 Download Presentation Ppt19, Thermochemistry

# Ppt19, Thermochemistry

Download Presentation ## 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. 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

3. NOTE: T is not the same as “heat”! • Things that are “hot” have a high T (relative concept) • 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 • Confusing because we “sense” heat flow but brain interprets it as “temperature”: • An object feels hot if heat transfers into our skin! • An object feels cold if heat energy transfers out! • Heat is a type of energy; T is not an “energy” Ppt19

4. 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 each ms 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

5. 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)

6. Quick “Quiz”(PS8a, Q1) • Answer: FALSE. It “takes” (absorbs) energy to break any chemical bond! No exceptions!! • Breaking a bond is like “pulling apart two magnets” or “lifting a book”!! • PE of system increases; energy (either in the form of heat or work) comes from outside of system (surroundings) True or false (correct if false): (i) When a chemical bond is broken, energy is released. Ppt19

7. Quick “Quiz”(PS8a, Q1) Answer: FALSE. Some chemical reactions absorb energy (called “endothermic”), and some reactions release energy (“exothermic”) • If only bond breaking occurs: endoCl2 2 Cl • If only bond making occurs: exoCl + Cl Cl-Cl • If some bonds are broken and some (other ones) are made, it could be endo or exo (it depends). True or false (correct if false): (ii) When a chemical reaction takes place, energy is released. Ppt19

8. 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

9. Go to PS8a first Ppt19

10. 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(minus  “opposite of”, not “negative!) • the amount of E that leaves the system equals the amount of E that enters the surroundings [and vice versa] • Energy can change forms • E.g., from KE to PE or vice versa Ppt19

11. What is Enthalpy? (Better question: what is the change in enthalpy?) • At constant P, the difference between DE and DH is the amount of work (w) that is done on (or by) the system: • DH = DE – w (P const.) • For typical chemical processes, w << DE, and DH  DE • Since DE = q + w, q = DE – w …. = DH! (IF P is constant) • More about “work” later • 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 important. Its change is. Ppt19

12. 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! Also Q5 on PS8b • 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. • When a process occurs in the system at constant pressure, the change in H equals the amount of heat transferred: Ppt19

13. 1st Law Applied, and convention for q • 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 • 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 = -qsurr  Key equation (&concept) Ppt19

14. DHsys often represents a conversion of PE into KE or KE into PE (PE in sys; KE into or from surroundings) • DHsys = qsys AND qsys = -qsurr •  DHsys = -qsurr  Key equation • 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 (warming up the surroundings)! For a chemical or physical process at constant P (and T),DHsys↔ DPEsys Ppt19

15. Figure 6.2 (Zumdahl): Exothermic Process DHsys = -qsurr Ppt19 15

16. Endothermic Processes Generally convert KE of surroundings into PE in system DHsys = -qsurr • If endothermic (DH > 0), the rearrangement in system 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

17. 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

18. 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

19. 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

20. Ppt19

21. Ppt19

22. Helpful Question—Is a chemical or physical process taking place, or “just” heat flow? • Thus, qsys = -qsurr reduces to: • Csys x msys x DTsys= -(Csurr x msurr x DTsurr) • (no chemical or physical change in system) • 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 = C x m x DTin both the sys and surr Ppt19

23. 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

24. Helpful Question—continued • Thus, qsys = -qsurr reduces to: • DHsys= -(Csurr x msurr x DTsurr) • (IS a chemical or physical change in system) 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 Ppt19

25. 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

26. 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

27. 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

28. 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? Q6 on PS8b (written) uses “Stoichiometry with energy” ideas/skills. Try it (and more Mastering Q’s) now! Ppt19

29. 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

30. Example—Calorimetry (case with a physical or chemical change) Qs 8 and 9 on PS8b (written) use similar ideas/skills. Try them (and more Mastering problems) now! #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. dwater ~ 1.0 g/mL Ppt19

31. 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

32. Figure 6.9 Ppt19

33. Reminder (DE vs DH)(Slide 11, recopied here) • At constant P, the difference between DE and DH is the amount of work (w) done on (or by) the system: • DH = DE – w (P const.) • For typical chemical processes, w << DE, and DH  DE • Since DE = q + w, q = DE – w …. = DH! (IF P is constant) • More about “work” later • 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 important. Its change is. Ppt19

34. 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

35. 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

36. 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

37. Example Q9 on PS8b (written) uses similar ideas/skills. Try it (and more Mastering Qs) now! #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

38. 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

39. 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 ; DH1 = 95.4 kJ H2+ N2H4  2 NH3 ; DH2 = -187.6 kJ Answer: The sum of these two, because the sum of these two equals the overall process! Ppt19

40. 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 ; DH1 = 95.4 kJ H2+ N2H4  2 NH3 ; DH2 = -187.6 kJ Answer: The sum of these two, because the sum of these two equals the overall process! Ppt19

41. Energy Diagram View2-step Synthesis of NH3 DHoverall= DH1 + DH2 = 95.4 kJ +(-187.6 kJ) = -92.2 kJ Ppt19

42. Hard Part? Getting the “steps” • It’s usually not so easy to see how to add up the given equations to get the target equation! • We don’t generally “give you” the equations in a “ready to sum” way. • You have to create that series of steps before you can sum. Ppt19

43. Example (from handout) Given the data below, calculate the DH for: Target equation (my terminology) ClF(g) + F2(g) ClF3(g) (1) 2 ClF(g) + O2(g) Cl2O(g) + F2O(g)DH1 = 167.4 kJ Given equations (my terminology) (2) 2 ClF3(g) + 2 O2(g) Cl2O(g) + 3 F2O(g) DH2 = 341.4 kJ (3) 2 F2(g) + O2(g) 2 F2O(g) DH3 = -43.4 kJ NOTE: If you add up the three equations as is, they will not sum up to the target equation! → You will need to figure out how to get a set of equations that will sum up. Then you can sum up the (new) DH’s get DHtarget. Ppt19

44. Generalized Procedurefor Creating a set of equations that sum to the equation of interest (“target”) I will also demonstrate this in a moment. • In a nutshell: • Use substances in the target equation, one by one (skipping some as needed) to locate a certain “given” equation. • Manipulate that given equation by doing one or both of the following, as needed: • (you may) Reverse the equation and/or • Multiply the equation through by a number • After this “step-by-step”, you’ll have your “set” See handout that describes this in detail Ppt19

45. Generalized Procedurefor 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 number x, the DH becomes ____ times the original value. opposite x When you manipulate an equation, you’ll also manipulate its DH! Ppt19

46. First step Given the data below, calculate the DH for: ClF(g)+ F2(g) ClF3(g) (1) 2 ClF(g) + O2(g) Cl2O(g) + F2O(g)DH1 = 167.4 kJ (1) 2 ClF(g)+ O2(g) Cl2O(g) + F2O(g)DH1 = 167.4 kJ (2) 2 ClF3(g) + 2 O2(g) Cl2O(g) + 3 F2O(g) DH2 = 341.4 kJ (3) 2 F2(g) + O2(g) 2 F2O(g) DH3 = -43.4 kJ • Consider the first substance in the target equation, and look for that substance in one of the given equations. • If the substance is present in more than one of the given equations, then ignore that subtance and start this procedure over with the next substance in the target equation. Not the case with ClF, so we’re good. Ppt19

47. Second step Given the data below, calculate the DH for: x ½ ClF(g)+ F2(g) ClF3(g) (1) 2 ClF(g) + O2(g) Cl2O(g) + F2O(g)DH1 = 167.4 kJ (1) 2 ClF(g)+ O2(g) Cl2O(g) + F2O(g)DH1 = 167.4 kJ (2) 2 ClF3(g) + 2 O2(g) Cl2O(g) + 3 F2O(g) DH2 = 341.4 kJ x ½ (3) 2 F2(g) + O2(g) 2 F2O(g) DH3 = -43.4 kJ (1’) 1ClF(g)+ ½ O2(g)½ Cl2O(g) + ½ F2O(g)DH1’ = ½ (167.4) kJ 2) If the coefficient for the substance of interest in the given equation is not the same as its coefficient in the target equation, multiply the given equation (i.e., all coefficients) by whatever number is needed to get the number to match the coefficient in the target equation. Here, multiply Eq. (1) by ½ to make the coefficient of ClF match Ppt19

48. Third and Fourth steps Given the data below, calculate the DH for: ClF(g)+ F2(g) ClF3(g) (1) 2 ClF(g)+ O2(g) Cl2O(g) + F2O(g)DH1 = 167.4 kJ (2) 2 ClF3(g) + 2 O2(g) Cl2O(g) + 3 F2O(g) DH2 = 341.4 kJ (3) 2 F2(g) + O2(g) 2 F2O(g) DH3 = -43.4 kJ (1’) 1ClF(g)+ ½ O2(g)½ Cl2O(g) + ½ F2O(g)DH1’ = ½ (167.4) kJ “modified given equation” 3) If the substance of interest is on the opposite side in the given equation as the side in which it appears in the target equation, reverse the given equation (put what’s on the right side on the left and vice versa). Not applicable here. Already on same side. Done with Steps 2) and 3) Ppt19

49. Fifth step Given the data below, calculate the DH for: ClF(g)+ F2(g) ClF3(g) (1) 2 ClF(g)+ O2(g) Cl2O(g) + F2O(g)DH1 = 167.4 kJ (2) 2 ClF3(g) + 2 O2(g) Cl2O(g) + 3 F2O(g) DH2 = 341.4 kJ (3) 2 F2(g) + O2(g) 2 F2O(g) DH3 = -43.4 kJ (1’) 1ClF(g)+ ½ O2(g)½ Cl2O(g) + ½ F2O(g)DH1’ = ½ (167.4) kJ “modified given equation” 5) Cross out the given equation that you just used in Steps 2 and 3. You should not use this equation again in this procedure! Only consider the “unused” or “remaining” given equations from here on out! Ppt19

50. Sixth step Given the data below, calculate the DH for: x ½ ClF(g) + F2(g) ClF3(g) (1) 2 ClF(g)+ O2(g) Cl2O(g) + F2O(g)DH1 = 167.4 kJ (2) 2 ClF3(g) + 2 O2(g) Cl2O(g) + 3 F2O(g) DH2 = 341.4 kJ (3) 2 F2(g)+ O2(g) 2 F2O(g) DH3 = -43.4 kJ (3) 2 F2(g) + O2(g) 2 F2O(g) DH3 = -43.4 kJ x ½ (1’) 1 ClF(g) + ½ O2(g) ½ Cl2O(g) + ½ F2O(g)DH1’ = ½ (167.4) kJ (3’) 1F2(g)+ ½ O2(g)1 F2O(g) DH3’= ½ (-43.4) kJ 6) Look at the next substance in the target equation and do the same as in (1) - (4) above, being careful not to reuse any given equations (i.e., use only “remaining” given equations; not crossed out ones). Ppt19