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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 - PowerPoint PPT Presentation


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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 H H | | H-C —C-H  H-C C-H + 2(H-H) | | H H Calculate H.

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slide1

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.

slide2

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

slide3

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

slide4

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

slide5

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

slide6

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

slide7

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

slide8

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

slide9

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

slide10

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 “+ (-) = - .

slide11

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

slide12

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.

slide13

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

slide14

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.

slide15

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

slide16

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

slide17

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

slide18

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

slide19

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

slide20

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

slide21

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

slide22

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

slide23

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.

slide24

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

slide25

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.

slide26

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

slide27

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

slide28

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

slide29

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

slide30

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

slide31

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

slide32

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

slide33

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

slide34

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.

slide35

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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