Chem Unit6

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Chem Unit6

  1. 1. Stoichiometry στοιχεῖον (stoicheion, meaning element]) & μέτρον (metron, meaning measure)
  2. 2. Balancing Chemical Equations <ul><li>2 Mg + O 2 -> 2 MgO </li></ul>
  3. 3. Balancing Chemical Equations <ul><li>2 Mg + O 2 -> 2 MgO </li></ul><ul><li>1.2 x 10 24 + 6 x 10 23 form 1.2 x 10 24 formula units </li></ul><ul><li>atoms O 2 molecules MgO </li></ul><ul><li>Mg </li></ul>
  4. 4. Balancing Chemical Equations <ul><li>2 Mg + O 2 -> 2 MgO </li></ul><ul><li>2 moles + 1 mole form 2 moles </li></ul><ul><li>Mg O 2 MgO </li></ul><ul><li>Mole Ratio </li></ul>
  5. 5. Balancing Chemical Equations <ul><li>2 Mg + O 2 -> 2 MgO </li></ul><ul><li>48.6 g + 32 g form 80.6 grams </li></ul><ul><li>Mg O 2 MgO </li></ul><ul><li>Mass </li></ul><ul><li>Balance </li></ul><ul><li>The total mass of the product(s) must equal the total mass of the reactants </li></ul>
  6. 6. Yield <ul><li>Building Bicycles </li></ul>
  7. 7. Yield <ul><li>Building Bicycles </li></ul><ul><li>1 Frame + 1 Seat + 1 Handlebar + 2 Pedals + 2 Wheels </li></ul>
  8. 8. Finding Product Quantities <ul><li>How much O 2 gas is formed when </li></ul><ul><li>49.89 g KClO 3 decomposes? </li></ul><ul><li>KClO 3 -> KCl + O 2 </li></ul>
  9. 9. Balance the Equation <ul><li>How much O 2 gas is formed when </li></ul><ul><li>49.89 g KClO 3 decomposes? </li></ul><ul><li>2 KClO 3 -> 2 KCl + 3 O 2 </li></ul>
  10. 10. Balance the Equation <ul><li>How much O 2 gas is formed when </li></ul><ul><li>49.89 g KClO 3 decomposes? </li></ul><ul><li>2 KClO 3 -> 2 KCl + 3 O 2 </li></ul><ul><li>1 mole </li></ul><ul><li>49.89 g KClO 3 x 122.45 g = 0.41 moles KClO 3 </li></ul><ul><li>3 moles O2 </li></ul><ul><li>2 moles KClO3 x 0.41 moles KClO 3 = 0.615 moles O 2 </li></ul>
  11. 11. Balance the Equation <ul><li>How much O 2 gas is formed when </li></ul><ul><li>49.89 g KClO 3 decomposes? </li></ul><ul><li>2 KClO 3 -> 2 KCl + 3 O 2 </li></ul><ul><li>49.89 g KClO 3 ? </li></ul><ul><li>0.615 moles O 2 x 32 g/molee = 19.68 grams O 2 </li></ul>
  12. 12. Standard Temperature & Pressure <ul><li>STP </li></ul><ul><li>Temperature = 0 °C (273.15 K; 32.00 °F) </li></ul><ul><li>Pressure = 101.325 kPa (1 atm; 760  Torr) </li></ul><ul><li>For Gases: Volume = 22.4 Liters/mole </li></ul>
  13. 13. Balance the Equation <ul><li>How much O 2 gas is formed when </li></ul><ul><li>49.89 g KClO 3 decomposes? </li></ul><ul><li>2 KClO 3 -> 2 KCl + 3 O 2 </li></ul><ul><li>49.89 g KClO 3 ? </li></ul><ul><li>0.615 moles O 2 x 22.4 L/mole = 13.776 Liters O 2 </li></ul>
  14. 14. Yield <ul><li>Building Bicycles </li></ul><ul><li>1 Frame + 1 Seat + 1 Handlebar + 2 Pedals + 2 Wheels </li></ul><ul><li>= 1 bicycle </li></ul>
  15. 15. Yield <ul><li>Theoretical Yield </li></ul><ul><li>How much O 2 gas is formed when </li></ul><ul><li>49.89 g KClO 3 decomposes? </li></ul><ul><li>2 KClO 3 -> 2 KCl + 3 O 2 </li></ul><ul><li>49.89 g KClO 3 6.12 Liters O 2 </li></ul><ul><li>What if only 5.5 Liters of O 2 are produced? </li></ul>
  16. 16. Yield <ul><li>Theoretical Yield </li></ul><ul><li>2 KClO 3 -> 2 KCl + 3 O 2 </li></ul><ul><li>49.89 g KClO 3 6.12 Liters O 2 </li></ul><ul><li>What if only 5.5 Liters of O 2 are produced? </li></ul><ul><li>5.5 Liters </li></ul><ul><li>6.12 Liters x 100% = 89.9% actual yield </li></ul>
  17. 17. Gas Yield <ul><li>Nitroglycerin decomposes to </li></ul><ul><li>produce N 2 , O 2 , CO 2 and H 2 O </li></ul>
  18. 18. Gas Yield <ul><li>4 C 3 H 5 N 3 O 9 -> 6 N 2 + O 2 + 12 CO 2 + H 2 O </li></ul><ul><li>227 g how much gas results? </li></ul>
  19. 19. Gas Yield <ul><li>4 C 3 H 5 N 3 O 9 -> 6 N 2 + O 2 + 12 CO 2 + H 2 O </li></ul><ul><li> 1 mole </li></ul><ul><li>227 g x 227 g = 1 mole nitroglycerin </li></ul>
  20. 20. Gas Yield <ul><li>4 C 3 H 5 N 3 O 9 -> 6 N 2 + O 2 + 12 CO 2 + H 2 O </li></ul><ul><li> 1.5 moles N 2 </li></ul><ul><li>1 mole nitroglycerin -> 0.25 mole O 2 </li></ul><ul><li> 3 moles CO 2 </li></ul><ul><li> 0.25 mole H 2 O </li></ul>
  21. 21. Gas Yield <ul><li>4 C 3 H 5 N 3 O 9 -> 6 N 2 + O 2 + 12 CO 2 + H 2 O </li></ul><ul><li>1.5 moles N 2 x 22.4 L/ mole = 33.6 L </li></ul><ul><li>0.25 mole O 2 x 22.4 L/mole = 5.6 L </li></ul><ul><li>3 moles CO 2 x 22.4 L/mole = 67.2 L </li></ul><ul><li>0.25 mole H 2 O x 22.4 L/mole = 5.6 L </li></ul><ul><li>112 Liters ! </li></ul>
  22. 22. Making Aspirin <ul><li>White willow tree bark = pain reliever </li></ul>
  23. 23. Making Aspirin <ul><li>Salicylic acid acetylsalicylic acid </li></ul><ul><li>+ acetic anhydride -> and acetic acid </li></ul><ul><li>C 7 H 6 O 3 + C 2 H 6 O 3 -> C 9 H 8 O 4 + C 2 H 4 O 2 </li></ul><ul><li>each aspirin tablet 325 mg </li></ul><ul><li>How much salicylic acid would be needed if we assume 90% yield? </li></ul>
  24. 24. Making Aspirin <ul><li>C 7 H 6 O 3 + C 2 H 6 O 3 -> C 9 H 8 O 4 + C 2 H 4 O 2 </li></ul><ul><li>0.325 g x 1 mole/180 g = 0.0018 moles of aspirin </li></ul><ul><li>90 % of 0.0018 = 0.9 x 0.0018 = 0.00162 moles </li></ul><ul><li>0.00162 moles x 138 g/mole = 0.224 grams of salicylic acid </li></ul>
  25. 25. Rocket Fuel <ul><li>N 2 H 4 (l) + N 2 O 4 (l) -> N 2 (g) + H 2 O (g) </li></ul><ul><li>Is the reaction balanced? </li></ul>
  26. 26. Rocket Fuel <ul><li>2 N 2 H 4 (l) + N 2 O 4 (l) -> 3 N 2 (g) + 4 H 2 O (g) </li></ul><ul><li>What is the mole ratio </li></ul><ul><li>of N 2 H 4 to N 2 ? </li></ul>
  27. 27. Rocket Fuel <ul><li>2 N 2 H 4 (l) + N 2 O 4 (l) -> 3 N 2 (g) + 4 H 2 O (g) </li></ul><ul><li>What is the mole ratio </li></ul><ul><li>of N 2 H 4 to N 2 ? </li></ul><ul><li>3 moles N 2 </li></ul><ul><li>2 moles N 2 H 4 </li></ul>
  28. 28. Rocket Fuel <ul><li>2 N 2 H 4 (l) + N 2 O 4 (l) -> 3 N 2 (g) + 4 H 2 O (g) </li></ul><ul><li>How many Liters of N 2 will be </li></ul><ul><li>produced from 16000 g of N 2 H 4 ? </li></ul><ul><li>16000 g x 1 mole = 500 moles N 2 H 4 </li></ul><ul><li> 32g </li></ul><ul><li> 3 moles N 2 </li></ul><ul><li>500 moles N 2 H 4 x 2 moles N 2 H 4 </li></ul><ul><li> = 750.0 moles N 2 </li></ul><ul><li> = 16800 Liters of N 2 </li></ul>
  29. 29. Rocket Fuel <ul><li>2 N 2 H 4 (l) + N 2 O 4 (l) -> 3 N 2 (g) + 4 H 2 O (g) ENERGY </li></ul><ul><li>How many Liters of N 2 will be </li></ul><ul><li>produced from 16000 g of N 2 H 4 ? </li></ul><ul><li>16000 g x 1 mole = 500 moles N 2 H 4 </li></ul><ul><li> 32g </li></ul><ul><li> 3 moles N 2 </li></ul><ul><li>500 moles N 2 H 4 x 2 moles N 2 H 4 </li></ul><ul><li> = 750.0 moles N 2 </li></ul><ul><li> = 16800 Liters of N 2 </li></ul>
  30. 30. Chemical Change <ul><li>Energy is absorbed or released </li></ul><ul><li>Exothermic vs. Endothermic </li></ul>
  31. 31. Enthalpy Δ H <ul><li>Total energy of a system: </li></ul><ul><ul><li>Internal energy + (pressure) x (Volume) </li></ul></ul>
  32. 32. Enthalpy Δ H <ul><li>Only way to measure Δ H is when heat is added or removed: </li></ul><ul><li>Molar Enthalpy Change: Δ H = C Δ T </li></ul><ul><li>(Recall Specific Heat Capacity: Q = m C Δ T) </li></ul>
  33. 33. Molar Enthalpy Change <ul><li>Δ H = C Δ T </li></ul><ul><li>What is Δ H if an aluminum can is cooled from 25° C to 4° C? </li></ul>
  34. 34. Temperature Scales
  35. 35. Molar Enthalpy Change <ul><li>Δ H = C Δ T </li></ul><ul><li>What is Δ H if an aluminum can is cooled from 25° C to 4° C? (C for Al = 24.2 J/K mole) </li></ul><ul><li>25° C = 298 K </li></ul><ul><li>4° C = 277 K </li></ul><ul><li>Δ H = C Δ T = 24.2 J/K mole (21 K) </li></ul><ul><li>= 508.2 J/mole </li></ul>
  36. 36. Enthalpy Change <ul><li>Negative = Exothermic </li></ul><ul><li>C(s) + 1/2 O 2 (g) -> CO(g) ∆H = - 110.5 kJ </li></ul><ul><li>Positive = Endothermic </li></ul><ul><li>C 2 H 6 (g) -> 2C(s) + 3H 2 (g) ∆H = + 4.83 kJ </li></ul>
  37. 37. Enthalpy Change (Endothermic) A + B ----> C + D
  38. 38. Enthalpy Change (Exothermic) W + X --> Y + Z
  39. 39. Enthalpy in Chemical Changes
  40. 40. Enthalpy in Chemical Changes
  41. 41. Enthalpy in Chemical Changes
  42. 42. Stoichiometric Enthalpy <ul><li>∆ H = ? When 4.8 g C reacts with O 2 ? </li></ul><ul><li>C + O 2 -> CO 2 ∆H = -393.5 kJ/mole </li></ul><ul><li>4.8 g x 1mole/12g = 0.4 moles </li></ul><ul><li>0.4 moles x (-393.5 kJ/mole) </li></ul><ul><li>= 157.4 kJ </li></ul>
  43. 43. Hess’ Law <ul><li>Total ∆H = Sum of ∆H for all steps </li></ul><ul><li>Example: </li></ul><ul><li>H 2 S(g) + 2O 2 (g) ->SO 3 (g) + H 2 O(l) </li></ul><ul><li>∆ H = ??????? </li></ul>
  44. 44. Hess’ Law <ul><li>H 2 S (g) + 2O 2 (g) -> H 2 SO 4 (l) ∆H= -628KJ </li></ul><ul><li>H 2 SO 4 (l) -> SO 3 (g) + H 2 O (g) ∆H= +164KJ </li></ul><ul><li>H 2 O (g) -> H 2 O (l) ∆H= -88KJ </li></ul><ul><li>H 2 S (g) + 2O 2 (g) ->SO 3 (g) + H 2 O (l) ∆H = ? </li></ul>
  45. 45. Hess’ Law
  46. 46. Germain Henri Hess <ul><li>Enthalpy change (∆H) depends only on the initial and final states of the reaction </li></ul><ul><li>(not on the intermediate stages) </li></ul><ul><li>Hess’ cycle </li></ul>
  47. 47. Hess’ Law <ul><li>CH 4 + 2 O 2 ----> CO 2 + 2 H 2 O ∆H = ?? </li></ul><ul><li>CH 4 + 2 O 2 -----> CO + 2 H 2 O + ½ O 2 </li></ul><ul><li> ∆H = -607 kJ </li></ul><ul><li>CO + 2 H 2 O + ½ O 2 ------> CO 2 + 2 H 2 O ∆H = -283 kJ </li></ul>
  48. 48. Standard Enthalpy of Formation <ul><li>Formation from the composing elements: </li></ul><ul><li>Glucose : </li></ul><ul><li>6 C + 6 H 2 + 3 O 2 --> C 6 H 12 O 6 </li></ul><ul><ul><li>∆H = –1273.3 kJ </li></ul></ul>
  49. 49. Hess’ Law <ul><li>CH 4 + 2 O 2 ----> CO 2 + 2 H 2 O ∆H = -890 kJ </li></ul><ul><li>CH 4 + 2 O 2 -----> CO + 2 H 2 O + ½ O 2 </li></ul><ul><li> ∆H = -607 kJ </li></ul><ul><li>CO + 2 H 2 O + ½ O 2 ------> CO 2 + 2 H 2 O ∆H = -283 kJ </li></ul>
  50. 50. Hess’ Cycle
  51. 51. Unit 6 Recap <ul><li>Stoichiometry </li></ul><ul><li> Mole Ratios </li></ul><ul><ul><li> Yield = actual/theoretical </li></ul></ul><ul><ul><li>Limiting and extra reactants </li></ul></ul><ul><ul><li>1 mole = 22.4 Liters at STP </li></ul></ul><ul><li>Enthalpy Change ∆H </li></ul><ul><li>endothermic vs. exothermic </li></ul><ul><li>∆ H = C∆T Hess’ Law </li></ul>

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