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Ppt19, Thermochemistry

Ppt19, Thermochemistry. Basic Ideas and Definitions KE(review), heat, temperature, potential energy, thermal energy, enthalpy, etc. Calorimetry—Obtaining energy changes by measuring T changes. Thermochemical Equations: Stoichiometry with Energy! Hess’s Law (and related Ideas)

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Ppt19, Thermochemistry

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  1. Ppt19, Thermochemistry • Basic Ideas and Definitions • KE(review), heat, temperature, potential energy, thermal energy, enthalpy, etc. • Calorimetry—Obtaining energy changes by measuring T changes. • Thermochemical Equations: Stoichiometry with Energy! • Hess’s Law (and related Ideas) • Using energy changes of known reactions to calculate energy changes of related ones • Standard Enthalpies of Formation • Using tabulated values to calculate energy changes

  2. Quick “Quiz” PS8a, Q1! True or false (correct if false): (i) When a chemical bond is broken, energy is released. (ii) When a chemical reaction takes place, energy is released. Ppt19

  3. Thermodynamics is the study of energy changes • Two basic KINDS of energy: • Kinetic (KE): energy of motion (of a particle) • Recall: For a sample of particles: • KEavg(per particle)  TKelvin • KEavg(x # particles) is called “thermal energy” • T is NOT an energy, but it is proportional to one kind of energy (thermal energy or avg. KE of particles) Thermo = “heat” (a type of energy) Dynamic = “motion” => “changes” Ppt19

  4. NOTE: T is not the same as “heat”! (relative concept) • Things that are “hot” have a high temperature • They have a high average KE per particle • They do not “have” a lot of “heat” in them • Heat is energy that transfers from a hotter sample to a colder one • An object feels hot to us if heat energy transfers into our skin when we touch it! • An object feels cold to us if heat energy transfers out of our skin when we touch it! • We’ll define heat in a moment, but for now, please note that: • Heat is a type of energy; T is not an “energy” Ppt19

  5. Follow up: Difference between T and heat • They are the same temperature!! • The belt buckle feels hotter to you because it conducts heat well, so the amount of heat that transfers into your skin in a given fraction of a second is much greater than the amount of heat that transfers in from the cloth! • You go to your car on a hot summer day after the car has been sitting out for several hours. • Which is hotter, the metal belt buckle or the cloth seat? Ppt19

  6. Potential Energy (PE): The Second Type of Energy • Chemical potential energy results from forces between atoms or molecules • It takes energy to pull bonded atoms apart • It takes energy to pull molecules in a liquid apart (to turn into a gas) • It takes energy to pull an electron away from a nucleus • When physical or chemical changes take place, positions of atoms or molecules change relative to one another  PE changes! • PE is energy of position • Results from forces (e.g., book in gravitational field)

  7. Dx means “change inx” • Dx = xf – xi “final minus initial” • If T goes from 35 to 45 ºC, then: • DT = 45 – 35 = +10. ºC • Did T increase or decrease? • It increased • A positive delta means an increase in the variable • If T goes from 45 to 35 ºC, then: • DT = 35 – 45 = -10. ºC • Did T increase or decrease? • It decreased • A negative delta means a decrease in the variable Ppt19

  8. Go to PS8a first Ppt19

  9. 1st Law: Energy is neither created nor destroyed. Euniverse is a constant • Energy can transfer from one “place” to another • Define a SYSTEM and a SURROUNDINGS • Universe = system + surroundings (See Board) • DEuniv= 0  DEsys + DEsurr= 0 •  DEsys = -DEsurr(the amount of energy that leaves the system equals the amount of energy that enters the surroundings [and vice versa]) • Energy can change forms • E.g., from KE to PE or vice versa Ppt19

  10. What is Enthalpy? (Better question: what is the change in enthalpy?) • Heat (q): energy that transfers from a hotter sample of matter to a cooler one (reminder) • There is no “heat” IN a sample of matter • A sample of matter “has” KE (of its particles) and PE associated with the relative positions of its particles • Internal Energy (Esys) = sum of all KE & PE of particles • Enthalpy (H) is a property of a system whose change is similar to the change in E for typical chemical processes • The formal definition of H is not really important. Its change is. • The difference between DE and DH is the amount of work (w) that is done on the system (or by the system) • DH = DE - w • For typical chemical processes, w << DE, and DH  DE • More about “work” later Ppt19

  11. The change in Hsys(DHsys) equals the amount of heat (flow) if a process occurs at constant P • DHsys=qP(subscript “p” means “at const. P”) • If DHsys > 0, heat flows INTO the system (to make H increase) •  “ENDOTHERMIC” • If DHsys < 0, heat flows OUT of the system (to make H decrease) •  “EXOTHERMIC” PS8a, Q5! • Note: if P is NOT constant during some process, the amount of heat (flow) is NOT equal to the change in H (DHsys)! “H” is “enthalpy”, NOT heat! The change in H just happens to equalqwhen a process is carried out at constant P. • The way that H is defined, it turns out that: When a process occurs in the system at constant pressure, the change in H equals the amount of heat transferred: Ppt19

  12. 1st Law Applied, and convention for q • qsys = -qsurr  Key equation (&concept) • E.g.: If 10 J flows from sys to surr: qsys = -10 J and qsurr = +10 J • E.g.: If 20 J flows into sys from surr: qsys = +20 J and qsurr = -20 J • qsys < 0 means heat flowed OUT of the system (and into the surroundings) & qsurr > 0 • qsys > 0 means heat flowed INTO the system (and out of the surroundings) & qsurr < 0 Ppt19

  13. DHsys often represents a conversion of PE into KE (often as heat transfer with surroundings) • For a chemical or physical process at constant P and T, DHsys↔ DPEsys • Thus, if DHsys < 0 (exothermic), chemical PE in the system ends up getting converted into KE in the surroundings, and the energy transfer occurs as heat: • DHsys = qsys AND qsys = -qsurr •  DHsys = -qsurr  Key equation Ppt19

  14. Figure 6.2 (Zumdahl): Exothermic Process DHsys = -qsurr Ppt19 14

  15. Same equation applies for endothermic processes • For an endothermic process: DHsys = -qsurr • If endothermic (DH > 0), then the rearrangement requires energy to occur, and that energy flows in from the surroundings (qsurr < 0) • [imagine the REVERSE of the process on prior slide] • The “-” sign means “opposite of”, not “negative”!!! Ppt19

  16. DHsys is not determined by the surroundings—it is “assessed” by it! • NOTE: The fact that the value of DH equals - qsurr should not be interpreted to mean that the value is determined by the surroundings—it is not!! • The value of DHsys is determined by the rearrangement (changes in position of atoms / molecules / ions) in the system. • i.e., it is determined by the PROCESS in the system • The surroundings is just a “reporter” of sorts Ppt19

  17. Reminder (PS 8a, Q3) “It takes energy to raise a substance’s T” How much energy? • If the only thing that happens to a substance (A) is that it changes T, then: • qA = Cs, A x mA x DTA • Cs is the specific heat (capacity) of a substance • amount of heat energy needed to raise 1 g of a substance by 1 C • A large(r) Cs means “hard(er) to change its T” • (Other abbreviations: s, S.H., c) Ppt19

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  20. II. Calorimetry A property of the surroundings; can be determined if Csurr is known via: qsurr = Cs, surr x msurr x DTsurr The “surroundings” is usually “reduced” to a calorimeter, a liquid, a solid, etc. (Assume no heat is lost to the “rest” of the surroundings) • Used to obtain changes in enthalpy (DHsys’s) by measuring (changes in) temperature of the “surroundings” (DTsurr’s). • Use: DHsys = -qsurr Ppt19

  21. Helpful Question—Is a chemical or physical process taking place, or “just” heat flow? • If there is NO process in the system (or surroundings): • Heat flow is a result of different initial T’s in sys & surr • T of both sys and surr changes because of heat flow • q is related to DT by: • q = s x m x DTin both the sys and surr • Thus, qsys = -qsurr reduces to: Csys x msys x DTsys = -(Csurr x msurr x DTsurr) (no chemical or physical change in system) Ppt19

  22. Example—Calorimetry(case with heat transfer only—no phys or chm change) #1 on Handout Sheet: If a 40.1 g piece of iron at 652 °C is dropped into a sample of 328 g of water at 32.4 °C, what will be the final temperature after thermal equilibrium is established? Assume that no heat is lost during the process. CFe = 0.45 J/(gC) Ppt19

  23. Helpful Question—continued NOTE: qsys is NOT equal to Csys x msys x DTsyshere!! The process dictates DHsys—surroundings responds to energy change in system • If there IS a process in the system (but not in the surroundings) and Tsys is kept constant (common): • q flow is ultimately caused by the process(see next slide) • BecausePEchange in the system (DHsys) converted to KE • qsurr is related to DT • Thus, qsys = -qsurr reduces to: DHsys = -(Csurr x msurr x DTsurr) (IS a chemical or physical change in system) Ppt19

  24. III. Short but important interlude—Meaning of Thermochemical Equations • The amount of DH associated with a process depends on the amount of the process that occurs • 2) The DH for a chemical equation is not the same as the DHsys associated with an actual chemical reaction. • Just like the coefficient in a chemical equation is not the same as the amount of moles of a substance that actually “reacts” or “forms” during an actual chemical reaction! • 3) “Stoichiometry with energy” idea Before we work with the calorimetry eqn on the prior slide, recall ideas from PS8a (Q2 & next slide): Ppt19

  25. Follow up from PS8a Q2 (diff rxn eqn) 2) If 8 mol C is to react: # of moles of CaO needed? Amt of energy absorbed? CaO(s) + 3 C(s)  CaC2(s) + CO(g); DH = 465 kJ • If 5 mol CO is formed: How many moles of C react? What is the DH of the rxn? Ppt19

  26. Stoichiometry with energy! (Example) Reminder: If there’s a “process”, q flow is ultimately caused by that process, with the amount being dependent on how much process occurs) #2 on Handout Sheet: How much heat (in kJ) is evolved or absorbed in the reaction of 233.0 g of carbon with enough CaO to produce calcium carbide? CaO(s) + 3 C(s)  CaC2(s) + CO(g); DH = 464.8 kJ (b) Is the process exothermic or endothermic? Ppt19

  27. Another Example # 3 on Handout Sheet: 85.8 kJ of energy is evolved (i.e., released) at constant pressure when 3.56 g of P4 is burned according to: P4(s) + 5 O2(g) → P4O10(s) What is the ΔH for the (thermo)chemical equation? Ppt19

  28. Return to Calorimetry DHsys is “caused” by the process in the system (“stoichiometry with energy”), but we can determine its value experimentally in a particular situation by measuring the T change of the surroundings. • Recall: qsys = -qsurr • If there IS a process in the system (but not in the surroundings), This reduces to DHsys = -(Csurr x msurr x DTsurr) (chemical or physical change in system) Ppt19

  29. Example—Calorimetry (case with a physical or chemical change) #4 on Handout Sheet: Instant cold packs contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the temperature because of the endothermic process: NH4NO3(s)  NH4NO3(aq); DH = +25.7 kJ What is the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 125 mL of water? Assume the specific heat capacity of the dissolved NH4NO3 is negligible compared to water, an initial temperature of 25.0 C, and no heat transfer between the cold pack and the environment. (Recall, dwater ~ 1.0 g/mL) Ppt19

  30. Exp 14 Part B • Dissolve some solid in some water • In an insulated cup • System is the solid (plus the small amount of water molecules that interact with the dissolved FUs of the solid) • Process that occurs in sys is “dissolution” • Assume (excess) H2O is the surroundings • Hsys = -qsurr becomes: DHdiss = -qwater • qwater = Cwater x mwater x DTwater(assume Twater=Tsol’n) Ppt19

  31. Figure 6.9 Ppt19

  32. Heat and Enthalpy (Slide 10 recopied here) • Heat (q): energy that transfers from a hotter sample of matter to a cooler one • There is no “heat” IN a sample of matter • A sample of matter “has” KE (of its particles) and PE associated with the relative positions of its particles • Internal Energy (Esys) = sum of all KE & PE of particles • Enthalpy (H) is a property of a system whose change is similar to the change in E for typical chemical processes • The formal definition of H is not really important. Its change is. • The difference between DE and DH is the amount of work (w) that is done on the system (or by the system) • DH = DE - w • For typical chemical processes, w << DE, and DH  DE • More about “work” later Ppt19

  33. Return to Internal Energy (E)—Heat and Work Matter here. • In equation form: DEsys = q + w (chemists) • Of course, if heat flows out of the system, or if the system does work, Esys decreases • There are two ways to increase the (internal) energy of a system: • Have heat (q) flow into it (qsys > 0) • Have the surroundings do work (w) on it (w > 0; chemists’ convention) Ppt19

  34. Return to Internal Energy (E)—Heat and Work Matter here. DEsys = q + w becomes • DEsys = DHsys + w and thus • DHsys = DEsys - w (as noted earlier) • If work is small, DH is approximately equal to DE Recall that qp = DH, so at constant P Ppt19

  35. Figure 6.7 If the system expands against an external pressure (i.e., piston moves upward), DVsys is positive and “w” is negative (system does work on surroundings). w = -PDV Ppt19

  36. Example #5 on Handout Sheet: Assume that a particular reaction produces 244 kJof heat and that 35 kJ of PV (expansion/contraction) work is done on the system. What are the values of DE and DH for the system? For the surroundings? Ppt19

  37. IV. Hess’s Law Value of H (or E) for a system depends ONLY on the state of the system (i.e., the P, T, moles of substances, states of substances, etc.) It doesn’t matter how you got to that state Called a “state function” The change in H in going from State 1 to State 2 does not depend on how you get there (i.e., “path”).  “Hess’s Law”: DHoverall= DH1 + DH2 + DH3 + etc. (1,2,3 are processes that “add up” to the overall) Ppt19

  38. Example 1 3 H2+ N2  2 NH3 ; DH = ??? What is the DH for the above equation if we know the following? 2 H2+ N2  N2H4 ; DH = 95.4 kJ H2+ N2H4  2 NH3 ; DH = -187.6 kJ Answer: The sum of these two, because the sum of these two equals the overall process! Ppt19

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  40. Generalized “Procedure”for Creating a set of equations that sum to the equation of interest (“target”) • Apply the following ideas (PS8a, Q2!): • If you reverse an equation, the sign of DH becomes the __________. • If you multiply an equation through by a whole number x, the DH becomes ____ times the original value. opposite x See handout sheet examples and boardwork Ppt19

  41. “Special” Example Find DH for CH4(g) + 2 O2(g)→ CO2(g) + 2 H2O(l)given: C(s) + 2 H2(g) → CH4(g); DH1 = -74.8 kJ O2(g) → O2(g); DH2 = kJ C(s) + O2(g) → CO2(g); DH3 = -393.5 kJ H2(g) + ½ O2(g) → H2O(l); DH4 = -285.8 kJ b/c nothing changed! ? 0 Ppt19

  42. “Special” Example Find DH for CH4(g) + 2 O2(g)→ CO2(g) + 2 H2O(l) CH4(g) → C(s) + 2 H2(g) ; DH1’ = - DH1 2 O2(g) → 2 O2(g); DH2’ = -2DH2 ------------------------------------------------------------------------------------------------------------------ C(s) + O2(g) → CO2(g); DH3’ = DH3 2 H2(g) + O2(g) → 2 H2O(l); DH4’ = 2DH4 Elements! Ppt19

  43. “Special” Example Find DH for CH4(g) + 2 O2(g)→ CO2(g) + 2 H2O(l)given: C(s) + 2 H2(g) → CH4(g); DH1 = O2(g) → O2(g); DH2 = C(s) + O2(g) → CO2(g); DH3 = H2(g) + ½ O2(g) → H2O(l); DH4 = Ppt19

  44. Q: Is there an “easy” way to calculate the DHeqn without doing calorimetry? • Yes and no • Calorimetry does have to be done, but it can be done ahead of time and not on the reaction of interest • Obtain an effective value of “H” for one mole of every substance (via calorimetry) • Effective “H” is actually called the “standard enthalpy of formation” of a substance X: DH°f(X) • Use these “H” values of reactants and products (substances) to calculate DHrxn for any chemical equation! Ppt19

  45. How to calculate DH°eqn(from tabulated data) DH°eqn= “H” of all products” - “H” of all reactants “Hfinal” “Hinitial” = [cDH°f(C) + dDH°f(D)] - [aDH°f(A) + bDH°f(B)] From Tro: For an equation: a A + b B  c C + d D; DH°eqn Ppt19

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  47. CH3Cl(g) -86.3 Find DH for : CH4(g) + Cl2(g)→ CH3Cl(g) + HCl(g) Find DH for : CH4(g) + 2 O2(g)→ CO2(g) + 2 H2O(l)

  48. How to calculate DH°eqn EXAMPLE Find DH for : CH4(g) + 2 O2(g)→ CO2(g) + 2 H2O(l) Ppt19

  49. Can we pick a “generalizable” path for ANY reaction? Yes! Always go through ELEMENTS Determine DH for making a substance from its ELEMENTS (called DHformation) Tabulate these “DHf’s” for ALL SUBSTANCES Use them to calculate DHrxn for any chemical equation! Ppt19

  50. NOTES • The DH°fof a substance is the: • Enthalpy change associated with forming one mole of a substance from its elements • As such, the value for any element is zero* • The DH°ffor a substance depends on the physical state of the substance in question (because it takes or releases energy to change a substance’s state) * In its standard state Ppt19

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