COMBUSTION of PROPANE

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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 C3H8 2 x C-C bonds
8 x C-H bonds
In O2
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 C3H8 2 x C-C bonds
8 x C-H bonds
In O2
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 O2 reacting with each mole of propane
Therefore, in 5 x O2
5 x O=O bonds
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
Remember, there are
2 x C-C bonds
8 x C-H bonds
5 x O=O bonds (after balancing the equation)
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
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 CO2 2 x C=O bonds
In H2O 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 CO2 being formed (for each mole of propane)
and
4 moles of H2O
therefore
In 3 x CO2
In 4 x H2O
3 x 2 x C=O bonds
4 x 2 x O-H bonds
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
Remember, there are
3 x 2 x C=O bonds
4 x 2 x O-H bonds
being formed (after balancing the equation)
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
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, DH
DH = 8554 kJ - 6690 kJ
= 1864 kJ
Let’s focus on the units now:
This is 1864 kJ per mole of propane, C3H8 undergoing
complete combustion
COMBUSTION of PROPANE
8.
What if we wish to determine the heat of
reaction, DH, per gram of propane, C3H8
undergoing complete combustion
In 1 mole of C3H8 there are 44 g of C3H8
Therefore, the DH value we calculated previously is
associated with 44 g of C3H8
DH = 1864 kJ per 44 g of propane
undergoing complete combustion
COMBUSTION of PROPANE
If 44 g of C3H8 releases 1864 kJ of energy,
then 1.0 g of C3H8 will release 1/44 of that amount
DH = 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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