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Chemistry 102(060) Summer 2014

Chemistry 102(060) Summer 2014. Instructor: Dr. Upali Siriwardane e-mail : upali@latech.edu Office : CTH 311 Phone 257-4941 Office Hours : M,Tu , W,Th,F 9:00-11:00 am ; Test Dates : 9:30-10:45 am., DAVH 209. July 21 , 2014 (Test 1): Chapter 13

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Chemistry 102(060) Summer 2014

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  1. Chemistry 102(060) Summer 2014 Instructor: Dr. UpaliSiriwardane e-mail: upali@latech.edu Office: CTH 311 Phone257-4941 Office Hours: M,Tu, W,Th,F 9:00-11:00 am ; Test Dates:9:30-10:45 am., DAVH 209 July 21, 2014 (Test 1): Chapter 13 July 28, 2014 (Test 2): Chapter 14 August 5, 2014 (Test 3): Chapter 15 &16 August 13, 2014 (Test 4) Chapter 17 August 14, 2014 (Make-up test) comprehensive: Chapters 13-17

  2. Chapter 6. Thermochemistry 6.1 Chemical Hand Warmers 231 6.2 The Nature of Energy: Key Definitions 232 6.3 The First Law of Thermodynamics: There Is No Free Lunch 234 6.4 Quantifying Heat and Work 240 6.5 Measuring for Chemical Reactions: Constant-Volume Calorimetry246 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure 249 6.7 Constant-Pressure Calorimetry: Measuring 253 6.8 Relationships Involving 255 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation 257 6.1 0 Energy Use and the Environment 263

  3. Chapter 17. Free Energy and Thermodynamics 17.1 Nature’s Heat Tax: You Can’t Win and You Can’t Break Even 769 17.2 Spontaneous and Nonspontaneous Processes 771 17.3 Entropy and the Second Law of Thermodynamics 773 17.4 Heat Transfer and Changes in the Entropy of the Surroundings 780 17.5 Gibbs Free Energy 784 17.6 Entropy Changes in Chemical Reactions: Calculating 788 17.7 Free Energy Changes in Chemical Reactions: Calculating 792 17.8 Free Energy Changes for Nonstandard States: The Relationship between and 798 17.9 Free Energy and Equilibrium: Relating to the Equilibrium Constant (K)

  4. Chapter 6. Thermochemistry 6.1 Chemical Hand Warmers 231 6.2 The Nature of Energy: Key Definitions 232 6.3 The First Law of Thermodynamics: There Is No Free Lunch 234 6.4 Quantifying Heat and Work 240 6.5 Measuring for Chemical Reactions: Constant-Volume Calorimetry 246 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure 249 6.7 Constant-Pressure Calorimetry: Measuring 253 6.8 Relationships Involving 255 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation 257 6.1 0 Energy Use and the Environment 263

  5. Method 1: Calculate DH for the reaction: SO2(g) + 1/2 O2(g) + H2O(g) ----> H2SO4(l) DH = ? Other reactions: SO2(g) ------> S(s) + O2(g) ; DH = 297kJ H2SO4(l)------> H2(g) + S(s) + 2O2(g); DH = 814 kJH2(g) +1/2O2(g) -----> H2O(g); DH = -242 kJ

  6. Calculate DH for the reaction SO2(g) ------> S(s) + O2(g); DH1 = 297 kJ - 1 H2(g) + S(s) + 2O2(g) ------> H2SO4(l); DH2 = -814 kJ - 2 H2O(g) ----->H2(g) + 1/2 O2(g) ; DH3 = +242 kJ - 3 ______________________________________ SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l); DH = ? • DH = DH1+ DH2+ DH3 • DH = +297 - 814 + 242 DH = -275 kJ

  7. 1) Calculate entropy change for the reaction: • 2 C(s) + 1/2 O2(g) + 3 H2(g) --> C2H6O(l); ∆H = ? (ANS -277.1 kJ/mol)Given the following thermochemical equations: • C2H6O(l) + 3 O2(g)---> 2 CO2(g) + 3 H2O(l); ∆H = - 1366.9 kJ/mol • 1/2 O2(g) + H2(g) ----> H2O(l); ∆H = -285.8 kJ/mol • C(s) + O2(g) ----> CO2(g); ∆H = -393.3 kJ/mol

  8. Calculate Heat (Enthalpy) of Combustion: 2nd method C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(l) ; DHo= ? DHf (C7H16) = -198.8 kJ/mol DHf (CO2) = -393.5 kJ/mol DHf (H2O) = -285.9 kJ/mol DHf O2(g) = 0 (zero) What method? DHo = S nDHfo products – S n DHfo reactants n = stoichiometric coefficients 2nd method

  9. Calculate DH for the reaction DH = [Sn ( DHof) Products] - [Sn (DHof) reactants] DH = [ 7(- 393.5 + 8 (- 285.9)] - [-198.8 + 11 (0)] = [-2754.5 - 2287.2] - [-198.8] = -5041.7 + 198.8 = -4842.9 kJ = -4843 kJ

  10. Why is DHofof elements is zero? DHof, Heat formations are for compounds Note: DHof of elements is zero

  11. 2) Calculate enthalpy change given the • ∆Hfo[SO2(g)] = -297 kJ/mole and • ∆Hfo [SO3(g)] = -396 kJ/mole • 2SO2 (g) + O2 (g) -----> 2 SO3(g); ∆H= ? ANS -198 kJ/mole)

  12. What is relation of DH of a reaction to covalent bond energy? • DH = S½bonds broken½- S½bonds formed½ • How do you calculate bond energy from DH? • How do you calculate DH from bond energy?

  13. 3) Use the table of bond energies to find the ∆Ho for the reaction: • H2(g) + Br2(g) 2 HBr(g); • H-H = 436 kJ, Br-Br= 193 kJ, H-Br = 366 kJ

  14. Calculation of standard entropy changes Example. Calculate the DSorxnat 25 oCfor the following reaction. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) Substance So(J/K.mol) CH4 (g) 186.2 O2 (g) 205.03 CO2 (g) 213.64 H2O (g)188.72

  15. Calculate the DS for the following reactions usingD So = SD So (products) - SD S o(reactants)a) 2SO2 (g) + O2 (g) ------> 2SO 3(g) D So [SO2(g)] = 248 J/K mole ; D So [O2(g)] = 205 J/K mole; D So [SO3(g)] = 257 J/K moleb) 2NH 3 (g) + 3N2O (g) --------> 4N2 (g) + 3 H2O (l) D So[ NH3(g)] = 193 J/K mole ; D So [N2(g)] = 192 J/K mole; D So [N2O(g)] = 220 J/K mole; D S[ H2O(l)] = 70 J/K mole

  16. 2SO2 (g) + O2 (g------> 2SO 3(g)D So [SO2(g)] = 248 J/K mole ; D So [O2(g)] = 205 J/K mole; D So [SO3(g)] = 257 J/K moleDSo 496 205 514DSo = SDSo (products) - SDS o(reactants)DSo = [514] - [496 + 205]DSo = 514 - 701DSo = -187 J/K mole

  17. Calculating DS for a Reaction 2 H2(g) + O2(g)  2 H2O(liq) DSo = 2 So (H2O) - [2 So (H2) + So (O2)] DSo = 2 mol (69.9 J/K•mol) – [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)] DSo = -326.9 J/K There is a decrease in S because 3 mol of gas give 2 mol of liquid. Based on Hess’s Law second method: DSo = So(products) - So(reactants)

  18. 4) Calculate the ∆S for the following reaction using: • a) 2SO2 (g) + O2 (g) ----> 2SO3(g) So [SO2(g)] = 248 J/K mole ; So[O2(g)] = 205 J/K mole; So[SO3(g)] = 257 J/K mole

  19. Free energy, DG The sign ofDG indicates whether a reaction will occur spontaneously. + Not spontaneous 0 At equilibrium - Spontaneous The fact that the effect of DS will vary as a function of temperature is important. This can result in changing the sign of DG.

  20. Standard free energy of formation, DGfo DGfo Free energy change that results when one mole of a substance if formed from its elements will all substances in their standard states. DG values can then be calculated from: DGo= S npDGfoproducts – S nrDGforeactants

  21. Standard free energy of formation Substance DGfo Substance DGfo C (diamond) 2.832HBr (g) -53.43 CaO (s) -604.04 HF (g) -273.22 CaCO3 (s)-1128.84 HI (g) 1.30 C2H2 (g) 209 H2O (l) -237.18 C2H4 (g) 86.12 H2O (g) -228.59 C2H6 (g) -32.89 NaCl (s) -384.04 CH3OH (l) -166.3 O (g) 231.75 CH3OH (g) -161.9 SO2 (g) -300.19 CO (g) -137.27 SO3 (g) -371.08 All have units of kJ/mol and are for 25 oC

  22. How do you calculate DG There are two ways to calculate DG for chemical reactions. i) DG = DH - TDS. ii) DGo = SDGof (products) - SDG of (reactants)

  23. Calculating DGorxn Method (a) : From tables of thermodynamic data we find DHorxn = +25.7 kJ DSorxn = +108.7 J/K or +0.1087 kJ/K DGorxn = +25.7 kJ - (298 K)(+0.1087 J/K) = -6.7 kJ Reaction is product-favored in spite of positive DHorxn. Reaction is “entropy driven” NH4NO3(s) + heat  NH4NO3(aq)

  24. Calculating DGorxn Combustion of carbon C(graphite) + O2(g) --> CO2(g) Method (b) : DGorxn = DGfo(CO2) - [DGfo(graph) + DGfo(O2)] DGorxn = -394.4 kJ - [ 0 + 0] Note that free energy of formation of an element in its standard state is 0. DGorxn = -394.4 kJ Reaction is product-favored DGorxn = SDGfo (products) - SDGfo (reactants)

  25. Calculation of DGo We can calculate DGo values from DHo and DSovalues at a constant temperature and pressure. Example. Determine DGofor the following reaction at 25oC Equation N2 (g) + 3H2(g) 2NH3 (g) DHfo, kJ/mol 0.00 0.00 -46.11 So, J/K.mol 191.50 130.68192.3

  26. Predict the spontaneity of the following processes from DH and DS at various temperatures.a)DH = 30 kJ, DS = 6 kJ, T = 300 Kb)DH = 15 kJ,DS = -45 kJ,T = 200 K

  27. a) DH = 30 kJ DS = 6 kJ T = 300 KDG = DHsys-TDSsys or DG = DH - TDS.DH = 30 kJDS = 6 kJ T = 300 KDG = 30 kJ - (300 x 6 kJ) = 30 -1800 kJDG = -1770 kJb) DH = 15 kJ DS = -45 kJ T = 200 KDG = DHsys-TDSsys or DG = DH - TDS.DH = 15 kJDS = -45 kJ T = 200 K DG = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJDG = 15 + 9000 kJ = 9015 kJ

  28. 5) Predict the spontaneity of the following processes from ∆H and ∆S at various temperatures. • a) ∆H = 30 kJ ∆S = 6 kJ T = 300 K • b) ∆H = 15 kJ ∆S = -45 kJ T = 200 K

  29. 6) Calculate the ∆Go for the following chemical reactions using given ∆Ho values, ∆So calculated above and the equation ∆G = ∆H - T∆S. • 2SO2 (g) + O2 (g) > 2 SO 3(g) ; ∆Go= • ∆Ho = -198 kJ/mole; ∆So = -187 J/K mole; T = 298 K ∆Gosystem ∆Sosystem ∆Hosystem T

  30. 7) Which of the following condition applies to a particular chemical reaction, the value of ∆H° = 98.8 kJ and ∆S° = 141.5 J/K. This reaction is

  31. Effect of Temperature on Reaction Spontaneity

  32. DGo = DHo - TDSo

  33. 8) At what temperature a particular chemical reaction, with the value of ∆H° = 98.8 kJ and ∆S° = 141.5 J/K becomes a) Equilibrium: b) Spontaneous:

  34. How do you calculate DG at different T and P DG = DGo + RT ln Q Q = reaction quotient at equilibrium DG = 0 0 = DGo + RT ln K DGo = - RT ln K If you know DGo you could calculate K

  35. Concentrations, Free Energy, and the Equilibrium Constant Equilibrium Constant and Free Energy DG = DGo + RT ln Q Q = reaction quotient 0 = DGo + RT lnKeq DGo = - RT lnKeq

  36. 9) Calculate the non standard ∆G for the following equilibrium reaction and predict the direction of the change using the equation: • ∆G= ∆Go + RT ln Q • Given ∆Gfo[NH3(g)] = -17 kJ/mole • N2 (g) + 3H2(g) → 2NH3(g); ∆G=? at 300K, PN2= 300, PNH3 = 75 and PH2 = 300

  37. 10) The Ka expression for the dissociation of acetic acid in water is based on the following equilibrium at 25°C: • HC2H3O2(l) + H2O ⇄ H+(aq) + C2H3O2-(aq) • What is ∆G° if Ka=1.8 x 10-5?

  38. Calculate the DG value for the following reactions using:D Go = SD Gof (products) - SD Gof (reactants)N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; DGo = ?D Gfo[ N2O5 (g) ] = 134 kJ/mole ; D Gfo [H2O(g)] = -237 kJ/mole; DGfo[ HNO3(l) ] = -81 kJ/mole N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; DGo = ?DGfo 1 x 134 1 x (-237) 2 (-81) 134 -237 -162DGo = DGof (products) - 3DGof (reactants)DGo = [-162] - [134 + (-237)]DGo = -162 + 103DGo = -59 kJ/moleThe reaction have a negative DG and the reaction is spontaneous or will take place as written.

  39. Free Energy and Temperature 2 Fe2O3(s) + 3 C(s) ---> 4 Fe(s) + 3 CO2(g) DHorxn = +467.9 kJ DSorxn = +560.3 J/K DGorxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does DGorxn change from (+) to (-)? Set DGorxn = 0 = DHorxn - TDSorxn

  40. Thermodynamics and Keq Keq is related to reaction favorability and so to Gorxn. The larger the (-) value of DGorxn the larger the value of K. DGorxn = - RT lnK where R = 8.31 J/K•mol

  41. -32.91 kJ -(0.008315 kJ.K-1mol-1)(298.2K) DGo -RT Free energy and equilibrium For gases, the equilibrium constant for a reaction can be related to DGo by: DGo = -RT lnK For our earlier example, N2 (g) + 3H2 (g) 2NH3 (g) At 25oC, DGo was -32.91 kJ so K would be: ln K = = ln K = 13.27; K = 5.8 x 105

  42. Calculate the D G for the following equilibrium reaction and predict the direction of the change using the equation:DG = D Go + RT ln Q ; [D Gfo[ NH3(g) ] = -17 kJ/moleN2 (g) + 3 H2 (g) 2 NH3 (g); D G = ? at 300 K, PN2 = 300, PNH3 = 75 and PH2 = 300N2 (g) + 3 H2 (g) 2 NH3 (g); DG = ?

  43. To calculate DGoUsing DGo = SDGof (products) - SDGof (reactants) DGfo[ N2(g)] = 0 kJ/mole; DGfo[ H2(g)] = 0 kJ/mole; DGfo[ NH3(g)] = -17 kJ/moleNotice elements have DGfo = 0.00 similar to DHfo N2 (g) + 3 H2 (g) 2 NH3 (g); DG = ?DGfo 0 0 2 x (-17) 0 0 -34 DGo = SDGof (products) - SDGof (reactants)DGo = [-34] - [0 +0]DGo = -34DGo = -34 kJ/mole

  44. To calculate QEquilibrium expression for the reaction in terms of partial pressure:N2 (g) + 3 H2 (g) 2 NH3 (g) p2NH3K = _________ pN2 p3H2 p2NH3Q = _________ ; pN2 p3H2 Q is when initial concentration is substituted into the equilibrium expression 752Q = _________ ; p2NH3= 752; pN2 =300; p3H2=3003300 x 3003Q = 6.94 x 10-7

  45. To calculate DGoDG = DGo + RT ln QDGo= -34 kJ/mole R = 8.314 J/K mole or 8.314 x 10-3kJ/Kmole T = 300 K Q= 6.94 x 10-7DG = (-34 kJ/mole) + ( 8.314 x 10-3 kJ/K mole) (300 K) ( ln 6.94 x 10-7)DG = -34 + 2.49 ln 6.94 x 10-7DG = -34 + 2.49 x (-14.18)DG = -34 -35.37DG = -69.37 kJ/mole

  46. Thermodynamics and Keq Calculate K (from G0) N2O4 --->2 NO2DGorxn = +4.8 kJ DGorxn = +4800 J = - (8.31 J/K)(298 K) ln K DGorxn = - RT lnK K = 0.14 DGorxn > 0 : K < 1 DGorxn < 0 : K > 1

  47. Concentrations, Free Energy, and the Equilibrium Constant The Influence of Temperature on Vapor Pressure H2O(l)=> H2O(g) Keq = pwatervapor pwatervapor = Keq = e- G'/RT

  48. DG as a Function of theExtent of the Reaction

  49. DG as a Function of theExtent of the Reactionwhen there is Mixing

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