Combustion Of Alcohols Essay Research Paper Planning

Combustion Of Alcohols Essay, Research Paper

Planning In this investigation I will be burning alcohols to heat up a beaker of water. I will be burning five alcohols, methanol, ethanol, propanol, butanol and pentanol. The aim is to find out how much energy is produced when burning these alcohols. ‘An alcohol is a series of organic homologous compounds, with the general formula Cn H n + 1OH´. Alcohols react with oxygen in the air to form water and carbon dioxide. The reaction that is involved in burning alcohols is exothermic because heat is given out. Form this reason the reactant energy is higher than that of the product.

The energy is given out when forming the bonds between the new water and carbon dioxide molecules. The amount of energy produced by such exothermic reactions can be calculated by using the formula Mass of the substance x rise in temp x SHC( specific heat capacity). The specific heat capacity is the number of joules required to heat one gram of water by 1ºC. I chose to use water because it is safe, easily found, and has a reliable specific heat capacity of 4.2.

The bond that are formed in an exothermic reaction can be of two types. The first could be ionic, where a metal is produced. Ionic bonding involves electrons transferring from one atom to the other consequently leaving an electrostatic force between them. The other form of bonding is covalent where atoms share electrons to complete their outer orbit. An example being Methane where four hydrogen atoms each share an electron with a carbon atom.

The method that I will use is as follows… ·Measure 100cm of water in to a glass beaker

·Place the beaker into the grasp of the clamp stand.

·Record the starting temperature of the water

·Weigh the spirit burner with the lid on.

·Put the chosen alcohol burner under the beaker allowing the flame to just touch the beaker.

·Leave to heat up until the temperature of the water is exactly 50ºC more than the original temperature.

·Weigh the spirit burner

·Record all results

The variables that must remain constant throughout the experiment are… ·Mass of the water 100cm

·Type of beaker, glass

·Temperature rise of 50ºC.

·Surrounding temperature of around 23ºC

·The height of the beaker from the wick

·Same set of scales

·Weigh the spirit burner with the lid on.

The variable that must be changed is…

·The type of alcohol used

The formulae of the alcohols that I will be using are…

·Methanol CH OH

·Ethanol C H OH

·Propanol C H OH

·Butanol C H OH

·Pentanol C H OHPrediction I predict that the more bonds there are holding the carbon, oxygen and hydrogen atoms together, more energy will be required to break them apart. For example Ethanol has the formula C H OH. In this formula you have five C-H bonds, one C-C bond, one C-O bond and one O-H bond. To separate these types of bonds you require a certain amount of energy which I will show in a table..

TYPE OF BONDENERGY REQUIRED TO BREAK THE BOND(j)

C-H410

C-O360

O-H510

O=O496

C=O740

C-C350

To separate C-H bond you need to apply 410 joules of energy. There are five such bonds in ethanol so you multiply 410 by five to get 2050 joules. You do these calculations for all the other types of bonds that make up ethanol, add them all together and you get 3270 joules. All of the other alcohols can be broken up in this way. Below is a table showing the energy required to break up the bonds in each alcohol. Type of alcoholEnergy required to break the bonds in the alcohol (j)

Methanol2100

Ethanol3270

Propanol3740

Butanol4560

Pentanol5380

As you can see a longer molecule takes more energy to break it´s bonds, in this case Pentanol. Compared to a smaller molecule, methanol which requires less energy to do so. I can come to predict that the longer the molecular structure in the alcohol the more energy it will take to remove the bonds. So when I come to predicting results I can safely say that pentanol will evolve more energy than methanol simply because it has more bonds to break. Type of alcoholBefore exp 1After exp 1Before exp 2After exp 2Before exp 3After exp 3Average temperature difference(ºC)Average increase in weight (g)

MethanolTemp(ºC)23732373237350

Weight(g)153.12152.12162.84159.56156.35154.09-2.18

EthanolTemp(ºC)26762373217150

Weight(g)169.78166.76169.21163.56176.00172.91-3.92

PropanolTemp(ºC)23732474237350

Weight(g)191.56189.21168.11157.00189.01187.32-5.05

ButanolTemp(ºC)23732171247450

Weight(g)159.76156.7198.1284.32187.56181.58-7.61

PentanolTemp(ºC)24742272237350

Weight(g)164.85157.23167.94155.38161.83153.19-9.61

Analysis

As you can see in the obtaining section I have recorded all the information from the results that I need to complete the calculations… ·Energy evolved

·Energy per gram

·Energy per mole First of all I will calculate the amount of energy evolved. I will achieve this by using this simple formula…

Energy evolved = Mass x Rise in temperature x SHC

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