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Open system matter & heat can enter/exit

Open system matter & heat can enter/exit. Isolated system neither matter nor heat can enter/exit e.g. The Universe. surroundings. system. Closed heat but not matter can enter/exit. system. The 1 st Law of Thermodynamics The energy of the universe is constant

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Open system matter & heat can enter/exit

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  1. Open system matter & heat can enter/exit Isolated system neither matter nor heat can enter/exit e.g. The Universe surroundings system Closed heat but not matter can enter/exit

  2. system The 1st Law of Thermodynamics The energy of the universe is constant DEuniv = 0 Heat (q) Work (w) DEsys = q + w surroundings Heat (q) and work (w) are the means of transferring energy to/from a system.

  3. Specific Heat (see page 461) The amount of heat that must be added to a substance to raise the T by 1ºC. units = J g-1 ºC-1 or cal g-1 ºC-1 symbol = c. q = m • c • DT Alternate form called heat capacity: CP ( J mol-1ºC-1) The Hope Diamond weighs 9.10 g. How much heat would be required change the T From 20.0ºC to 100.ºC?

  4. thermometer Insulating lid Open system to add materials. System is closed during DT. Water 100˚C 51.190g Insulating walls allow little heat to pass A Calorimeter measures q. qCu = qH2O + qcal Cu = 0.092 cal g-1 K-1 H2O= 1.00 cal g-1 K-1 (150 g) Ccal = ????? Ti = 17.0 ˚C Tf = 19.2 ˚C 23.0 cal ºC-1 If the Hope diamond is boiled and placed in this same calorimeter, what is Tf?

  5. Whitten 8 – 15:63 - pg 597 5.1 g gold jewelry heated to 100.0 ºC placed in calorimeter (Ccal = 1.54 J/ºC) with 16.9 g water at 22.5 ºC. Tequil = 23.2 ºC. What is cmetal? If cAU = 0.129 J/g/ºC, is the jewelry pure gold? qh = m • c • DT = qc = m • c • DT + Ccal • DT 5.1 • c • (100.0 – 23.2) = 16.9 • 4.184 • (23.2 – 22.5) + 1.54 • (23.2 – 22.5) 391.7c = 50.6 c = 0.129 … consistent with pure gold??? Actually c = 0.1 since actual sig figs reduced to 1 by (Tf – Tc)

  6. Thermochemical Equations …….. CH4 + 2O2 CO2 + 2H2O + ??? energy heat What else does this reaction produce? Chemical potential E in bonds What form does this energy take? Where does this energy come from? Standard conditions: P = 1atm & T = 25˚C Enthalpy (H) = qP or DHrx (Internal Energy (E) = qV) CH4 + 2O2 CO2 + 2H2O + 890 kJ mol-1 or DH˚rx = -890 kJ mol-1 exothermic: heat is a product DH˚rx < 0 endothermic: heat is a reactant DH˚rx > 0 Thermochemical equations and stoichiometry How much heat is produced by the combustion of 47g of CH4?

  7. DHrx = qrx/mol A J mol-1 or kJ mol-1 qrx = qH2O + qcal A calorimeter can be used to determine the heat given off by a chemical reaction or DH˚rx. A + B  C + D + qP What happens to Twater if reaction is exothermic?

  8. Constant-Volume Calorimetry CH4 + 2O2 CO2 + 2H2O DH˚rx = -890 kJ mol-1

  9. CH4 + 2O2 CO2 + 2H2O DH˚rx = -890 kJ mol-1 Standard Heat of Formation: DHf DHrx for the formation of a compound from it’s elements as they would exist in their standard state ―1 atm & 25C (298K). Elements in standard states: (DHºf = 0) solid nonmetals: Cgr (graphite) – Si, P, S, I2 Gases H2, N2, O2, Cl2, F2, etc. Solid metals: most Liquid metals: Hg, Ga, Write the reactions representing the formation of …. CO2 – CH4– O2 – H2O Cgr CO2DHf = - 393.5 kJ mol-1 Cgr + 2H2 CH4DHf = - 74.85 kJ mol-1 H2 + ½O2 H2O(l)DHf = - 285.8 kJ mol-1

  10. Constant-Volume Calorimetry Cgr+ O2 CO2DH˚rx = -393.5 kJ mol-1 Cgr + ½O2 CO DH˚f = ? Hess’ Law The heat of a reaction is the sum of the heats of formation of all the products minus the heats of formation of all the reactants. or ….. DH˚rx = SiniDH˚f,i (products) - SiniDHºf.i (reactants) where ni is the stoiciometriccoefficient CO + ½O2→ CO2DHºrx = -283.0 kJ/mol CO + ½O2→ CO2 DHºrx = DHºf,CO2 - DHºf,CO2- ½DHºf,O2 = -283.0 kJ/mol

  11. Reactions for which DH˚ can be measured in calorimeter CO + ½O2 CO2DH˚rx = -283.0 kJ mol-1 C(gr) + O2 CO2DH˚f = -393.5 kJ mol-1 Since O2 is the standard state for oxygen then DH˚f = 0 This is true for any element in its standard state, e.g. C(gr) Cgr + ½O2 CO DH˚f = ??? kJ mol-1 Hess’ Law: DH˚rx (CO → CO2) = DH˚f (CO2) - DH˚f (CO) - ½DH˚f (O2) -283.0 = -393.5 - x - 0 x = -393.5 – 0 - (-283.0) = -110.5 kJ mol-1 see appendix K # 6.80– page 266

  12. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = ? kJ/mol Fuel Values of sugar (glucose) -1274.5 0 -393.5 -285.8 See pages A-8 to A-12 DH = -2801 kJ/mol Is this consistent with food label values? 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J

  13. AVG Bond Energy in kJ mol-1 C – C 346 C = C 602 H – H 436 C – Cgr 717 C = O 732 O – H 463 C – H 413 C = N 615 N – H 391 C – O 358 N = N 418 C – N 305 O = O 498 O – O 146 C – F 485 N – N 163 C – Cl 339 C ≡ C 835 Cl – Cl 242 C – Br 285 C ≡ N 887 F – F 155 C – S 272 N ≡ N 945 CH4 + 2O2 CO2 + 2H2O DH˚rx = -890 kJ mol-1 4 C-H + 2 O=O  2 C=O + 4 O-H 4 • 413 + 2 • 498 - 2 • 732 - 4 • 463 = - 668 kJ/mol DH˚rx ~ Sum of reactant bond energies – Sum of Product bond energies Not as accurate as Thermodynamic tables – because C-H bonds (etc.) not the same in all compounds

  14. A calorimeter contains 75.0g of water at 16.95 °C. A 93.3 g sample of iron (Fe) at 65.58 °C was placed in the cup. The final temperature was 19.68 °C. Calculate the heat capacity of the calorimeter, C. (include units) Data: cH2O = 4.184 J/g/°C cFe = 0.444 J/g/°C mFE • cFE • DT = mH2O • cH2O • DT + Ccal • DT 93.3 • 0.444 • (65.58-19.68) = 75.0 • 4.184 • (19.68-16.95) + Ccal • (19.68-16.95)

  15. a) For the reaction: 2HCl(g) + ½O2(g) → H2O(l) + Cl2(g) what is DH°rx ? DH°f HCl(g) -92.3 KJ/mol H2O(l) -285.8 KJ/mol If finished go on to do the following given ….. ½F2(g) + ½H2(g) → HF(g)DH°f = -600.0 KJ/mol H2(g) + ½O2(g) → H2O(l)DH°f = -285.8 KJ/mol b) For the reaction: 2HCl(g) + F2(g) → 2HF(g) + Cl2(g) what is DH°rx ?

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