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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-HH-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-HH-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-HH-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-HH-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-HH-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-HH-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-HH-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-HH-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-HH-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-HH-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-HH-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