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Chemistry 102(01) spring 2009

Chemistry 102(01) spring 2009. Instructor: Dr. Upali Siriwardane e-mail : upali@chem.latech.edu Office : CTH 311 Phone 257-4941 Office Hours : M,W, 8:00-9:00 & 11:00-12:00 a.m.; Tu,Th,F 9:00 - 10:00 a.m.  

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Chemistry 102(01) spring 2009

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  1. Chemistry 102(01) spring 2009 • Instructor: Dr. Upali Siriwardane • e-mail: upali@chem.latech.edu • Office: CTH 311 Phone 257-4941 • Office Hours: M,W, 8:00-9:00 & 11:00-12:00 a.m.; Tu,Th,F 9:00 - 10:00 a.m.   • Test Dates: March 25, April 26, and May 18; Comprehensive Final Exam: May 20,2009 9:30-10:45 am, CTH 328. • March 30, 2009 (Test 1): Chapter 13 • April 27, 2009 (Test 2): Chapters 14 & 15 • May 18, 2009 (Test 3):Chapters 16, 17 & 18 • Comprehensive Final Exam: May 20,2009:Chapters 13, 14, 15, 16, 17 and 18

  2. Review of Chapter 6. Energy and Chemical Reactions 6.1 The Nature of Energy 6.2 Conservation of Energy 6.3 Heat Capacity 6.4 Energy and Enthalpy 6.5 Thermochemical Equations 6.6 Enthalpy change for chemical Rections 6.7 Where does the Energy come from? 6.8 Measuring Enthalpy Changes: Calorimetry 6.9 Hess's Law 6.10 Standard Enthalpy of Formation 6.11 Chemical Fuels for Home and Industry 6.12 Food Fuels for Our Bodies

  3. Thermochemistry Heat changes during chemical reactions Thermochemical equation. eg. H2 (g) + O2 (g) ---> 2H2O(l) DH =- 256 kJ; DH is called the enthalpy of reaction. if DH is + reaction is called endothermic if DH is - reaction is called exothermic

  4. universe system surroundings Why is it necessary to divide Universe into System and Surrounding Universe = System + Surrounding

  5. Types of Systems • Isolated system • no mass or energy exchange • Closed system • only energy exchange • Open system • both mass and energy exchange

  6. Why is it necessary to divide Universe into System and Surrounding Universe = System + Surrounding

  7. What is the internal energy change (DU) of a system? DU is associated with changes in atoms, molecules and subatomic particles Etotal = Eke + E pe + DU DU = heat (q) + w (work) DU = q + w DU = q -P DV; w =- P DV

  8. What forms of energy are found in the Universe? mechanical thermal electrical nuclear mass: E = mc2 others yet to discover

  9. What is 1st Law of Thermodynamics Eenergy is conserved in the Universe All forms of energy are inter-convertible and conserved Energy is neither created nor destroyed.

  10. What exactly is DH? Heat measured at constant pressure qp Chemical reactions exposed to atmosphere and are held at a constant pressure. Volume of materials or gases produced can change. Volume expansion work = -PDV DU = qp + w; DU = qp -PDV qp = DU + PDV; w = -PDV DH = DU + PDV; qp = DH(enthalpy )

  11. How do you measure DU? Heat measured at constant volume qv Chemical reactions take place inside a bomb. Volume of materials or gases produced can not change. ie: work = -PDV= 0 DU = qv + w qv = DU + o; w = 0 DU = qv = DU(internal energy )

  12. What is Hess's Law of Summation of Heat? To heat of reaction for new reactions. Two methods? 1st method: new DH is calculated by adding DHs of other reactions. 2nd method: Where DHf ( DH of formation) of reactants and products are used to calculate DH of a reaction.

  13. 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

  14. 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

  15. 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

  16. 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

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

  18. Chapter 18. Thermodynamics: Directionality of Chemical Reactions 18.1 Reactant-Favored and Product-Favored Processes 18.2 Probability and Chemical Reactions 18.3 Measuring Dispersal or Disorder: Entropy 18.4 Calculating Entropy Changes 18.5 Entropy and the Second Law of Thermodynamics 18.6 Gibbs Free Energy 18.7 Gibbs Free Energy Changes and Equilibrium Constants 18.8 Gibbs Free Energy, Maximum Work, and Energy Resources 18.9 Gibbs Free Energy and Biological Systems 18.10 Conservation of Gibbs Free Energy 18.11 Thermodynamic and Kinetic Stability

  19. Chemical Thermodynamics spontaneous reaction – reaction which proceed without external assistance once started chemical thermodynamics helps predict which reactions are spontaneous

  20. Thermodynamics Will the rearrangement of a system decrease its energy? If yes, system is favored to react — a product-favoredsystem. Most product-favored reactions are exothermic. Often referred to as spontaneousreactions. “Spontaneous” does not imply anything about time for reaction to occur. Kinetic factors are more important for certain reactions.

  21. Thermodynamics and Kinetics Diamond  graphite Thermodynamically product favored Slow Kinetics Paper burns Thermodynamically product favored Fast Kinetics In this chapter we only look into thermodynamic factors

  22. Bases on Energy: Product-Favored Reactions In general, product-favored reactions are exothermic. (Negative DH) In general, reactant-favored reactions are endothermic. (Positive DH)

  23. Product-Favored Reactions But many spontaneous reactions or processes are endothermic or even have DH = 0. NH4NO3(s)  NH4NO3(aq); DH = +

  24. Direction of Reaction Product favored reactions are always a transformation of a reactants favored reaction. Product Favored Reaction 2Na(s) + 2Cl2(g)=> 2NaCl(s) Reactant Favored Reaction 2NaCl(s)=> 2Na(s) + 2Cl2(g) However, non spontaneous reactions could be made to take place by coupling with energy source: another reaction or electric current Electrolysis 2NaCl(l)=> 2Na(s) + 2Cl2(g)

  25. Expansion of a Gas The positional probability is higher when particles are dispersed over a larger volume Matter tends to expand unless it is restricted

  26. Gas Expansion and Probability

  27. Entropy, S The thermodynamic property related to randomness is ENTROPY, S. Product-favored processes: final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. Reaction of K with water

  28. S[H2O(l)] > S[H2O(s)] at 0° C.

  29. S (gases) > S (liquids) > S (solids) Entropies of Solid, Liquidand Gas Phases

  30. Entropy, S Entropies of ionic solids depend on coulombic attractions. So (J/K•mol) MgO 26.9 NaF 51.5

  31. Entropy and Molecular Structure

  32. Entropy and Dissolving

  33. Qualitative Guidelines for Entropy Changes Entropies of gases higher than liquids higher than solids Entropies are higher for more complex structures than simpler structures Entropies of ionic solids are inversely related to the strength of ionic forces Entropy increases when making solutions of pure solids or pure liquids in a liquid solvent Entropy decrease when making solutions of gases in a liquid

  34. Entropy of a Solution of a Gas

  35. Phase Transitions H2O(s)=> H2O(l)DH > 0; DS > 0 H2O(l)=> H2O(g)DH > 0; DS > 0 spontaneous at high temperatures H2O(l)=> H2O(s)DH < 0; DS < 0 H2O(g)=> H2O(l)DH < 0; DS < 0 spontaneous at low temperatures

  36. Entropy Changes for Phase Changes For a phase change, DSSYS = qSYS/T (q = heat transferred) Boiling Water H2O (liq)  H2O(g) DH = q = +40,700 J/mol

  37. Phase Transitions Heat of Fusion energy associated with phase transition solid-to-liquid or liquid-to-solid DGfusion = 0 = DHfusion - T DSfusion 0 = DHfusion - T DSfusion DHfusion = T DSfusion Heat of Vaporization energy associated with phase transition gas-to-liquid or liquid-to-gas DHvaporization = T DSvaporization

  38. Qualitative prediction of DS of Chemical Reactions • Look for (l) or (s) --> (g) • If all are gases: calculate Dn Dn = Sn (gaseous prod.) - S n(gaseous reac.) N2 (g) + 3 H2 (g) --------> 2 NH3 (g) Dn = 2 - 4 = -2 If Dn is -DS is negative (decrease in S) If Dn is +DS is positive (increase in S)

  39. Entropy Change Entropy (DS) normally increase (+) for the following changes: i) Solid ---> liquid (melting) + ii) Liquid ---> gas + iii) Solid ----> gas most + iv) Increase in temperature + v) Increasing in pressure(constant volume, and temperature) + vi) Increase in volume +

  40. Predict DS! 2 C2H6(g) + 7 O2(g)--> 4 CO2(g) + 6H2O(g) 2 CO(g) + O2(g)-->2 CO2(g) HCl(g) + NH3(g)-->NH4Cl(s) H2(g) + Br2(l) --> 2 HBr(g)

  41. Based on Hess’s Law second method: DSo =  So (products) -  So (reactants) 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.

  42. Third Law of Thermodynamics Provides reference point for absolute entropy Entropy of a perfectly crystalline substance at absolute zero (T= 0 K) is zero. Unlike DH entropy values are positive above temperatures above absolute zero.

  43. Standard Molar Entropy Values

  44. Standard Entropies at 25oC Substance So (J/K.mol) Substance So (J/K.mol) C (diamond) 2.37 HBr (g) 198.59 C (graphite) 5.69 HCl (g) 186.80 CaO (s) 39.75 HF (g) 193.67 CaCO3 (s) 92.9 HI (g) 206.33 C2H2 (g) ` 200.82 H2O (l) 69.91 C2H4 (g) 219.4 H2O (g) 188.72 C2H6 (g) 229.5 NaCl (s) 72.12 CH3OH (l) 127 O2 (g) 205.03 CH3OH (g) 238 SO2 (g) 248.12 CO (g) 197.91 SO3 (g) 256.72

  45. Entropy & Spontaneity • How can water boil and freeze spontaneously? • Enthalpy change can not predict spontaneity! • Some endothermic processes are spontaneous • Need another thermodynamic property.

  46. Laws of Thermodynamics First : The total energy of the universe is constant Second : The total entropy (S) of the universe is always increasing Third : The entropy(S) of a pure, perfectly formed crystalline substance at absolute zero is zero

  47. 2nd Law of Thermodynamics Dissolving NH4NO3 in water—an entropy driven process. NH4NO3(s)  NH4NO3(aq); DH = +

  48. Second Law of Thermodynamics In the universe the ENTROPY cannot decrease for any spontaneous process The entropy of the universe strives for a maximum in any spontaneous process, the entropy of the universe increases for product-favored process DSuniverse = ( Ssys + Ssurr) > 0 DSuniv = entropy of the Universe DSsys = entropy of the System DSsurr = entropy of the Surrounding DSuniv = DSsys + DSsurr

  49. Entropy of the Universe DSuniv = DSsys + DSsurr DSunivDSsysDSsurr + + + + +(DSsys>DSsurr) - + - + (DSsurr>DSsys)

  50. 2nd Law of Thermodynamics 2 H2(g) + O2(g)  2 H2O(liq) DSosys = -326.9 J/K Entropy Changes in the Surroundings Can calc. that DHorxn = DHosystem = -571.7 kJ = +1917 J/K

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