Combustion:
When a hydrocarbon is burnt is a plentiful supply of air it reacts with the oxygen in the air to form carbon dioxide and hydrogen. This is an example of combustion:
Hydrocarbon + oxygen → carbon dioxide + water
For e.g. when methane (natural gas) is burnt:
CH4 (g) + 2O2 → CO2 (g) 2H2O (l)
Where the energy does comes from?
When a fuel is burnt the reaction can be considered to take place in two stages. In the first stage covalent bonds between the atoms in the fuel molecules and oxygen atoms combine together and new covalent bonds are formed for e.g. consider the combustion of propane:
Propane + oxygen → carbon dioxide + water
The alcohols i will be using:
Hypothesis:
Below is a table of values, showing the energy values of each of the bonds that are needed to work out my predictions. Each pair of bonds has its own particular value.
Name of Alcohol: Methanol
Balanced chemical equation: CH3OH + 1.5O2 → CO2 + 2H2O
STEP 1
The equation simplified and represented as bonds
1x + (1.5 x O = O) → (1 x O = C =O) + (2 x H – O –H)
STEP 2
As you can see when the equation is simplified some of the bonds are repeated. Take CH3OH as an example.
It has 3 (C –H) bonds
1 (O – H) bond
and 1 (C – O) bond
So if this equation was represented in terms of the number of bonds:
(3 x C – H) + (1 x C – O) + (1 x O – H) + (1.5 x O = O) → (2 x C = O) + (4 x O – H)
STEP 3
As I have mentioned before each of the bonds has a particular value, listed in the table in the previous page. So by replacing the bonds by values I would get:
(3 x 413) + (1 x 358) + (1 x 464) + (1.5 x 497) → (2 x 745) + (4 x 464)
STEP 4
Now what I have to do is calculate the values on both sides of the equation
3x 413= 1239
1x 358= 358
1x 464= 464
1.5x 497= 745.5
2x 745= 1490
4x 464= 1856
1239+ 358+ 464+ 745.5= 2806.5
1490+ 1856= 3346
2806.5 → 3346
STEP 5
Now I have values for both sides of the equation, and to find out how much heat energy is released I have to find the difference of the values.
3346 – 2806.5 = 539.5
HEAT ENERGY RELEASED BY METHANOL= -539.5 kj/mol
Name of Alcohol: Ethanol
Balanced chemical equation: C2H5OH + 3O2 → 2CO2 +3H2O
STEP 1
This equation can be represented as:
1x + (3 x O = O) → (2 x O = C =O) + (3 x H – O –H)
STEP 2
Represented in terms of bonds:
(5 x C – H) + (1 x C – C) + (1 x C – O) + (1 x O – H) + (3 x O = O) → (4 x C = O) + (6 x O – H)
STEP 3
Replaced by values:
(5 x 413) + (1 x 346) + (1 x 358) + (1 x 464) + (3 x 497) → (4 x 745) + (6 x 464)
STEP 4
After doing calculations:
4742 → 5764
STEP 5
How much heat energy is released:
5764 – 4742 = 1040
HEAT ENERGY RELEASED BY ETHANOL: 1040 kj/mol
Name of alcohol: Propan-1-ol
Balanced chemical equation: C3H7OH + 4.5O2 → 3CO2 +4H2O
STEP 1
This equation can be represented as:
1x + (4.5x O = O) → (3x O = C =O) + (4x H – O –H)
STEP 2
Represented in terms of bonds:
(7x C – H) + (2x C – C) + (1x C – O) + (1x O – H) + (4.5x O = O) → (6x C = O) + (8x O – H)
STEP 3
Replaced by values:
(7x 413) + (2x 346) + (1 x 358) + (1 x 464) + (4.5x 497) → (6x 745) + (8x 464)
STEP 4
After doing calculations:
6641.5 → 8182
STEP 5
How much heat energy is released:
8182 – 6641.5 = 1540.5
HEAT ENERGY RELEASED BY PROPAN-1-OL= 1540.5 kj/mol
Name of alcohol: Butan-1-ol
Balanced chemical equation: C4H9OH + 6O2 → 4CO2 +5H2O
STEP 1
This equation can be represented as:
1x + (6x O = O) → (4x O = C =O) + (5x H – O –H)
STEP 2
Represented in terms of bonds:
(9x C – H) + (3x C – C) + (1x C – O) + (1x O – H) + (6x O = O) → (8x C = O) + (10x O – H)
STEP 3
Replaced by values:
(9x 413) + (3x 346) + (1 x 358) + (1 x 464) + (4.5x 497) → (8x 745) + (10x 464)
STEP 4
After doing calculations:
8559 → 10600
STEP 5
How much heat energy is released:
10600 – 8559 = 2041
HEAT ENERGY RELEASED BY BUTAN-1-OL= 2041 kj/mol
Name of alcohol: Pentan-1-ol
Balanced chemical equation: C5H11OH + 6.5 O2 → 5CO2 +6H2O
STEP 1
This equation can be represented as:
H H H H
| | | |
1x H -C-C-C-C-O-H + (7.5x O = O) → (5x O = C =O) + (6x H – O –H)
| | | |
H H H H
STEP 2
Represented in terms of bonds:
(11x C – H) + (3C – C) + (1x C – O) + (1x O – H) + (7.5 = O) → (10C = O) + (12 O – H)
STEP 3
Replaced by values:
(11x413) + (3x 346) + (1 x 358) + (1 x 464) + (7.5x 497) → (10x745) + (12x 464)
STEP 4
After doing calculations:
10130.5 → 13018
STEP 5
How much heat energy is released:
13018 - 10130.5= 2887.5
HEAT ENERGY RELEASED BY PENTAN-1-OL=2887.5Kj/mol
From doing these calculations, I can say that:
“As the alcohol increases the number of carbons more energy is produced to break the bonds” so I predict that pentan-1-ol will produce the most energy to break its bonds since it has the longest carbon chain and methanol will produce the lowest amount of heat energy as it has the shortest carbon chain.
I think it will happen for the following scientific reasons:
As the size of the carbon chain grows, more bonds are added to the structure of the alcohol. This means that each time, more energy from the surroundings must be extracted in order to break these starting bonds (endothermic stage). Yet, the more energy that is taken to break the bonds, the more energy is used to form the product's bonds and this makes the energy of the products greater than that of the reactants. When the products are formed, excess energy is released which is mainly due to the amount of C = O bonds being made. The products of methanol have 2 C = O bonds and 4 O - H bonds. Since a C = O bond has a high energy value of 745, many of them will ensure that the energy of the products is greater than that of the reactants.
Below you can see the graph showing the how much energy is given off with the number of carbons. You can see that the energy increases as the number of carbons increases.
Apparatus:
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5 Spirit lamps – each containing one of the alcohols I will use.
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Copper can – to act as a calorimeter and to contain the water. The heat will pass through this and heat the water.
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Clamp, boss, and stand – to hold the copper can into position above the spirit lamp.
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Heat proof mats – to be used as draught shielding
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Thermometer – to measure the temperature of the water.
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Stirrer- to stir the water while it is being heated
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Measuring cylinder (100cm3) – to measure the volume of water, 100cm3
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Electronic balance –to two decimal places, to weigh the mass of the spirit lamps with the lid on at the beginning and end of the experiment.
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Water 100cm3 – the liquid that is being used for the alcohols to heat up.
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4 blocks- to change the height of the burner
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splinters- to light the spirit lamps
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safety goggle- to protect you and prevent any contact with alcohols
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stop watch- to record how long it took for complete combustion to take place for each alcohol
Preliminary work:
The preliminary work was done by fuels which were not alcohols. The fuels used were splints, paper, straws, and a solid. I did the preliminary work to a have a brief understanding of what I will have to do for the actual experiment. By doing the preliminary work I can find out what measurement would be the best to use. The experiment was done to see which mass of water be best to use.
Preliminary method:
- collect all the apparatus and wear goggles
- place the gauze on top of the tripod and place the tripod on top of the heat proof mat
- put a bowl with one of the fuels in it under the tripod
- make sure that the stop clock is set up
- measure 50cm3of water and put it in the copper can
- take a note of the temperature of the water
- place the copper can on top of the tripod
- burn the fuel and start stop clock
- use heat proof mats to cover round the tripod so maximum heat goes up to the can
- once it reaches to maximum temperature, stop the clock and make a note of the temperature of water and the time taken for the fuel to burn fully
- do this with all the fuels
Before burning:
Mass of water 50 cm3
Mass fuel 2 grams
Results:
The amount of heat transferred from the fuels to water in Kj can be found using the relationship:
Energy= mass of water times temperature rise times 4,2j/g/c
(In KJ) (In Kg) (In *C)
- it can be assumed that at room temperature 1cm3 of water has the mass of 1g
Preliminary Results:
Changes:
For the actual experiment I am going to use 100 cm3 of water because 50cm3 mass of water was too large a rise as this caused too much heat to be lost to the environment, and 150cm3 was too small. Therefore the 100cm3 value was used, as this temperature was right for the enthalpy calculations. Also the amount of wick was investigated, and I found out that if it were too large, there was more heat to be lost in the environment, and if it were too small, most heat given out is lost in the can, so 5cm is thought to be the optimum length. This will be kept constant throughout, as well as the mass of water and the temperature will be kept constant, to ensure that it is a fair test. I am also going to use alcohols instead of fuels.