Chem Unit6

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Transcript of Chem Unit6

  • 1.Stoichiometry (stoicheion, meaning element]) & (metron, meaning measure)

2. Balancing Chemical Equations

  • 2 Mg +O 2 ->2 MgO

3. Balancing Chemical Equations

  • 2 Mg +O 2 ->2 MgO
  • 1.2 x 10 24+6x 10 23 form1.2 x 10 24formula units
  • atomsO 2moleculesMgO
  • Mg

4. Balancing Chemical Equations

  • 2 Mg +O 2 ->2 MgO
  • 2 moles+1 moleform2 moles
  • Mg O 2 MgO
  • Mole Ratio

5. Balancing Chemical Equations

  • 2Mg + O 2 ->2MgO
  • 48.6 g+32 gform80.6 grams
  • Mg O 2 MgO
  • Mass
  • Balance
  • The total mass of the product(s) must equal the total mass of the reactants

6. Yield

  • Building Bicycles

7. Yield

  • Building Bicycles
  • 1 Frame + 1 Seat + 1 Handlebar + 2 Pedals + 2 Wheels

8. Finding Product Quantities

  • How much O 2gas is formed when
  • 49.89 g KClO 3decomposes?
  • KClO 3 ->KCl+O 2

9. Balance the Equation

  • How much O 2gas is formed when
  • 49.89 g KClO 3decomposes?
  • 2KClO 3 ->2KCl+3O 2

10. Balance the Equation

  • How much O 2gas is formed when
  • 49.89 g KClO 3decomposes?
  • 2KClO 3 ->2KCl+3O 2
  • 1 mole
  • 49.89 g KClO 3x 122.45 g = 0.41 moles KClO 3
  • 3 moles O2
  • 2 moles KClO3 x0.41 moles KClO 3 =0.615 moles O 2

11. Balance the Equation

  • How much O 2gas is formed when
  • 49.89 g KClO 3decomposes?
  • 2KClO 3 ->2KCl+3O 2
  • 49.89 g KClO 3?
  • 0.615 moles O 2x 32 g/molee= 19.68 gramsO 2

12. Standard Temperature & Pressure

  • STP
  • Temperature = 0C (273.15K; 32.00F)
  • Pressure = 101.325kPa (1atm; 760 Torr)
  • For Gases: Volume = 22.4 Liters/mole

13. Balance the Equation

  • How much O 2gas is formed when
  • 49.89 g KClO 3decomposes?
  • 2KClO 3 ->2KCl+3O 2
  • 49.89 g KClO 3?
  • 0.615 moles O 2x 22.4 L/mole=13.776 Liters O 2

14. Yield

  • Building Bicycles
  • 1 Frame + 1 Seat + 1 Handlebar + 2 Pedals + 2 Wheels
  • = 1 bicycle

15. Yield

  • Theoretical Yield
  • How much O 2gas is formed when
  • 49.89 g KClO 3decomposes?
  • 2KClO 3 ->2KCl+3O 2
  • 49.89 g KClO 36.12 Liters O 2
  • What if only 5.5 Liters of O 2are produced?

16. Yield

  • Theoretical Yield
  • 2KClO 3 ->2KCl+3O 2
  • 49.89 g KClO 36.12 Liters O 2
  • What if only 5.5 Liters of O 2are produced?
  • 5.5 Liters
  • 6.12 Litersx 100%=89.9% actual yield

17. Gas Yield

  • Nitroglycerin decomposes to
  • produce N 2 , O 2 , CO 2and H 2 O

18. Gas Yield

  • 4 C 3 H 5 N 3 O 9-> 6 N 2+ O 2+ 12 CO 2+ H 2 O
  • 227 ghow much gas results?

19. Gas Yield

  • 4 C 3 H 5 N 3 O 9-> 6 N 2+ O 2+ 12 CO 2+ H 2 O
  • 1 mole
  • 227 g x227 g=1 mole nitroglycerin

20. Gas Yield

  • 4 C 3 H 5 N 3 O 9-> 6 N 2+ O 2+ 12 CO 2+ H 2 O
  • 1.5 moles N 2
  • 1 mole nitroglycerin->0.25 mole O 2
  • 3 moles CO 2
  • 0.25 mole H 2 O

21. Gas Yield

  • 4 C 3 H 5 N 3 O 9-> 6 N 2+ O 2+ 12 CO 2+ H 2 O
  • 1.5 moles N 2x22.4 L/ mole=33.6 L
  • 0.25 mole O 2x 22.4 L/mole= 5.6 L
  • 3 moles CO 2x 22.4 L/mole= 67.2 L
  • 0.25 mole H 2 O x 22.4 L/mole= 5.6 L
  • 112 Liters !

22. Making Aspirin

  • White willow tree bark = pain reliever

23. Making Aspirin

  • Salicylic acidacetylsalicylic acid
  • + acetic anhydride -> and acetic acid
  • C 7 H 6 O 3+ C 2 H 6 O 3->C 9 H 8 O 4+ C 2 H 4 O 2
  • each aspirin tablet 325 mg
  • How much salicylic acid would be needed if we assume 90% yield?

24. Making Aspirin

  • C 7 H 6 O 3+ C 2 H 6 O 3->C 9 H 8 O 4+ C 2 H 4 O 2
  • 0.325 gx1 mole/180 g=0.0018 moles of aspirin
  • 90 % of 0.0018= 0.9 x 0.0018 = 0.00162 moles
  • 0.00162 moles x 138 g/mole=0.224 grams of salicylic acid

25. Rocket Fuel

  • N 2 H 4 (l) +N 2 O 4 (l) -> N 2 (g) +H 2 O (g)
  • Is the reaction balanced?

26. Rocket Fuel

  • 2N 2 H 4 (l) +N 2 O 4 (l) ->3N 2 (g) +4H 2 O (g)
  • What is the mole ratio
  • of N 2 H 4to N 2 ?

27. Rocket Fuel

  • 2N 2 H 4 (l) +N 2 O 4 (l) ->3N 2 (g) +4H 2 O (g)
  • What is the mole ratio
  • of N 2 H 4to N 2 ?
  • 3 moles N 2
  • 2 moles N 2 H 4

28. Rocket Fuel

  • 2N 2 H 4 (l) +N 2 O 4 (l) ->3N 2 (g) +4H 2 O (g)
  • How many Liters of N 2 will be
  • produced from 16000 g of N 2 H 4 ?
  • 16000 g x 1 mole=500 moles N 2 H 4
  • 32g
  • 3 moles N 2
  • 500 moles N 2 H 4x2 moles N 2 H 4
  • = 750.0 moles N 2
  • =16800 Liters of N 2

29. Rocket Fuel

  • 2N 2 H 4 (l) +N 2 O 4 (l) ->3N 2 (g) +4H 2 O (g)ENERGY
  • How many Liters of N 2 will be
  • produced from 16000 g of N 2 H 4 ?
  • 16000 g x 1 mole=500 moles N 2 H 4
  • 32g
  • 3 moles N 2
  • 500 moles N 2 H 4x2 moles N 2 H 4
  • = 750.0 moles N 2
  • =16800 Liters of N 2

30. Chemical Change

  • Energy is absorbed or released
  • Exothermic vs. Endothermic

31. Enthalpy H

  • Total energy of a system:
    • Internal energy+(pressure) x (Volume)

32. Enthalpy H

  • Only way to measure H is when heat is added or removed:
  • Molar Enthalpy Change: H = C T
  • (Recall Specific Heat Capacity: Q = m C T)

33. Molar Enthalpy Change

  • H=C T
  • What is H if an aluminum can is cooled from 25 C to 4 C?

34. Temperature Scales 35. Molar Enthalpy Change

  • H=C T
  • What is H if an aluminum can is cooled from 25 C to 4 C?(C for Al = 24.2 J/K mole)
  • 25 C=298 K
  • 4 C=277 K
  • H=C T =24.2 J/K mole (21 K)
  • =508.2 J/mole

36. Enthalpy Change

  • Negative=Exothermic
  • C(s) + 1/2 O 2 (g) -> CO(g)H = - 110.5 kJ
  • Positive=Endothermic
  • C 2 H 6 (g) -> 2C(s) + 3H 2 (g)H = + 4.83 kJ

37. Enthalpy Change (Endothermic) A+B---->C+D 38. Enthalpy Change (Exothermic) W + X-->Y+Z 39. Enthalpy in Chemical Changes 40. Enthalpy in Chemical Changes 41. Enthalpy in Chemical Changes 42. Stoichiometric Enthalpy

  • H = ? When 4.8 g C reacts with O 2 ?
  • C+O 2 ->CO 2 H = -393.5 kJ/mole
  • 4.8 g x 1mole/12g = 0.4 moles
  • 0.4 moles x (-393.5 kJ/mole)
  • =157.4 kJ

43. Hess Law

  • Total H = Sum of H for all steps
  • Example:
  • H 2 S(g)+2O 2 (g) ->SO 3 (g)+H 2 O(l)
  • H = ???????

44. Hess Law

  • H 2 S (g)+2O 2 (g)-> H 2 SO 4 (l)H= -628KJ
  • H 2 SO 4 (l) -> SO 3 (g)+H 2 O (g)H= +164KJ
  • H 2 O (g)->H 2 O (l)H= -88KJ
  • H 2 S (g)+2O 2 (g)->SO 3 (g)+H 2 O (l)H =?

45. Hess Law 46. Germain Henri Hess

  • Enthalpy change (H) dependsonlyon the initial and final states of the reaction
  • (not on the intermediate stages)
  • Hess cycle

47. Hess Law

  • CH 4 +2 O 2 ----> CO 2+ 2 H 2 OH = ??
  • CH 4 +2 O 2 ----->CO + 2 H 2 O + O 2
  • H = -607 kJ
  • CO + 2 H 2 O + O 2------>CO 2 +2 H 2 OH = -283 kJ

48. Standard Enthalpy of Formation

  • Formation from the composing elements:
  • Glucose :
  • 6 C+6 H 2 +3 O 2 --> C 6 H 12 O 6
    • H = 1273.3 kJ

49. Hess Law

  • CH 4 +2 O 2 ----> CO 2+ 2 H 2 OH = -890 kJ
  • CH 4 +2 O 2 ----->CO + 2 H 2 O + O 2
  • H = -607 kJ
  • CO + 2 H 2 O + O 2------>CO 2 +2 H 2 OH = -283 kJ

50. Hess Cycle 51. Unit 6 Recap

  • Stoichiometry
  • Mole Ratios
    • Yield = actual/theoretical
    • Limiting and extra reactants
    • 1 mole = 22.4 Liters a