The Energy Content Of Different Fuels

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Chemistry Coursework

The Energy Content Of Different Fuels

PLAN:

Introduction:

Different fuels produce different amounts of energy when they are burnt in oxygen. I

am going to investigate the different energy contents of alcohols. I am going to use

just alcohols because when I find out the energy per mole produced by different

alcohols I can compare them. I am going to use Methanol, Ethanol, Propanol,

Butanol and Pentanol.

When fuels are burnt in oxygen, water and carbon dioxide are formed. The

combustion also produces heat. This is because when the alcohol burns energy

from the heat is used to break the bonds and when they form carbon dioxide and

water they are making bonds and energy is released. This energy becomes heat. I

am going to investigate how much energy is produced by each alcohol I burn and

compare the results when I have finished.

?

Prediction:

For my prediction I am going to calculate how much energy should be produced by

each alcohol. I am going to do this by using a secondary source of information. This

is a book called 'AS Chemistry' published by 'Collins' by 'Nicholls and Radcliffe'. I

will uses this by taking the values given in this book of how many kJ/mol are

required to break/make certain bonds. The information that I am going to use is:

To break a C?H (Carbon to Hydrogen) bond it requires 413 Kjmol-1 of energy.

To break a C?O (Carbon to Oxygen) bond it requires 360 Kjmol-1 of energy.

To break an O?H (Oxygen to Hydrogen) bond it requires 463 Kjmol-1 of energy.

To break a C?C (carbon to Carbon) bond it requires 347 Kjmol-1 of energy.

To break an O=O (double oxygen) bond it requires 498 Kjmol-1 of energy.

To make a C=O (Carbon to Oxygen double) bond it produces -743Kjmol-1 of

energy.

To make an O?H (oxygen to Hydrogen) bond it produces - 463 kjmol-1 of energy.

I am now going to use this information to calculate the total energy required to

break the bonds on the left hand side of each equation and then how much energy is

produced when it makes the bonds on the right hand side and therefore predict how

much energy per mole is going to be produced at the end.

Methanol:

Methanol + Oxygen = Carbon dioxide + water

2CH3OH + 3O2 ? 2CO2 + 4H2O

On the Left-Hand Side of the equation:

2 x ?3 x C?H (413 kjmol-1)

2 x ?1 x C?O (360 kjmol-1)

2 x ?1 x O?H (463 kjmol-1)

3 x O=O (498 kjmol-1)

Total energy required to break the bonds=

(6 x 413) + (2 x 360) + (2 x 463) + (3 x 498) =

2478 + 720 + 926 + 1494 = 5618 kjmol-1

On the Right hand side of the equation:

2 x? 2 x C=O (-743 kjmol-1)

4 x ?2 x O?H (-463 kjmol-1)

Total energy produced to make the bonds=

(4 x ?743) + (8 x ?463) =

(?2972) + (?3704) = ? 6676 kjmol-1

Therefore the expected enthalpy of combustion of Methanol is=

(-6676 + 5618) = ? 1058 kjmol -1 for 2 moles because the balanced equation had 2

moles of Methanol. Therefore, the prediction is

-1058 / 2 = -529 kjmol-1

Ethanol:

Ethanol + Oxygen = Carbon dioxide + water

C2H5OH + 3O2 = 2CO2 + 3H2O

On the left-hand side of the equation:

5 x C?H (413 kjmol-1)

x C?C (347 kjmol-1)

x C?O (360 kjmol-1)

x O?H (463 kjmol-1)

3 x ?1 x O=O (498 kjmol-1)

Total energy required to break the bonds:

(5 x 413) + (1 x 347) + (1 x 360) + (1 x 463) + (3 x 498) =

2065 + 347 + 360 + 463 + 1494 = 4729 kjmol-1

On the right-hand side of the equation:

2 x ?2 x C=O (-743 kjmol-1)

3 x ?2 x O?H (-463 kjmol-1)

Total energy produced by making the bonds =

(4 x -743) + (6 x -463) =

-2972 + -2778 = -5750 Kjmol-1

Therefore the expected enthalpy of combustion of Ethanol is:

(-5750) + 4729 = -1021 kjmol-1

Propanol:

Propanol + Oxygen = Carbon dioxide + water

C3H7OH + 41/2O2 = 3CO2 + 4H2O

= 2C3H7OH + 9O2 = 6CO2 + 8H20

On the left-hand side of the equation:

2 x ?7 x C-H (413 kjmol-1)

2 x ?2 x C-C (347 kjmol-1)

2 x ?1 x C-O (360 kjmol-1)

2 x ?1 x O-H (463 kjmol-1)

9 x O=O (498 kjmol-1)

Total energy required to break the bonds:

(14 x 413) + (4 x 347) + (2 x 360) + (2 x 463) + (9 x 498) =

5782 + 1388 + 720 + 926 + 4482 = 13298 kjmol-1

On the right-hand side of the equation:

6 x ?2 x C=O (-743 kjmol-1)

8 x ?2 x O-H (-463 kjmol-1)

Total energy produced by making bonds:

(12 x -743) + (16 x -463) =

-8916 + -7408 = -16324 kjmol-1

Therefore the expected enthalpy of combustion for Propanol is:

(-16324) + 13298 = -3026 kjmol-1, for 2 moles because the balanced equation had

2 moles of Propanol. Therefore, the prediction is =

-3026 / 2 = -1513 kjmol-1

Butanol:

Butanol + Oxygen = Carbon dioxide + water

C4H9OH + 6O2 = 4CO2 + 5H2O

On the left-hand side of the equation:

9 x C-H (413 kjmol-1)

3 x C-C (347 kjmol-1)

x C-O (360 kjmol-1)

x O-H (463 kjmol-1)

6 x O=O (498 kjmol-1)

Total energy required to break the bonds:

(9 x 413) + (3 x 347) + (1 x 360) + (1 x 463) + (6 x 498) =

3717 + 1041 + 360 + 463 + 2988 = 8569 kjmol-1

On the right-hand side of the equation:

4 x ?2 x C=O (-743 kjmol-1)

5 x ?2 x O-H (-463 kjmol-1)

Total energy produced by making bonds:

(8 x -743) + (10 x -463) =

-5944 + -4630 = -10574 kjmol-1

Therefore the expected enthalpy of combustion for Butanol is:

(-10754) + 8569 = -2005 kjmol-1

Pentanol:

Pentanol + Oxygen = Carbon dioxide + water

C5H11OH + 71/202 = 5CO2 + 6H2O

= 2C5H110H + 15O2 = 10CO2 + 12H2O

On the left-hand side of the equation:

2 x ?11 x C-H (413 kjmol-1)

2 x ?4 x C-C (347 kjmol-1)

2 x ?1 x C-O (360 kjmol-1)

2 x ?1 x O-H (463 kjmol-1)

5 x O=O (498 kjmol-1)

Total energy required to break the bonds:

(22 x 413) + (8 x 347) + (2 x 360) + (2 x 463) + (15 x 498) =

9086 + 2776 + 720 + 926 + 7470 = 20978 kjmol-1

On the right-hand side of the equation:

0 x ?2 x C=O (-743 kjmol-1)

2 x ?2 x O-H (-463 kjmol-1)

Total energy produced by making bonds:

(20 x -743) + (24 x -463) =

( -14860) + (-11112) = -25972 kjmol-1

Therefore the expected enthalpy of combustion for Pentanol is:

(-25972) + 20978 = -4994 kjmol-1 , for 2 moles because the balanced equation

had 2 moles of Propanol. Therefore, the prediction is =

-4984 / 2 = -2497 kjmol-1

This is therefore a table of all my calculations together:

Fuel

Expected Enthalpy of

Combustion

Methanol

-529 kjmol-1

Ethanol

-1021 kjmol-1

Propanol

-1513 kjmol-1

Butanol

-2005 kjmol-1

Pentanol

-2447 kjmol-1

With these results there is a special pattern. If you find the differences between the

different enthalpy of combustions and relate them to the amount of Carbons I have

been able to find a formula that relates the amount of Carbons to the enthalpy of

combustion. To do this I found that there is exactly 492 kjmol-1 between each one as

a carbon is added. This means that the formula had something to do with 492. I

found that if you times the number of carbons by 492 and then add 37 you get the

expected enthalpy of combustion. This can therefore be written as:

492n + 37 = enthalpy of combustion. ('n' is the number of Carbons in the fuel.)
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Although my calculations are correct I do not expect my results to come out like this

because I will not be able to get the experiment totally accurate. Mainly Because

some of the heat will have been lost to the surrounding air and my method of finding

the amount of energy produced may not be completely accurate. The

measurements that I take may also be not exact due to human error.

The amount of Carbons in the fuel changes its amount of energy that it will produce.

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