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The 1 st Law of Thermodynamics The energy of the universe is constant D E univ = 0

system. The 1 st Law of Thermodynamics The energy of the universe is constant D E univ = 0. Heat (q). Work (w). D E sys = q + w. surroundings. Heat (q) is a form of energy: E = internal energy – measured at constant volume. H = enthalpy – energy measured at constant pressure.

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The 1 st Law of Thermodynamics The energy of the universe is constant D E univ = 0

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  1. 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) is a form of energy: E = internal energy – measured at constant volume. H = enthalpy – energy measured at constant pressure.

  2. Constant-Volume Calorimetry • CH4 + 2O2 CO2 + 2H2O + heat DH˚rx = -890 kJ mol-1 • ˚ means standard conditions: P = 1 atm • T = 298 K This reaction is? a) exothermic b) endothermic Exothermic: DH˚rx = - (heat is product) Endothermic: DH˚rx = + (heat is reactant) Enthalpy (H) is the measure of heat energy we will use

  3. 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). The standard state for oxygen is? a) O(l) b) O(g) c) O2(l) d) O2(g) The standard state for carbon is? a) graphite(Cgr)b) diamond (Cdia) c) bucky balls (C60)

  4. 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+ O2(g) CO2(g)DHf = - 393.5 kJ mol-1 Cgr + 2H2(g) CH4(g)DHf = - 74.85 kJ mol-1 H2(g) + ½O2(g) H2O(l)DHf = - 285.8 kJ mol-1 Which is the worst greenhouse gas? a) CO2 b) CH4 c) O2 d) H2

  5. Constant-Volume Calorimetry Cgr+ O2 CO2DH˚f= -393.5 kJ mol-1 Cgr + ½O2 CO DH˚f = ? could this reaction be done in calorimeter? a) yes b) no CO + ½O2→ CO2DHºrx = -283.0 kJ/mol 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 stoichiometric coefficient CO + ½O2→ CO2 DHºrx = DHºf,CO2 - DHºf,CO2- ½DHºf,O2 = -283.0 kJ/mol

  6. Hess’ Law DH˚rx = SiniDH˚f,i (products) - SiniDHºf.i (reactants) 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 Hess’ Law: DH˚rx (CO → CO2) = DH˚f (CO2) - DH˚f (CO) - ½DH˚f (O2) -283.0 = -393.5 - x - 0 Cgr + ½O2 CO DH˚f = ??? kJ mol-1 a) -676.5 kJ mol-1 b) – 110.5 kJ mol-1 c) + 110.5 kJ mol-1 x = -393.5 – 0 - (-283.0) = -110.5 kJ mol-1 see appendix K

  7. Hess’ Law DH˚rx = SiniDH˚f,i (products) - SiniDHºf.i (reactants)

  8. 2NO2↔ N2O4 DH°rx = ??? 2NO2↔ N2O4 DH°rx = 9.66 – 2(33.85) = -58.04 kJ/mol NO + ½O2↔ NO2 a) exothermic b) endothermic Which is more stable (less reactive)? a) Cgr and O2 b) CO c) CO2. Can you get extra energy by holding a piece of 2.0 g of Mg ribbon over a Bunsen burner? a) yes b) no

  9. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = ? kJ/mol Fuel Values of sugar e.g. glucose (FW = 180) -1274.5 0 -393.5 -285.8 See pages A-8 to A-12 DH = -2801 kJ/mol How many calories come from each gram of sugar? kJ/mol to Cal/mol to Cal/g a) ~ 1 b) ~2 c) ~4 d) ~9 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J = 4.184 kJ

  10. 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 Bond Energies Energy is given off (-) whenever a bond is formed The energy in a molecule is due to the energy stored in its bonds DH˚rx ~ Sum of reactant bond energies – Sum of Product bond energies Note the signs are opposite of Hess’ Law for heats of formation!

  11. 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 H2(g)+ ½O2(g) H2O(g) DH˚rx = -242 kJ mol-1 1H-H + ½O=O  2O-HDH˚rx = ??? 436 + 498/2 – 2(463) = -241 kJ/mol There is typically more error associated with bond energy calculations! 1H-H + ½O=O  2O-HDH˚rx = ???

  12. 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(g) + 2O2(g)  CO2(g) + 2H2O(g) DH˚rx = -802 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

  13. 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 ?

  14. Spontaneous – Merriam-Webster 1: proceeding from natural feeling or native tendency without external constraint 2: arising from a momentary impulse 3 : controlled and directed internally: SELF-ACTING 4: produced without being planted or without human labor: INDIGENOUS 5: developing or occurring without apparent external influence, force, cause, or treatment 6: not apparently contrived or manipulated Will Na2CO3 dissolve in water? Will Na2CO3 dissolve in hexane? Why? Will water flow downhill? Will it happen? Explain why ice melts when heated? (2 pts XC – by email by Friday)

  15. H2O (s) H2O (l)DH0 ~ +6.01 kJ/mol at 0°C Spontaneity Factors • Reduce Enthalpy (H) seek lowest energy • release heat - exothermic • form bonds - that are stronger than the ones broken • processes for which DH is (-) are ‘favored’. Observation — ice melts at 25°C — spontaneous Does melting ice (at 25°C) lower energy? Are the bonds in water of a different nature than those in ice? 1st Law of Thermodynamics The Energy of the Universe is constant. DEsys+ DEsurr= 0 or DEsys = q + w • Increase Entropy (S) • (make a system more disordered)

  16. 1st Law of Thermodynamics The Energy of the Universe is constant. DEsys+ DEsurr= 0 or DEsys = q + w • Increase Entropy (S) • (make a system more disordered) Entropy (S) is the thermodynamic function that indicates the degree of disorder inherent in a system. 2nd Law of Thermodynamics ― The Entropy of the Universe is increasing. DSsys + DSsurr > 0

  17. q in q out ice As ice  water the DSsys↑ but DSsurr↓. The 2nd Law says that DSsys + DSsurr > 0. The process is spontaneous if T is > Tnmp water As water  ice the DSsys↓ but DSsurr↑. The process is spontaneous, DSsys + DSsurr > 0, only if Tsurr < Tnmp, in which case the increased order of ice is compensated by decreased order of the surroundings.

  18. As you heat a substance its entropy value will … a) increase b) decrease c) stay same 3rd Law of Thermodynamics The Entropy of a pure substance is 0 at 0K. A substance will be perfectly ordered at 0K. As T↑, the substance will have increased motion which means it has increased disorder and S↑. Data regarding the heat capacity can be used to determine the entropy that a substance will have at standard T (298K). This allows values of standard entropies to be added to tables of standard enthalpies of formation.

  19. Which picture represents a higher entropy state… a) b) Which picture represents a higher entropy state… a) b)

  20. Enthalpy of Reaction DH˚rx = S n DH˚f,products – S n DH˚f,reactants Entropy of Reaction DS˚rx = S n S˚products – S n S˚reactants CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH˚rx = -890 kJ mol-1 DSrx will be ….. a) (+) b) (-) c) 0 ( - ) Because 3 moles gas → 1 mole gas and liquid is more ordered than gas

  21. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH˚rx = -890 kJ mol-1 DS˚rx = ??? DS˚rx = 214 + 2(70) – 186 – 2(205) = -242 J/mol(note units) products are more ordered since DS = (-) more moles of gas = more disorder for reaction Which is more ordered? a) graphite b) diamond Which is more ordered? a) CO b) CO2

  22. # 6.80– page 266 H2(g) → 2H(g)DH°rx = 436.4 kJ/mol Br2(g) → 2Br(g)DH°rx = 192.5 kJ/mol H2(g) + Br2(g) → 2HBr(g)DH°rx = -72.4 kJ/mol H(g) + Br(g) → HBr(g)DH°rx = ??? kJ/mol Does the reaction (in red) above represent DH°f for HBr? a) yes b) no What is DH°ffor HBr(g)? a) -72.4 kJ/mol b) -36.2 kJ/mol c) + 72.4 kJ/mol Is DH°rxfor 2H(g) → H2(g)? a) exothermic b) endothermic DH°rxfor 2H(g) → H2(g)= -436.4 kJ/mol a) -282.4 kJ/mol b) -350.7 kJ/mol c) + 464.3 kJ/mol H(g) + Br(g) → HBr(g)DH°rx = -36.2 – (436.4/2) – 192.5/2 = -350.7 kJ/mol

  23. Spontaneity Factors • Reduce Enthalpy (H) • (release heat) • (form bonds that are stronger than the ones broken) • (processes for which DH is (-) are ‘favored’.) DG = DH - TDS • Increase Entropy (S) • (make a system more disordered) Problem – stronger bonds usually make molecules more ordered!So H & S often ‘oppose’ each other. Free Energy (G) – Chemical ‘potential’ energy The indicator of spontaneity for a process Any process will only proceed if G decreases. Therefore: DG < 0 for any spontaneous process

  24. DG = DH - TDS Temperature Trends for chemical and physical processes ↑T favors disorder As T↑ then entropy change becomes a greater factor in determining spontaneity. Phase changes solid  liquid  gas Weaker bonding with more disorder favored as T↑. In solid bond formation wins out over disorder. In gas, disorder wins out over bond formation.

  25. DG = DH - TDS CH4(g) + 2O2(g) CO2(g) + 2H2O(l) DH˚rx = -890 kJ mol-1 DS˚rx = -0.242 kJ mol-1 DG˚rx = DG˚rx(using DG = DH – TDS) = -890 – 298 (-0.242) = -818 kJ mol-1 DG˚rx(using table values – rounded) = 2(-237) + (-394.4) – (-50.8) – 2(0) = -817 difference due to rounding errors Entropy favors reactants (more disordered) so as T↑ DG˚rx will go down since DS˚rxis negative.

  26. DG˚= DH˚- TDS˚ 2NO2(g) N2O4(g) DH˚rx = (9.16) – 2(33.18) = -57.2 kJ mol-1 DS˚rx = 0.304 - 2(.240) = -0.176 kJ mol-1 DG˚rx = -57.2 - 298 (-0.176) = -4.75 kJ mol-1 or… DG˚rx = 97.89 - 2 (51.31) = -4.73 kJ mol-1

  27. DG˚ = DH˚ - TDS˚ DG = DH - TDS DG (not DG˚) is the indicator of spontaneity for a chemical process. Will a chemical reaction or physical change occur? If DG < 0 (-) yes The [products] will increase and [reactants] will decrease If DG = 0 the system is at equilibrium There will be no change in [reactants] or [products] If DG > 0 (+) no The reaction will proceed in the opposite direction as written The [reactants] will increase and [products] will decrease

  28. DG = DH - TDS • DG˚rxn is DGrxn when all [ ]s of reactants/products = 1M. • (usually this is not the equilibrium state) • DG˚rxnis a constant (but changes with T) • DGrxnvaries with the amounts of reactants and products such that ….. DG = DG˚ + RT ln Q When equilibrium is reached Q = a) 1 b) 0 c) Keq d) DG˚ When Q = 1, DG = a) 1 b) 0 c) Keq d) DG˚ When equilibrium is reached DG = a) 1 b) 0 c) Keq d) DG˚ Therefore… DG˚ = - RT lnKeq

  29. DGº = DHº - TDSº DG = DG˚ + RT ln Q At equilibrium DG = 0 = DG˚ + RT lnK and …. DG˚ = -RT ln K

  30. DG˚= DH˚- TDS˚ 2NO2(g) N2O4(g) DH˚rx = (9.16) – 2(33.18) = -57.2 kJ mol-1 DS˚rx = 0.304 - 2(.240) = -0.176 kJ mol-1 K-1 DG˚rx = -57.2 - 298 (-0.176) = -4.75 kJ mol-1 DG˚ = -RT lnKeq DG˚ = -RT lnKeq = -4.75 = -0.008314 • 298 • lnKeq Keq = exp(1.917) = 6.80 A large – value of DG˚ indicates a large value of K And indicates the reaction favors products

  31. 2NO2(g) N2O4(g) DG˚rx= -4.75 kJ mol-1 DG˚ = -RT lnKeq Keq = 6.80 DG = DG˚ + RT ln Q What is DG if you have 5 atm N2O4 and 0.5 atm NO2? Q = a) 1 b) 5/(0.5) c) 5/0.52 d) 0.52/5 Q = 5/0.52 = 20 18.28 DG = -4.75 + 0.008314 • 298 ln20 = +2.67 18.25

  32. 18.28 H2(g) + CO2(g) → H2O(g) + CO(g)KP = 4.40 at 2000K DG˚ = -RT lnKeq a) Calculate DG° at 2000K. DG° = a) - b) + c) 0 b) Calculate DG if …. PH2 = 0.25 atm, PCO2 = 0.78 atm, PH2O = 0.66 atm, PCO = 1.20 atm Q = 0.66 • 1.20 = 4.06 0.25 • 0.78 DG = DG˚ + RT lnQ -24.6 + 0.008314 • 2000 • ln 4.06 This reaction is going … a) forward b) backward

  33. 18.25 Fe(OH)2(s)↔ Fe2+(aq) + 2OH-(aq) KSP= 1.6 x 10-14 = [Fe2+] • [OH-]2 DG˚ = -RT ln KSP DG˚ = -0.008314 • 298 ln (1.6 x 10-14) = 78.7 kJ/mol Note that if K < 1 then DG˚ is (+) and the smaller K the more + DG˚ gets. If K > 1 then DG˚ is (-).

  34. 18.29 CaCO3(s) ↔ CaO(s) + CO2(g)DH°rx= 177.8 kJ mol-1 DS°rx= 160.5 J mol-1 K-1 DG°298 = DH°rx- TDS°rx = 177.8 – 298 (0.1605) = 130.0 kJ mol-1 DG°298= -RT ln K K = PCO2 K = e(-DG°/RT) = 1.6 x 10-23 and PCO2= 1.6 x 10-23atm DG°1073= DH°rx- TDS°rx = 177.8 – 1073 (0.1605) = 5.6 kJ mol-1 K = e(-DG°/RT) = 0.534 and PCO2 = 0.534 atm The higher T favors more disordered products If T > 1107K then DG° becomes negative

  35. CaCO3(s) ↔ CaO(s) + CO2(g) DG°1073= DH°rx- TDS°rx = 177.8 – 1073 (0.1605) = 5.6 kJ mol-1 K = e(-DG°/RT) = 0.534 and PCO2 = 0.534 atm The higher T favors more disordered products If T > 1107K then DG° becomes negative DG = DG °1073 + RT lnQ Which direction will reaction proceed at 1073K if 0.30 atm of CO2 is present withCaCO3(s)and CaO(s)? a) forward b) reverse DG = 5.6 + 0.008314 • 1073 • ln0.30 = -5.14 kJ/mol Even though DG° is + the reaction proceeds since CO2 < 0.534 atm. Reaction would go backward if PCO2 = 1.0 atm or > 0.534 atm.

  36. H2O H2O If P < 0.032 then … DG = (-) & H2O evaporates If P > 0.032 then … DG = (+) & H2O condenses Pequil = 0.032 atm 24 torr DGº = +8.56 KJ/mol DG = DG˚ + RT ln Q H2O(l) H2O(g)QP = PH2O

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