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Example 2 C 2 H 6  C 2 H 2 + 2H 2 C-C = 347, C-H = 413, C C = 837, H-H = 436

Example 2 C 2 H 6  C 2 H 2 + 2H 2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C —C-H  H-C C-H + 2(H-H) | | H H Calculate H.

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Example 2 C 2 H 6  C 2 H 2 + 2H 2 C-C = 347, C-H = 413, C C = 837, H-H = 436

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  1. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Decide – and state - which bonds are broken  Work out energy – bond breaking is endothermic BEnd  Decide – and state -which bonds are formed  Work out energy – bond forming is exothermic FEx Remember exothermic energy is negative.  Work out H = SUM of  +  : remember  is negative.

  2. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Decide – and state - which bonds are broken Bonds broken

  3. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Decide – and state - which bonds are broken Bonds broken 6 × C-H C-C

  4. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Work out energy – bond breaking is endothermic BEnd Bonds broken H 6 × C-H C-C

  5. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Work out energy – bond breaking is endothermic BEnd Bonds broken H 6 × C-H 6 × (+413) = +2478 C-C + 347 +2825

  6. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Decide – and state -which bonds are formed Bonds broken HBonds formed 6 × C-H 6 × (+413) = +2478 C-C + 347 +2825

  7. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Decide – and state -which bonds are formed Bonds broken HBonds formed 6 × C-H 6 × (+413) = +2478 2 × C-H C-C + 347 C C +28252 × H-H

  8. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Work out energy – bond forming is exothermic FEx Bonds broken HBonds formed H 6 × C-H 6 × (+413) = +2478 2 × C-H C-C + 347 C C +28252 × H-H

  9. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H Calculate H.  Work out energy – bond forming is exothermic FEx Bonds broken HBonds formed H 6 × C-H 6 × (+413) = +2478 2 × C-H 2 × (-413) = -826 C-C + 347 C C -837 +28252 × H-H 2 × (-436) = -872 -2535

  10. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H  Work out energy – bond forming is exothermic FEx Remember exothermic energy is negative. Bonds broken HBonds formed H 6 × C-H 6 × (+413) = +2478 2 × C-H 2 × (-413) = -826 C-C + 347 C C -837 +28252 × H-H 2 × (-436) = -872 -2535  Work out H = SUM of  +  : remember  is negative and “+ (-) = - .

  11. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 H H | | H-C—C-HH-C C-H + 2(H-H) | | H H  Work out energy – bond forming is exothermic FEx Remember exothermic energy is negative. Bonds broken HBonds formed H 6 × C-H 6 × (+413) = +2478 2 × C-H 2 × (-413) = -826 C-C + 347 C C -837 +28252 × H-H 2 × (-436) = -872 -2535  Work out H = SUM of  +  : remember  is negative and “+ (-) = - . H = +2825 + (-2535) = 2825 – 2535 = 290 kJ/mol

  12. Example 2 C2H6 C2H2 + 2H2 C-C = 347, C-H = 413, C C = 837, H-H = 436 Bonds broken HBonds formed H 6 × C-H 6 × (+413) = +2478 2 × C-H 2 × (-413) = -826 C-C + 347 C C -837 +28252 × H-H 2 × (-436) = -872 -2535  Work out H = SUM of  +  : remember  is negative. H = +2825 + (-2535) = 2825 – 2535 = 290 kJ/mol Is the reaction exothermic or endothermic? Endothermic WHY is it endothermic? (His positive is NOT the answer!) Endothermicbecause the energy taken into break bonds is MORE than the energy given out in forming bonds. Now complete the energy diagram on the sheet and then check your answer with the next slide.

  13. Energy diagram for an ENDOthermic reaction Endothermicbecause the energy neededto break bonds is MORE than the energy given out in forming bonds. H ATOMS Energy of bond formingless than energy of bond breaking so bond forming arrow shorter than bond breaking arrow. Bond Forming Chemical Energy Bond Breaking PRODUCTS H REACTANTS

  14. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.

  15. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.  Decide – and state - which bonds are broken Bonds broken

  16. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.  Decide – and state - which bonds are broken Bonds broken 2 × 3 × C-C 2 × 10 × C-H 13 × O=O

  17. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.  Work out energy – bond breaking is endothermic BEnd Bonds broken H 2 × 3 × C-C 2 × 10 × C-H 13 × O=O

  18. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.  Work out energy – bond breaking is endothermic BEnd Bonds broken H 2 × 3 × C-C 6 × (+347) = +2082 2 × 10 × C-H 20×(+413) = +8260 13 × O=O 13×(+498) = +6474 +16816

  19. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.  Decide – and state -which bonds are formed Bonds broken HBonds formed 2 × 3 × C-C 6 × (+347) = +2082 2 × 10 × C-H 20×(+413) = +8260 13 × O=O 13×(+498) = +6474 +16816

  20. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.  Decide – and state -which bonds are formed Bonds broken HBonds formed 2 × 3 × C-C 6 × (+347) = +2082 16 × O=C 2 × 10 × C-H 20×(+413) = +8260 20×O-H 13 × O=O 13×(+498) = +6474 +16816

  21. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.  Work out energy – bond forming is exothermic FEx Remember exothermic energy is negative. Bonds broken HBonds formed H 2 × 3 × C-C 6 × (+347) = +208216 × O=C 2 × 10 × C-H 20×(+413) = +8260 20×O-H 13 × O=O 13×(+498) = +6474 +16816

  22. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H.  Work out energy – bond forming is exothermic FEx Remember exothermic energy is negative. Bonds broken HBonds formed H 2 × 3 × C-C 6 × (+347) = +2082 16 × O=C 16(-743) = -11888 2 × 10 × C-H 20×(+413) = +8260 20×O-H 20(-464) = -9280 13 × O=O 13×(+498) = +6474= -21168 +16816

  23. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H. Bonds broken HBonds formed H 2 × 3 × C-C 6 × (+347) = +2082 16 × O=C 16(-743) = -11888 2 × 10 × C-H 20×(+413) = +8260 20×O-H 20(-464) = -9280 13 × O=O 13×(+498) = +6474= -21168 +16816  Work out SUM of  +  : remember  is negative.

  24. Example 3) 2C4H10 + 13O2 8CO2 + 10H2O C-C = 347, C-H = 413, O=O = 498, H H H H C=O = 743, H-O = 464 | | | | 2 H-C—C—C—C-H+ 13 (O=O) 8 (O=C=O) + 10(H-O-H) | | | | H H H H Calculate H. Bonds broken HBonds formed H 2 × 3 × C-C 6 × (+347) = +2082 16 × O=C 16(-743) = -11888 2 × 10 × C-H 20×(+413) = +8260 20×O-H 20(-464) = -9280 13 × O=O 13×(+498) = +6474= -21168 +16816  Work out SUM of  +  : remember  is negative. H = +16816 + (-21168) = 16816 – 21168 = -4352 kJ/mol

  25. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.

  26. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.  Decide – and state - which bonds are broken Bonds broken

  27. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.  Decide – and state - which bonds are broken Bonds broken 2 × C-C C=C 8 × C-H Cl-Cl

  28. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.  Work out energy – bond breaking is endothermic BEnd Bonds broken H 2 × C-C C=C 8 × C-H Cl-Cl

  29. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.  Work out energy – bond breaking is endothermic BEnd Bonds broken H 2 × C-C 2 × (+347) = +694 C=C +612 8 × C-H 8×(+413) = +3304 Cl-Cl = +243 +4853

  30. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.  Decide – and state -which bonds are formed Bonds broken HBonds formed 2 × C-C 2 × (+347) = +694 C=C +612 8 × C-H 8×(+413) = +3304 Cl-Cl = +243 +4853

  31. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.  Decide – and state -which bonds are formed Bonds broken HBonds formed 2 × C-C 2 × (+347) = +694 3 × C-C C=C +612 8 × C-H 8 × C-H 8×(+413) = +3304 2×C-Cl Cl-Cl = +243 +4853

  32. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.  Work out energy – bond forming is exothermic FEx Remember exothermic energy is negative. Bonds broken HBonds formed H 2 × C-C 2 × (+347) = +694 3 × C-C C=C +612 8 × C-H 8 × C-H 8×(+413) = +3304 2×C-Cl Cl-Cl = +243 +4853

  33. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H.  Work out energy – bond forming is exothermic FEx Remember exothermic energy is negative. Bonds broken HBonds formed H 2 × C-C 2 × (+347) = +694 3 × C-C 3 x (-347) = -1041 C=C +612 8 × C-H 8×(-413) = -3304 8 × C-H 8×(+413) = +3304 2×C-Cl 2 x (-338) = -676 Cl-Cl = +243-5021 +4853

  34. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H. Bonds broken HBonds formed H 2 × C-C 2 × (+347) = +694 3 × C-C 3 x (-347) = -1041 C=C +612 8 × C-H 8×(-413) = -3304 8 × C-H 8×(+413) = +3304 2×C-Cl 2 x (-338) = -676 Cl-Cl = +243-5021 +4853  Work out SUM of  +  : remember  is negative.

  35. Example 4) C4H8 + Cl2 C4H8Cl2 C-C = 347, C=C = 612, C-H = 413, Cl-Cl = 243 C-C = 347, C-H = 413, C-Cl = 338 H H H H H H H H | | | | | | | | H—C—C=C—C—H+ Cl—Cl H—C—C—C—C—H | | | | | | H H H Cl Cl H Calculate H. Bonds broken HBonds formed H 2 × C-C 2 × (+347) = +694 3 × C-C 3 x (-347) = -1041 C=C +612 8 × C-H 8×(-413) = -3304 8 × C-H 8×(+413) = +3304 2×C-Cl 2 x (-338) = -676 Cl-Cl = +243-5021 +4853  Work out SUM of  +  : remember  is negative. H = +4853 + (-5021) = 4853 – 5021 = -168 kJ/mol

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