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1. UNIT 4: THERMO 1

2. UNIT 4: THERMODYNAMICS I Text: Chapter 6 Thermodynamics is Thermochemistry is TERMS: “the system” = object being studied (or focused on) “the surroundings” = everything else in the universe the study of energy and its interconversions the study of energy changes associated with physical & chemical changes

3. H2O DEMO: NaOH(s) -------> Na+(aq) + OH-(aq) pellets dissolved in water heat released by “the system” (NaOH and the water) heat absorbed by “the surroundings” (glass beaker, your hand, air in room) “the system” released energy into “the surroundings” = • [energy lost by system = energy gained by surroundings]

4. What IS energy? that which is necessary SI unit for ENERGY is the JOULE. The English unit is the CALORIE (1 cal = 4.184J) ENERGY is NOT measured directlylike mass or volume. We can onlymeasure ________________ in terms of ___________ (or the capacity) to do work “w” or to produce heat “q” changes in energy work and heat

5. ENERGY: 1) potential = “stored” energy due to ________ or ____________ 2) kinetic = energy of “motion” KEparticle = Forms: 9.8m/s2 gravity position PE = m•g•h composition : stored in bonds between particles mv2 2 • mechanical • nuclear • electrical • thermal(heat) • electromagnetic(light) • sound(sonar) • chemical

6. random particle motion in matter THERMAL(HEAT) ENERGY reflects the and is affected by: 1) TEMPERATURE: as T _____, heat energy _____ 2) QUANTITY of the substance: as MASS _____, heat energy ____ Heat involves due to (from -----> ) ↑ ↑ ↑ ↑ the transfer of energy between two objects temperature differences warmer object cooler object

7. PHYSICAL CHANGE: ex: warm hand melts ice • H2O(s) ---> H2O(l)endothermic for the ice (“the system”) • CHEMICAL CHANGE: • atoms (or ions) “rearranging” to make products • breaking bonds _______ energy (_____thermic) • 2) making bonds ________ energy (____thermic) endo absorbs releases exo

8. 2 COMBUSTION of methane is EXOTHERMIC because energy needed to energy needed to break bondsmake bonds (absorbed) (released) CH4(g) + O2(g) -------> CO2(g) + H2O(g) CH4(g) + O2(g) -----> CO2(g) + H2O(g) + energy(heat & light) 2 - - - - - - - - - - - - - - - - higher PE due to weaker bonds (more reactive, less “stable”) • PE heat released 2 - - - - - - - - - - - - - lower PE due to stronger bonds (less reactive, more “stable”) 2 2 higher PE = lower PE +energy

9. SYNTHESIS of nitrogen monoxide is ENDOTHERMIC because energy needed to energy needed to break bondsmake bonds (absorbed) (released) NO(g) N2(g) + O2(g) ----------> energy + N2(g) + O2(g) -----> NO(g) • 2 • - - - - - - - - - - - - - • higher PE, weaker bonds • (less “stable”) • PE • heat absorbed • - - - - - - - - - - - - - - - - - • lower PE, stronger bonds • (more “stable”) 2 • energy+lower PE =higher PE

10. 1st Law of Thermodynamics: the energy of the universe is constant  The E (internal energy) is defined as Esystem = PEsystem + KEsystem Ecan be changed by a “flow” of heat(q) or work(w) or both!! DE = q + w “heating water” will have a positive q (adding heat to the system) “condensing a vapor” will have a negative q (subtracting heat from the system) the sum of the kinetic & potential energies of all particles in the system. +q -q • ∙∙ • ∙ • ∙ • ∙ ∙ • ∙ ∙ • ∙ • ∙

11. We focus on “the system”. heat added to subtracted from • +q • -q • system added to subtracted from • -w • +w work • surroundings

12. 1st Law of Thermodynamics: really means energy is neither created nor destroyed in ordinary chemical and physical changes. Thermodynamic “STATE” of a system is defined as a set of conditions which includes Temperature Pressure Volume Physical state (solid, liquid, gas) Composition (identity of substances and # of moles)

13. These are considered properties of a system called “state functions” [heat (q) & work (w) are NOT state functions, but part of the pathway] The value of “state function” depends only on STATE of that system not on HOW it got to that state (pathway)!! A change in “STATE” describes the difference between 2 states (independent of the pathway) “FINAL” - “INITIAL” = change in state function Ex: DT = Tfinal – Tinitial or DV = Vfinal– Vinitial

14. E(internal energy) includes all energywithin a substance: •KE of particles •attractive forces “between” p+s and e-s (ionization energies) •intermolecular forces between atoms, molecules or ions, etc E = (Efinal – Einitial) = (Eproducts – Ereactants) = q + w chemical change

15. The ONLY type of work involved in most chemical & physical changes is pressure-volume work. (work = force x distance) P = force = f V = (L x W x H) so V = d3 aread2 P x V = f x d3= f x d = work d2 Work done “on” or “by” a SYSTEM depends on the external pressure and the volume. DE = q + w DE = q + -PDV d d d

16. A. When gas expands-PDV = -P(V2>V1) = -w B. When gas contracts-PDV = -P(V2<V1) = +w C. In constant volume reactions, no PDV is done (because nothing moves through a distance) Solids & liquids do NOTexpand/contract significantly with pressure so DV~0

17. Ex: 2NH4NO3(s) ------> 2N2(g) + 4H2O(g) + O2(g) solid ALL gases Dn = nproduct gases - nreactant gases Dn = so DV is ____ w = -PDV so expansion means -w !! (work comingout of system) 7mol of gases 0 mol of gas = 7 mol– 0 mol +7 mol (+) = -P(+)

18. Ex: 2SO2(g) + O2(g) ------> 2SO3(g) w = -PDV • 3 molgases • 2molgases Dn= 2 - 3 • = -1 • = -P(-) • so contractionmeans +w!! • (work beingdone on system)

19. Ex: H2(g) + Cl2(g) ------> 2HCl(g) • 2 mol gases • 2 mol gases • Dn = 2 - 2 w = -PDV = 0 = -P(0) • so constant volumemeans w = 0 • (no work being done)

20. ENTHALPY, H “heat content” (from Day 2) • DE = q + w and therefore • DE = q+ -PDV • +PDV +PDV then • DH = DE + PDVsince then • DH = (q - PDV) + PDV therefore • DH = qp • Denthalpyequals heat gained or lost@ “constant pressure” • “change of heat for a reaction” or “enthalpy change”: • DHreaction = Hproducts - Hreactants

21. Ex: When KOH(s) is dissolved in water, heat is released : KOH(s) -----> K+(aq) + OH-(aq) DH = -43kJ/mol(exo!) Problem: How much heat is released when 14.0g of KOH is dissolved in water? Enthalpyis an “extensive property” (depends on amount of substance present). • H2O (means: for every 1 mol KOH dissolved, 43 kJ of heat will be released) • “direction • of heat flow” K- 39.10 O- 16.00 H – 1.01 56.11g/mol -11kJ • 1mol_ • 56.11g x-43kJ= 1mol 14.0g x -10.7289.. Ans: 11kJ released

22. For the reverse reaction, the enthalpy is opposite in signbut equal in magnitude. Ex: CH4(g)+ 2O2(g)---> CO2(g)+2H2O(g)DH = -890 kJ CO2(g)+2H2O(g)--->CH4(g)+ 2O2(g)DH = +890 kJ (exo) (endo)

23. A “thermochemical equation” expresses the energy change, DH, for a reaction “as written”: • Ex: the combustion of ethanol: • C2H5OH(l)+ 3O2(g)----> 2CO2(g)+3H2O(g)DH = -1367 kJ • So DH = -1367 kJ= -1367 kJ= -1367 kJ= -1367 kJ • 1molC2H5OH 3molO22molCO2 3molH2O • How much heat is released when 275.0g CO2 are produced? • C- 12.01 • O- 2(16.00) • 44.01g/mol • 1mol_ • 44.01g 275.0g x • x-1367kJ= • 2molCO2 -4271kJ -4270.90.. 4271kJ released

24. (lesser H) - (greater H) So DH = (-) exo heat coming “out of”or being “subtracted from” the reactants

25. (greater H) - (lesser H) • So DH = (+) endo • heat going “into” or being • “added to” the reactants

26. 1 g/mL spec. heat: 4.18J goC DT = Tfinal – Tinitial initial final = 85.0oC - 75.0oC = 10.0oC • (4.18J) • (goC) = (500.g) (10.0oC) 20900 q = 20900J = 20.9kJ absorbed

27. DT = (78.2J) = (45.6g) (c) (13.3oC) (c) = • (78.2J) 0.12894 (45.6g) (13.3oC) c = 0.129J/goC

28. M x VL = mol (1.0M)(0.050L)= 0.050molHCl DHneut=kJ/mol = 2.7kJ 0.050mol need total mass of solution100.g DHneut= -54kJ/mol exo • (4.18J) • (goC) = (100.g) 6.5oC 2717 6.5oC= DT • q = 2700J = 2.7kJ

29. DT = Tfinal – Tinitial Tf=x (x – 80.0oC) (x - 55.0oC) (4.18J) (goC) -(50.0g) (4.18J) (goC) = (20.0g) – – -50.0x = 20.0x + 4000oC) - 1100oC) +50.0x +50.0x + 1100oC) + 1100oC) = 70.0x 5100oC 72.857 Tfinal= 72.9oC

30. (x – 100.0oC) (x – 23.0oC) (0.444J) (goC) -(15.0g) (4.18J) (goC) = (55.0g) (x – 100.0oC) (x – 23.0oC) = 230.J oC -6.66J oC = 230.x -6.66x + 666oC – 5290oC +6.66x + 5290oC + 5290oC) +6.66x = 237x 5956oC 25.130 Tfinal= 25.1oC

31. cm3 q = mcDT • q = DHvapm need J/g  425oC 3 2 356.6oC - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1 need g Hg:  25oC -38.8oC - - - - - - - - - 125cm3 x 13.5g = cm3 1687.5 1690g

32. 25oC(l)-->356.6oC(l) • q = m c DT  • (331.6oC) • = (1690g) (0.139J) (goC) • 356.6oC • -25.000…oC • 77,896.15 q = 77,900J • DT = 331.6oC 356.6oC(l)-->356.6oC(g) • q = DHvapm • (1690g) 59.1kJ mol x _1mol_ 200.59g x1000J 1kJ = 295J g • 498,550 = 295J/g q = 499,000J 294.63

33. 356.6oC(g)-->425oC(g) • q = m c DT • 425oC • -356.6oC • = (1690g) • (0.102J) • (goC) • (68oC) • 68.4 q = 12,000J • 11,721.8 • DT = 68oC • 77,900J • 499,000J • +12,000J 588,900J = 588.9kJ absorbed

34. We can only measure the “change in enthalpy” USING HEATS of REACTION: HESS’S LAW statesthatthe enthalpy change for a reaction is the samewhether it occurs in one stepor a series of steps. (Law of Heat Summation) one step reactants----------------------> products energy difference ---> ---> ---> ---> will besame! series of steps DHreaction = Hproducts - Hreactants

35. Example: oxidation of nitrogen to nitrogen dioxide: One step:N2(g)+ 2O2(g)-----> 2NO2(g)DH = 68kJ (endo)! Two steps: N2(g)+ O2(g)-----> 2NO(g)DH = 180kJ (endo)! 2NO(g)+ O2(g) -----> 2NO2(g)DH = -112kJ (exo)! cancel common terms, ___________________________________ __________________ then add N2(g) + 2O2(g)-----> 2NO2(g)DH = 68kJ (endo)!

36. Rules for calculating enthalpy changes: Take the reactions given, rearrange them and combine them so that they add up to the net reactionyou are asked to find. 1) if an equation is reversed, then the sign on DH is reversed (bec. heat flow is opposite) 2) the magnitude of DH is directly proportional to the moles of reactants and products in the equation. If the coefficients in the equation are multiplied by an integer, then DHis multiplied by the same integer. Thus DHreaction= DH1+ DH2 + DH3…….

37. flip 2 2 DH = -34kJ SiH4(g) ---> Si(s) + 2H2(g) • Si(s) + O2(g)--->SiO2(s) DH = -911kJ DH = -484kJ • 2H2(g) + O2(g)--->2H2O(g) • SiH4(g) + 2O2(g)---> SiO2(s) + 2H2O(g) DH = -1429kJ

38. 2 • 2 • flip 2 • a. 2S(s)+ 3O2(g) --->2SO3(g) • DH = -790.4kJ • b. 2SO3(g) ---> O2(g) + 2SO2(g) • DH = +198.2kJ • 2S(s)+ 2O2(g) --->2SO2(g) • DH = -592.2kJ 2 2 • DH = -296.1kJ • S(s)+ O2(g) --->SO2(g)

39. STANDARD ENTHALPY OF FORMATION The standard enthalpy of formation DHofof a compound is defined as the change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their “standard states”. (also called standard molar enthalpy of formation or just heat of formation) The degree symbol on a thermodynamic function, as in DHofindicates that the corresponding process has been carried out under “standard conditions” (not STP!) The THERMOCHEMICAL“STANDARD STATE” of a substance is its most stable state at standard pressure (1atm) and at room temp (25oC, 298K) unless otherwise specified.

40. DHof refers to “reactants in their standard states” ---> “products in their standard states” STANDARD STATE: For a Compound: 1. the standard state of a gaseous compound is a pressure of exactly 1atm. 2. the standard state of a liquid or solid compound is the pure liquid or solid. 3. the standard state of a solution is a concentration of exactly 1 Molar. For an Element: the standard state is the form in which the element exists at 1atm and 25oC. ex: O2(g)Br2(l)Fe(s) Note: the enthalpy of formation DHof for any element in its standard state is zero!!!

41. The enthalpy change for a given reaction can be calculated by subtracting the enthalpies of formation of the reactants from the enthalpies of formation of the products. DHoreaction= SnpDHof(products) - SnrDHof(reactants) S (sigma) means “sum of” np means moles of products nrmeans moles of reactants Elements are not included in the calculation because elements require no change in form.

42. USING THE TABLE FOR “HEAT OF FORMATIONS”DHof Problem #1: Calculate the DHrxn 2NH3(g) + 3O2(g) + 2CH4(g) ------> 2HCN(g) + 6H2O(g) DHof: ____kJ/mol___kJ/mol_____kJ/mol_____kJ/mol_____kJ/mol 135.1 -75 -46 0 -242 DHorxn = products - reactants 2mol(135.1kJ) mol • +6mol(-242kJ) • mol 2mol(-46kJ) + mol 3mol(0kJ) + mol 2mol(-75kJ) mol 270.2kJ -92kJ • = + -1452kJ + -150kJ -242kJ -1181.8kJ -939.8 • DHorxn = -940.kJ

43. Problem #2: If DHrxn = 85kJ, find DHoffor NaHCO3 2NaHCO3(s) -------> Na2CO3(s) + CO2(g) + H2O(l) DHof: _?__kJ/mol_______kJ/mol_____kJ/mol_____kJ/mol baking soda • -394 • -286 • -1131 1mol cancels mol • DHorxn= [products] – [reactants] • (-286kJ) • (-394kJ)+ • (-1131kJ) + 85kJ = 2(x) 85kJ = • (-1811kJ) - • 2x -85kJ • +2x -85kJ • +2x • 2x = -1896kJ • x = -948kJ • DHof= -948kJ/mol

44. C + E --> C C + C--> C E + E --> C only! 2 ways to find DHrxn for this synthesis 1) DHrxn= 2mol (-826kJ) = -1652kJ (mol) 2) DHrxn = SnprodDHof-SnreactDHof - + 0kJ (mol) = 0kJ (mol) 2mol(-826kJ) (mol)

45. synthesis D DHorxn = +1652kJ 2Fe2O3(s) ---> 4Fe(s) + 3O2(g) g-->mol 25.0gFe x 1mol x 1652kJ = 55.85g 4molFe 185kJ absorbed 184.87