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# Chapter 6

Chapter 6. Thermochemistry: Energy Flow and Chemical Change. If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time. Review of Physics. Télécharger la présentation ## Chapter 6

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1. Chapter6 Thermochemistry: Energy Flow and Chemical Change If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

2. Review of Physics Velocity and acceleration: v = d/t, a = Dv/t Force and work: force is what is required to overcome inertia or to change the velocity of an object; Newton = kg*m/s2 Work = force acting through distance, w = f*d in Joules or kg*m2/s2 Energy is capacity to do work or produce heat, unit is Joules Potential (PE) and nonpotential energy (nPE) Heat: energy transferred as result of temperature difference; transfer always from warmer body to colder body

3. LEARNING OBJECTIVES: 1. Distinguish between heat & work; temperature, thermal energy & heat 2. Relate q, DT, cp, thru heat transfer equations q = m * cp * DT and q = DHphase * amount 3. Calculate DHrxn from calorimeter data 4. Understand how DHfo values are established for compounds 5. Calculate DHrxn from either heats of formation or Hess' law 6. Understand all the symbols: DHrxnoDHfo DHvapoDHfusoDHcombo

4. Figure 6.1 from 4th ed, not in Prin the system A chemical system and its surroundings. the surroundings The system is the chemical reaction being studied. The surroundings are everything else in the universe!

5. Definitions Thermodynamics is the study of heat and its transformations Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes Fundamental premise When energy is transferred from one object to another, it appears as work and/or as heat For our work we must define a system to study; everything else then becomes the surroundings The system is composed of particles with their own internal energies (E or U). Therefore the system has an internal energy. When a change occurs, the internal energy changes

6. MORE ON ENERGY First law of thermodynamics or the Law of Conservation of Energy: In an isolated system energy can't be created or destroyed, so the total energy can't change Energy is in two major forms - nonpotential (often kinetic) and potential Energy can be transferred between the two forms, but the total, nPE + PE, has to remain the same

7. ENERGY CONTINUED: 1. POTENTIAL ENERGY - relative to position (stored) 2. TYPES OF NONPOTENTIAL ENERGY: KINETIC ENERGY - MOVING OBJECT = ½mv2 HEAT - CHANGING TEMPERATURE – FROM ENERGY TRANSER Thermal energy from the motion of atoms or molecules in s, l or g Heat involves the transfer of energy between two objects due to a temperature difference OTHER FORMS OF nPE: Radiant energy from the sun Electrical energy generated by mechanical devices or by redox reactions

8. ENERGY CONTINUED: Chemical reactions either release or absorb energy as: heat - thermal energy** light - radiant energy electrical current - electrical energy

9. What does high thermal energy mean? One relative measure of thermal energy is temperature, but temperature does not equal thermal energy Compare a cup of coffee at 102°F to a tub of water at 102°F Which contains more thermal energy? WHY?

10. MORE THERMOCHEMISTRY: SYSTEM = the substance being evaluated for energy content in a thermodynamic process SURROUNDINGS = everything outside the system in the process ENDOTHERMIC PROCESS = a thermodynamic process in which energy transfers into a system from its surroundings EXOTHERMIC PROCESS = from system to surroundings ENTHALPY CHANGE = energy transferred at a constant pressure, DH

11. "A bit of physics again" There are 2 ways to transfer energy between system and surroundings: work or heat Work = action of a force that causes displacement within the system in a measurable dimension as “-DX” if linear or as “DV” if volume Work = -F*Dh Since Area*Dh = DV and Pressure = F/Area Then w = - (F/Area) * Area*Dh = - PDV for work done at constant pressure

12. Figure 6.5 Pressure-volume work. When the volume of a system increases by an amount DV against an external pressure, P, they system pushes back and does PV work on the surr (w=-PDV).

13. Heat Transfer and the Meaning of Enthalpy Change, DH As previously discussed, if no force is involved, then energy flows spontaneously as heat from regions of higher T to regions of lower T. Most rxns take place at constant pressure. We also know DErxn = Ef - Ei = q + w, and that w = -PDV. At constant P, DErxn = q + w DErxn = qP - PDV At constant P, DHrxn = DErxn + PDV substituting: DHrxn = (qp - PDV) + PDV DHrxn = qp Therefore, Enthalpy of Rxn = Heat transfer DHrxn = qp

14. Heat Transfer and the Meaning of Enthalpy Change, DH If DH < 0 rxn is exothermic, if >0 rxn is endothermic For chemical reactions: Hf = final enthalpy of products Hi = initial enthalpy of reactants DHrxn = Hproducts - Hreactants

15. Figure 6.1 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. When internal energy of a system decreases, DE is lost to the surroundings and is negative. When internal energy of a system increases, DE is gained by the surroundings and is positive. DE = Efinal - Einitial = Eproducts - Ereactants

16. A system transferring energy as heat only. Figure 6.2 Ice water gains energy as heat, q, from surr until Tsys = Tsurr; DE>0 so q is pos. Hot water transfers energy as heat, q, to the surr until Tsys=Tsurr; DE<0 so q is neg. See notes in box below or in textbook.

17. Energy, E work done on surroundings DE<0 A system losing energy as work only. Figure 6.3 Zn(s) + 2H+(aq) + 2Cl-(aq) H2(g) + Zn2+(aq) + 2Cl-(aq) The internal energy of the system decreases as the reactants form products because the H2(g) does work, w, on the surr by pushing back the piston. The vessel is insulated, so q = 0. Here DE<0 and sign of w is positive.

18. Table 6.1 The Sign Conventions* for q, w and DE (from perspective of the system) + = DE q w + + + + - depends on sizes of q and w - + depends on sizes of q and w - - - * For q: + means system gains heat; - means system loses heat. * For w: + means work done on system; - means work done by system.

19. DEuniverse = DEsystem + DEsurroundings Units of Energy Joule (J) 1 J = 1 kg*m2/s2 Calorie (cal) = energy to heat exactly 1 g H2O by exactly 1.00°C at 14.5°C 1 cal = 4.184 J 1 kcal = 1000 cal = 1 Cal (nutritional) (British Thermal Unit 1 Btu = 1055 J)

20. Figure 6.5 in 4th ed. only Some interesting quantities of energy.

21. PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (DE) in J, kJ, and kcal. kcal kJ -776J -0.776kJ 4.184kJ 103J Sample Problem 6.1 Determining the Change in Internal Energy of a System PLAN: Define system and surroundings, assign signs to q and w and calculate DE. The answer should be converted from J to kJ and then to kcal. SOLUTION: q = - 325 J w = - 451 J DE = q + w = -325 J + (-451 J) = -776 J = -0.776kJ = -0.185 kcal

22. Two different paths for the energy change of a system. Figure 6.4 The change in internal energy when a given amount of octane burns in air is the same no matter how the energy is transferred. On left, fuel is burned in open can and energy is lost almost entirely as heat. On right, fuel is burned in car engine so a portion of energy is lost as work to move the car and less energy is lost as heat.

23. Figure 6.5 Pressure-volume work. When the volume of a system increases by an amount DV against an external pressure, P, they system pushes back and does PV work on the surr (w=-PDV).

24. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g) Enthalpy, H Enthalpy, H heat out heat in Figure 6.6 Enthalpy diagrams for exothermic and endothermic processes. CH4 + 2O2 H2O(g) Hfinal Hinitial DH < 0 DH > 0 CO2 + 2H2O H2O(l) Hfinal Hinitial A Exothermic process B Endothermic process

25. PROBLEM: In each of the following cases, determine the sign of DH, state whether the reaction is exothermic or endothermic, and draw an enthalpy diagram. These are thermochemical equations! (a) H2(g) + 1/2O2(g) H2O(l) + 285.8 kJ (DH = -285.8 kJ) (b) 40.7 kJ + H2O(l) H2O(g) (or DH = +40.7 kJ H2(g) + 1/2O2(g) (reactants) H2O(g) (products) H2O(l) (products) H2O(l) (reactants) Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of DH PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction SOLUTION: (a) The reaction is exothermic. (b) The process is endothermic. EXOTHERMIC DH = -285.8kJ ENDOTHERMIC DH = +40.7kJ

26. Phase Changes: Latent Heat THREE MAJOR EXPRESSIONS FOR CHANGE IN STATE: DHfus for change from solid to liquid (= -DHfreeze from liquid to solid) DHvap for change from liquid to gas (= -DHcond from gas to liquid) DHsubl for change from solid to gas (= -DHdep from gas to solid) There is no temperature change, so we call it a latent heat transfer: q = DHphase * quantity

27. Enthalpy of Phase Change Substance DHfusDHvap Benzene, C6H6 126 J/g 395 J/g Bromine, Br2 67.8 J/g 187 J/g Ethanol, CH3CH2OH 104 J/g 854 J/g Mercury, Hg 11.6 J/g 296 J/g Sodium Chloride 517 J/g 3100 J/g Water 333 J/g 2257 J/g

28. Practice heat transfer 1. For 237 mL of ice at 0.00oC find the heat required to melt it and then the heat required to warm it to 25.0oC. Finally find the total of the two heat transfers. Dice is 0.917 g/mL. qphase = DHfus * quantity = 333 J/g * 237 mL * 0.917g/mL = 72,371 J qwarm = m*cp*DT = 237 mL * 0.917 g/mL *4.184J/goC*(25.0-0.00)oC = 22,730 J qtotal = 9.51x104 J

29. Practice heat transfer now you do it! 2. Find the total heat transfer energy required to heat 1.00 mole of ice from below the freezing point at -10.0oC to above the boiling point as steam at 110.0oC. (That includes two phase changes!) Given this data: cp(ice) = 37.7 J/moloC; cp(water) = 75.3 J/moloC; cp(steam) = 36.4 J/moloC; DHfus = 333 J/g; DHvap = 2257 J/g. (Total q ~ 55000 J)

30. TO BEGIN TO UNDERSTAND CALORIMETRY: Heat capacity, cp, of a substance = quantity of heat that can be transferred through the substance per quantity/per change in temperature Cp = quantity of heat transferred DT * gram or mole Specific heat capacity is the heat energy required to warm up 1.00 g of a substance by 1.00°C Ex: Water’s specific heat is 4.184 J/g-degree Glass’ specific heat is 0.80 J/g-degree

31. TO BEGIN TO UNDERSTAND CALORIMETRY: Specific heat capacity Cp is a physical property of a substance Cp can also be expressed as molar heat capacity by multiplying by molar mass For example, for water: 4.184 J/g-oC * 18.015 g/mol =75.37 J/mol-oC

32. Thermal energy Thermal energy transferred is called sensible heat transfer, q, uses specific heat capacity in this heat transfer equation: q = m*cp*(Tf - Ti) The direction of heat transfer is determined by which is warmer body: heat flows from hot to cold Heat has to be transferred from something to something. The warmer body loses heat; the colder body gains heat. qlost = -qgained

33. Thermal energy example You drink 1 cup of coffee, 250. mL, at 60.0°C & your body is at 37.0°C. Assume 0.997 g/mL density, and water’s cp can be used for both coffee and your body with 60.0 kg mass. Heat is transferred from the coffee to your body. What is your body’s final temperature? q(coffee)= 249.25 g * 4.184 J/g-degree (Tf-60.0) q(body)= 60.0 kg (103 g/kg) 4.184 J/g-degree (Tf – 37.0) Set q(coffee) = -q(body) 249.25 g * 4.184 J/g-degree (Tf-60.0) = -[60.0 kg (103 g/kg) 4.184 J/g-degree (Tf – 37.0) Solve for Tf Tf = 37.1°C in sig figs

34. Substance Specific Heat Capacity (J/g*K) Substance Specific Heat Capacity (J/g*K) Elements Materials aluminum, Al 0.900 wood 1.76 graphite,C 0.711 cement 0.88 iron, Fe 0.450 glass 0.84 copper, Cu 0.387 granite 0.79 gold, Au 0.129 steel 0.45 Compounds water, H2O(l) 4.184 ethanol, C2H5OH(l) 2.46 ethylene glycol, (CH2OH)2(l) 2.42 carbon tetrachloride, CCl4(l) 0.864 Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials

35. Figure 6.7 Coffee-cup calorimeter. Still used in school labs to measure heat at constant pressure, qp. Surroundings actually include water, cups, thermometer, and stirrer.

36. PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and DHrxn (in kJ/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g*K) Sample Problem 6.4 Determining the Heat of a Reaction PLAN: Find mass of solution: m = 75.0 mL * 1.00 g/mL = 75.0 g Find DT = Tf – Ti = 27.21oC – 25.00oC = 2.21oC qsol’n = m*cp*DT = 75.0 g * 4.184 J/g-deg * 2.21oC = 693 J

37. HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + OH-(aq) H2O(l) 0.500 M x 0.0500 L = 0.0250 mol OH- For NaOH For HCl 0.500 M x 0.0250 L = 0.0125 mol H+ Sample Problem 6.4 Determining the Heat of a Reaction continued SOLUTION: HCl is the limiting reactant. 0.0125 mol of H2O will form during the rxn. Find DHrxn = - qsol’n = -693 J (opposite of q from previous slide) (-693 J/0.0125 mol H2O)(kJ/103 J) = -55.4 kJ/ mol H2O formed

38. CALORIMETRY: The bomb calorimeter is used for combustible substances that are solid or liquid SYSTEM = SUBSTANCE + OXYGEN DHcalor = qbomb + qwater SURROUNDINGS = BOMB + WATER DHrxn = - DHcalor There is no DV, therefore qrxn = -(qbomb + qwater) Note: bomb and water have different heat capacities even if change in temperature is the same. Must know or determine heat capacity of calorimeter

39. Figure 6.8 A bomb calorimeter

40. Combustion of octane in bomb calorimeter: 1. Balanced eq'n*: C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g) Given this data: 1.00 g of octane, 1.20 kg water, initial T of bomb + water is 25.00°C final T is 33.20°C Heat capacity of water is 4.184 J/g-deg Heat capacity of calorimeter is determined to be 837 J/deg 2. Calculations qbomb = Cbomb * DT = 837 J/deg * (33.20 - 25.00) = +6.86 x 103 J qwater = Cw * m * DT = 4.184 J/g-deg * 1200 g * (33.20 - 25.00) = +41.2 x 103 J total q = -(qbomb + qw) = -(6.86 + 41.2) kJ = -48.1 kJ/gram octane 3. Find molar DH: -48.1 kJ/gram * 114.2 g/mol = -5490 kJ/mol Note: in thermochemistry the substance of interest has a coefficient of one, and other coefficients can be fractions!

41. Practice Work on problems 23, 25 and 32. 23: Calc q when 0.10 g ice is cooled from 10.0oC to -75.0oC, given cp of ice is 2.087 J/g-degree. 25: A 27.7 g sample of ethylene glycol loses 688 J of heat. Given Tf is 32.5oC and cp of 2.42 J/g-degree, determine Ti. 32: When 25.0 mL of 0.500 M H2SO4 is added to 25.0 mL of 1.00M KOH in a calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate DH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.)

42. AMOUNT (mol) of compound A AMOUNT (mol) of compound B HEAT (kJ) gained or lost Figure 6.9 Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction. molar ratio from balanced equation DHrxn (kJ/mol)

43. DH is function of quantity of reactants based on mass or moles Burning 441 g propane in backyard grill, how much heat is transferred to surroundings, including meat, air, grill, etc.? 1. Always have a balanced chemical equation: C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) 2. Look up DHcomb : -2220 kJ/mol propane 3. How many moles consumed? 441 g/44.09 g/mol = 10.0 mol 4. How much heat? -2220 kJ/mol * 10.0 mol = -22,200 kJ

44. Writing a Thermochemical Equation and an example A thermochemical equation is a balanced chemical equation that includes the enthalpy of reaction.

45. Intro to Hess’ Law Given the combustion reaction: 2 H2(g) + O2(g) 2 H2O(g)DHrxn= -484 kJ If we cut in half the reaction in half to make the formation reaction of one mol of H2O: H2(g) + 1/2 O2(g)  H2O(g) Then we have to cut the value of the first reaction in half: DHrxn = - 242 kJ/mol H2O = DHf If we do some multiple of a known reaction we have to apply the same multiplier to the DHrxn

46. Intro to Hess’ Law Given the vaporization of liquid water to steam: H2O(l) H2O(g)DHvap = 44 kJ/mol Then what is DHcond ? H2O(g) H2O(l) Reverse of vaporization is condensation: DHcond = -44 kJ/mol = -DHvap If a reaction is reversed the sign of DH is changed: DHrevrxn = - DHfwdrxn

47. Intro to Hess’ Law We can use the formation of water and the condensation of steam data to determine for the combustion reaction if the product is liquid water. Rxn 1: H2(g) + 1/2 O2(g)  H2O(g) DHrxn = - 242 kJ/mol Rxn 2: H2O(g) H2O(l)DHcond = -44 kJ/mol Add the two reactions to obtain this reaction: H2(g) + ½ O2(g) H2O(l) DHrxn = -242kJ + -44 kJ = -286 kJ/mol H2O(l)

48. Intro to Hess’ Law HESS' LAW OF CONSTANT HEAT SUMMATION: If a reaction is the sum of 2 or more other reactions, then DH for overall process must be the sum of DH of constituent reactions

49. Hess’ Law Example Find DHrxn for 2 SO2(g) + O2(g)2 SO3(g), given the info for the following two rxns: SO2(g) S(s) + O2(g)DHrxn = +297 kJ 2 S(s) + 3 O2(g) 2 SO3(g)DHrxn = -791 kJ Double rxn 1 and add to rxn 2, cancel common species. Sum of DH's is -197kJ For practice see problems 45-48!

50. Enthalpy of Formation: DHo is standard quantity: 1 atm pressure, 1.000 M, 298.15 K DHfo is standard enthalpy of formation of compound, ion, etc., at standard conditions from the elements in their standard states (recall diatomics, etc.) Elements by definition have DHfo = 0.0

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