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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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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.


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


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


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


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


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


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


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


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


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 “+ (-) = - .


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


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.


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


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.


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


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


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


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


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


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


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


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


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.


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


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.


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


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


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


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


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


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


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


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


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.


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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