COMBUSTION of PROPANE 1.What does the reaction involve? (reactants and products) Reaction needs to be balanced.

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COMBUSTION of PROPANE 1.What does the reaction involve? (reactants and products) Reaction needs to be balanced

COMBUSTION of PROPANE 2.Envision the “structural formulas” for the reactants and the products (from your knowledge of Lewis structures)

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

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,

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

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

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

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

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

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

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

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

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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