COMBUSTION of PROPANE

COMBUSTION of PROPANE

1. What does the reaction involve? (reactants and products)

Reaction needs to be balanced

+C3H8 + O2 CO2 H2O

+C3H8 + O2 CO2 H2O3 45


2. Envision the “structural formulas” for the reactants and the products (from your

knowledge of Lewis structures)

C C C HH

H H H

HHH

O O CO O OH H


3. List the bonds broken (between which atoms)

C C C HH

H H H

HHH

O O

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


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

in 5 x O2 5 x O=O bonds

Therefore,


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/moleO=O bonds: 498 kJ/mole


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 kJ5 x 498 kJ/mole = 2490 kJ

Remember, there are2 x C-C bonds8 x C-H bonds5 x O=O bonds (after balancing the equation)

TOTAL: 6690 kJ


5. We now need to look at the products and the new bonds formed as a result of the reaction

CO O OH H

These are the new bonds which are formed


new bonds formed within the products

In CO2 2 x C=O bondsIn 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

In 3 x CO2 3 x 2 x C=O bondsIn 4 x H2O 4 x 2 x O-H bonds

therefore


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


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 are3 x 2 x C=O bonds4 x 2 x O-H bonds

being formed (after balancing the equation)

TOTAL: 8554 kJ


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


8. What if we wish to determine the heat of reaction, DH, per gram of propane, C3H8 undergoing complete combustion

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

In 1 mole of C3H8 there are 44 g of C3H8


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 togethermight further clarify the type of calculation involved in converting the amount of energy from per mole to per gram

kJ

1 mole of C3H8

1 mole of C3H8

44 g

1864x

COMBUSTION of PROPANE

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