 kJ H2O(g) ---> H2O(liq) + 44 kJ ----------------------------------------------------------------------- H2(g) + 1/2 O2(g) --> H2O(liq) + 286 kJ Example of HESS’S LAW— If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns. © 2006 Brooks/Cole - Thomson 53 54 ChemTours GO TO: http://www.wwnorton.com/college/che mistry/chemistry3/ch/05/studyplan.asp x View animation on (a) State functions and path functions Write key points on this page © 2006 Brooks/Cole - Thomson 55 Hess’s Law & Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. Active Figure 6.18 © 2006 Brooks/Cole - Thomson 56 Hess’s Law & Energy Level Diagrams Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed. Active Figure 6.18 © 2006 Brooks/Cole - Thomson 57 ∆H along one path = ∆H along another path • This equation is valid because ∆H is a STATE FUNCTION • These depend only on the state of the system and not how it got there. • V, T, P, energy — and your bank account! • Unlike V, T, and P, one cannot measure absolute H. Can only measure ∆H. © 2006 Brooks/Cole - Thomson Standard Enthalpy Values Most ∆H values are labeled ∆Ho Measured under standard conditions P = 1 atm = 101.3 kPa = 760 mmHg Concentration = 1 mol/L T = usually 25 oC with all species in standard states e.g., C = graphite and O2 = gas © 2006 Brooks/Cole - Thomson 58 Enthalpy Values Depend on how the reaction is written and on phases of reactants and products H2(g) + 1/2 O2(g) --> H2O(g) ∆H˚ = -242 kJ 2 H2(g) + O2(g) --> 2 H2O(g) ∆H˚ = -484 kJ H2O(g) ---> H2(g) + 1/2 O2(g) ∆H˚ = +242 kJ H2(g) + 1/2 O2(g) --> H2O(liquid) © 2006 Brooks/Cole - Thomson ∆H˚ = -286 kJ 59 Standard Enthalpy Values NIST (Nat’l Institute for Standards and Technology) gives values of ∆Hfo = standard molar enthalpy of formation — the enthalpy change when 1 mol of compound is formed from elements under standard conditions. See Table 5.3 and Appendix C of OUR TB and AP reference Tables © 2006 Brooks/Cole - Thomson 60 61 Reference Tables for Thermochemistry • Text Book: Table 5.3 Standard enthalpies of formation and Appendix C: Thermodynamic Quantities for Selected Substances at 298.15 K (25 °C) Use these for problems from our textbook. • AP test reference tables • Regents Reference tables © 2006 Brooks/Cole - Thomson ∆Hfo, standard molar enthalpy of formation H2(g) + 1/2 O2(g) --> H2O(g) ∆Hfo (H2O, g)= -241.8 kJ/mol By definition, ∆Hfo = 0 for elements in their standard states. © 2006 Brooks/Cole - Thomson 62 Using Standard Enthalpy Values Use ∆H˚’s to calculate enthalpy change for H2O(g) + C(graphite) --> H2(g) + CO(g) (product is called “water gas”) © 2006 Brooks/Cole - Thomson 63 Using Standard Enthalpy Values H2O(g) + C(graphite) --> H2(g) + CO(g) From reference books we find • H2(g) + 1/2 O2(g) --> H2O(g) ∆Hf˚ of H2O vapor = - 242 kJ/mol • C(s) + 1/2 O2(g) --> CO(g) ∆Hf˚ of CO = - 111 kJ/mol © 2006 Brooks/Cole - Thomson 64 Using Standard Enthalpy Values H2O(g) --> H2(g) + 1/2 O2(g) ∆Ho = +242 kJ C(s) + 1/2 O2(g) --> CO(g) ∆Ho = -111 kJ -------------------------------------------------------------------------------- H2O(g) + C(graphite) --> H2(g) + CO(g) ∆Honet = +131 kJ To convert 1 mol of water to 1 mol each of H2 and CO requires 131 kJ of energy. The “water gas” reaction is ENDOthermic. © 2006 Brooks/Cole - Thomson 65 Using Standard Enthalpy Values 66 In general, when ALL Calculate ∆H of reaction? enthalpies of formation are known, ∆Horxn = ∆Hfo (products) - ∆Hfo (reactants) Remember that ∆ always = final – initial © 2006 Brooks/Cole - Thomson Using Standard Enthalpy Values Calculate the heat of combustion of methanol, i.e., ∆Horxn for CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn = ∆Hfo (prod) - ∆Hfo (react) © 2006 Brooks/Cole - Thomson 67 Using Standard Enthalpy Values CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Horxn = ∆Hfo (prod) - ∆Hfo (react) ∆Horxn = ∆Hfo (CO2) + 2 ∆Hfo (H2O) - {3/2 ∆Hfo (O2) + ∆Hfo (CH3OH)} = (-393.5 kJ) + 2 (-241.8 kJ) - {0 + (-201.5 kJ)} ∆Horxn = -675.6 kJ per mol of methanol © 2006 Brooks/Cole - Thomson 68 69 Calorimetry ChemTours GO TO: http://www.wwnorton.com/college/che mistry/gilbert2/contents/ch05/studyplan .asp View animation on calorimetry and Write key points in margins of this page © 2006 Brooks/Cole - Thomson CALORIMETRY Measuring Heats of Reaction © 2006 Brooks/Cole - Thomson 70 CALORIMETRY 71 Measuring Heats of Reaction Constant Volume “Bomb” Calorimeter • Burn combustible sample. • Measure heat evolved in a reaction. • Derive ∆E for reaction. © 2006 Brooks/Cole - Thomson 72 Calorimetry Some heat from reaction warms water qwater = (sp. ht.)(water mass)(∆T) Some heat from reaction warms “bomb” qbomb = (heat capacity, J/K)(∆T) Total heat evolved = qtotal = qwater + qbomb © 2006 Brooks/Cole - Thomson Measuring Heats of Reaction CALORIMETRY Calculate heat of combustion of octane. C8H18 + 25/2 O2 --> 8 CO2 + 9 H2O • Burn 1.00 g of octane • Temp rises from 25.00 to 33.20 oC • Calorimeter contains 1200 g water • Heat capacity of bomb = 837 J/K © 2006 Brooks/Cole - Thomson 73 Measuring Heats of Reaction CALORIMETRY Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/g•K)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ © 2006 Brooks/Cole - Thomson 74 Activities and Problem set 8 (due date_______) TextBook ch. 5 – all sections required for regents (in part), SAT II and AP exams POGIL activities on Thermochemistry and Calorimetry, Hess’s Law, Internal Energy and Enthalpy Vocab from Norton flash cards Lab activities: (specific heat capacity and heat of neutralization) Online practice quiz ch 5 due by_____ © 2006 Brooks/Cole - Thomson • 75 Ch 5 Problems TO DO: Print out questions, include your answers & show work • Study the sample exercises. • Complete the reading Guide and all associated exercises (RG follows your textbook) • Do Visualizing concepts problems • Do end of chapter 5 exercises in your e-text: 9,13, 18, 25, 28 , 29, 39, 40,42 , 43, 49, 51, 53, 54, 57, 60, 61, 63, 65, 67, 71 a,c, 74, 77, 85, 89, 90.