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COMBUSTION of PROPANE 1.What does the reaction involve? (reactants and products) Reaction needs to be balanced
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COMBUSTION of PROPANE 2.Envision the “structural formulas” for the reactants and the products (from your knowledge of Lewis structures)
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COMBUSTION of PROPANE 3.List the bonds broken (between which atoms) bonds within the reactants are expected to be broken In C 3 H 8 2 x C-C bonds 8 x C-H bonds In O 2 1 x O=O bond
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COMBUSTION of PROPANE The numbers and types of bonds within the reactants which are expected to be broken are… In C 3 H 8 2 x C-C bonds 8 x C-H bonds In O 2 1 x O=O bond However, we need to check the coefficient in front of each reactant (look at the balanced equation) There are 5 moles of O 2 reacting with each mole of propane in 5 x O 2 5 x O=O bonds Therefore,
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COMBUSTION of PROPANE 4.Find the Bond Dissociation Energies (BDE) associated for each bond (from Table 4.4) BDE’s for C-C bonds: 356 kJ/mole C-H bonds: 436 kJ/mole O=O bonds: 498 kJ/mole
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COMBUSTION of PROPANE Here is the energy required to break all the bonds mentioned above: 2 x 356 kJ/mole = 712 kJ 8 x 436 kJ/mole = 3488 kJ 5 x 498 kJ/mole = 2490 kJ Remember, there are 2 x C-C bonds 8 x C-H bonds 5 x O=O bonds (after balancing the equation) TOTAL: 6690 kJ
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COMBUSTION of PROPANE 5.We now need to look at the products and the new bonds formed as a result of the reaction These are the new bonds which are formed
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COMBUSTION of PROPANE new bonds formed within the products In CO 2 2 x C=O bonds In H 2 O 2 x O-H bonds However, we need to check the coefficient in front of each reactant (look at the balanced equation) There are 3 moles of CO 2 being formed (for each mole of propane) and 4 moles of H 2 O In 3 x CO 2 3 x 2 x C=O bonds In 4 x H 2 O 4 x 2 x O-H bonds therefore
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COMBUSTION of PROPANE 6.Find the Bond Dissociation Energies (BDE) associated for each bond being formed (from Table 4.4) BDE’s for C=O bonds: 803 kJ/mole O-H bonds: 467 kJ/mole
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COMBUSTION of PROPANE Here is the energy released when all the bonds mentioned above are formed: 6 x 803 kJ/mole = 4818 kJ 8 x 467 kJ/mole = 3736 kJ Remember, there are 3 x 2 x C=O bonds 4 x 2 x O-H bonds being formed (after balancing the equation) TOTAL: 8554 kJ
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COMBUSTION of PROPANE 7.Subtract the total energy released from the total energy required to break the bonds to determine the heat of reaction, H H = 8554 kJ - 6690 kJ = 1864 kJ Let’s focus on the units now: This is 1864 kJ per mole of propane, C 3 H 8 undergoing complete combustion
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COMBUSTION of PROPANE 8.What if we wish to determine the heat of reaction, H, per gram of propane, C 3 H 8 undergoing complete combustion Therefore, the H value we calculated previously is associated with 44 g of C 3 H 8 H = 1864 kJ per 44 g of propane undergoing complete combustion In 1 mole of C 3 H 8 there are 44 g of C 3 H 8
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COMBUSTION of PROPANE If 44 g of C 3 H 8 releases 1864 kJ of energy, then 1.0 g of C 3 H 8 will release 1/44 of that amount H = 1864/44 kJ per 1.0 g of propane = 42.4 kJ/g of propane viewing the following units together might further clarify the type of calculation involved in converting the amount of energy from per mole to per gram
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