Combustion reaction
C3H6 + 9/2 O2 = 3 CO2 + 3 H2O
6a)
Bonds destroyed (C3H6):
1 x C=C mean bond enthalpy = 612 x 1 = 612
1 x C-C mean bond enthalpy = 348 x 1 = 348
6 x C-H mean bond enthalpy = 412 x 6 = 2472
Bonds destroyed (O2):
9/2 x O=O mean bond enthalpy = 496 x 9/2 = 2232
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Total enthalpy of bonds destroyed = 5664
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Bonds created (CO2)
3 x 2 x C=O mean bond enthalpy = 743 x 6 = 4458
Bonds created (H2O)
3 x 2 x O-H mean bond enthalpy = 463 x 6 = 2778
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Total enthalpy of bonds created = 7236
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Since the formation of a bond is an exothermic process the individual atoms have a higher enthalpy than the atoms bonded together.
So the enthalpy of the reactants relative to the individual non-bonded atoms is:
-5664
(The negative sign means that to obtain non-bonded atoms I you would have to supply heat ("enthalpy") to increase the enthalpy of the system to zero. In this calculation the enthalpy of the individual unbonded atoms (or of the elements in their standard states, see answer to 6b) below) is considered to be zero.
In the same way the enthalpy of the products of the combustion reaction is given by:
-7236
Overall change in enthalpy on burning C3H6
= enthalpy of products - enthalpy of reactants
= -7236 - (-5664) = -7236 + 5664
= -1572 enthalpy units / mol C3H6 combusted
6b) For enthalpy change calculations involving chemical reaction the enthalpy of elements in their standard state is considered by definition to be zero. The bond enthalpies given in the question are presumably also relative to the elements C, H, O in their standard states. The standard state for a gaseous element is usually taken to be the element at 1 bar pressure and 25°C or some other temperature. It is important that all the enthalpy data used in a calculation is relative to the same standard states of the elements to get the right answer.
6c) Calculation of enthalpy of combustion from enthalpy of formation data
Combustion reaction
C3H6 + 9/2 O2 = 3 CO2 + 3 H2O
Enthalpies of formation of reactants:
1 x C3H6(g) = 1 x 20 = 20
9/2 x O2(g) = 9/2 x 0 (by definition) = 0 (by definition)
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Sum of enthalpies of formation of reactants = 20
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Enthalpies of formation of products:
3 x CO2(g) = 3 x -394 = - 1182
3 x H2O(g) = 3 x -242 = -726
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Sum of enthalpies of formation of products = -1908
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Change in enthalpy between reactants and products = enthalpy of combustion
= enthalpy of formation of products - enthalpy of formation of reactants
= -1908 - 20
=-1928 enthalpy units / mol C3H6 combusted
6d) This value is more accurate than that calculated in 6a) because the bond enthalpies are average values over many different compounds and the actual bond enthalpy will vary depending on the other atoms that near the bond under consideration. The enthalpies of formation given for C3H6, H2O and CO2 are however specific for these compounds (and probably originally calculated from measurements of the enthalpy of combustion!)