"4 Most fuels are compounds made from two main elements, carbon (symbol C) and hydrogen (symbol H).Let us consider the combustion of these two main constituents.Some basic information first. Carbon, a solid at normal temperatures, has a relative atomic mass of 12 and hydrogen, a gas at normal temperatures, has a relative atomic mass of 1.Carbon has the ability to combine with itself and with hydrogen to form many compounds known as hydrocarbons.Some of these combinations are gases (e.g. methane, CH ), 4 some are liquids (e.g. octane, C H ) and a few are solids (e.g. waxes which 8 18 have high numbers of carbon and hydrogen atoms combined together such as C H ). 30 62 As we have already stated, to burn a fuel we need oxygen (symbol O), which has a relative atomic mass of 16.Most elemental gases, such as hydrogen and oxygen, exist as two atoms combined together to form a molecule and we therefore write these materials in equations as H and O , the subscript 2 2 indicating the two atoms joined together.When carbon (a solid at normal temperatures) is completely burnt with sufficient oxygen (a gas) the product formed is carbon dioxide (CO – a gas) according to the balanced chemical 2 equation CO+?CO (s) 2 (g) 2 (g) This represents one molecule (atom) of carbon solid reacting with 1 molecule of oxygen gas to give us one molecule of carbon dioxide gas.We often simplify this to one mole (where a mole is the atomic/molecular (or formula) mass of the material in grams) of carbon solid reacts with 1 mole of oxygen gas to give us one mole of carbon dioxide gas. [Note, 1 kilomole (kmol) is the molecular mass in kilograms and is the more common unit in industry]. Teesside University Open Learning © Teesside University 2011 (Engineering)5 On a relative mass basis, since C has an atomic mass of 12 and O an atomic mass of 16 (thus O has a molecular or formula mass of2× 16=32 and 2 CO has a molecular mass of 12+2× 16=44) we can also write 2 CO+?CO (s) 2 (g) 2 (g) 1 mole 1 mole 1 mole 12 g 32 g 44 g 4 Similarly, when hydrogen (a gas at normal temperatures) completely burns with sufficient oxygen (a gas) the product formed is water vapour (H O – a 2 gas) according to the balanced chemical reaction H+? 0.5 O H O 2 (g) 2 (g) 2 (g) 1 mole 0.5 mole 1 mole Note that here we only require 0.5 mole of oxygen for every mole of hydrogen burnt and water formed. On a relative mass basis, since H has an atomic mass of 1 (H thus being 2× 1=2 and H O being 2× 1+16=18) we can 2 2 also write H+? 0.5 O H O 2 (g) 2 (g) 2 (g) 2 g 0.5×= 32 16 g 18 g Note that because we only require 0.5 mole of oxygen, the molecular mass involved in the reaction is halved.However, with gases we can add a further statement to this equation.1 mole of any gas at the same temperature and pressure occupies the same volume so we can state on a relative volume basis, if all the gases are at the same temperature and pressure, H+? 0.5 O H O 2 (g) 2 (g) 2 (g) 33 3 1 m 0.5 m 1 m Teesside University Open Learning © Teesside University 2011 (Engineering)6 [Note that in all the above equations: (i) The amount of the individual carbon, hydrogen and oxygen atoms on each side of the equation are the same, i.e. they balance. (ii) The total mass of material on each side of the equation balances. (iii) BUT the volumes and number of moles on each side do not balance, and do not need to (and rarely do!) balance.This is because of the way in which the atoms are combined in the fuel and products.] If we consider fuels which consist of only carbon and hydrogen, then the products of complete combustion are always carbon dioxide gas and water vapour.For example, methane (natural gas) has the formula CH .When burnt completely 4 with sufficient oxygen the products are carbon dioxide and water vapourCH +? O CO + H O 4 (g) 2 (g) 2 (g) 2 (g) You may notice that in this equation the number of oxygen and hydrogen atoms on either side of the arrow do not balance, though the carbon does.This is not acceptable and the first thing to do is to balance the equation for the reaction equation to be of use to us. Let's do this and start by looking at the carbon first.There is one on the right and one on the left – good, a balance!Now the hydrogen (H).There are four on the left of the arrow in CH but only 2 on the right in water (H O).To 4 2 balance we cannot, in this case, change the numbers of hydrogens on the left as methane always has 4 hydrogens.Nor can we change the number of hydrogens in water.We can, however, double the amount of water formed in the reaction by putting a 2 in front of the H O to give us a hydrogen balance.Note this 2 2 in front of the H O means that the amount of oxygen is also doubled. 2 Teesside University Open Learning © Teesside University 2011 (Engineering)7 This gives usCH +? O CO + 2H O 4 (g) 2 (g) 2 (g) 2 (g) The carbon and hydrogen now balance but the oxygens do not [there are 2 on left (O ) and 4 on right – 2 in CO and 2 in 2H O].We can balance the oxygen 2 2 2 by putting a 2 in front of the oxygen, i.e. doubling the amount of oxygen gas required for the reaction, to give us the balanced equation CH+?22 O CO+ H O 4 (g) 2 (g) 2 (g) 2 (g) What this tells us is that one mole of methane requires two moles of oxygen to produce one mole of carbon dioxide and two moles of water vapour.CH+? 2O CO+ 2H O 4 (g) 2 (g) 2 (g) 2 (g) 1mmole 2 moles 1 mole 2 moles 1 mole of methane consists of one atom of carbon (atomic mass 12) and 4 atoms of hydrogen (each of atomic mass 1) which gives a molecular mass of 12+4× 1=16.On a relative mass basis and on a relative volume basis, since they are all gaseous compounds, we can state for methane CH +?22 O CO + H O 4 (g) 2 (g) 2 (g) 2 (g) 12+× 4 1=16 g 23× 2==+ 64 g 12 21× 6= 44 g 2 21×+16= 36 g () 33 333 12m m 1 m 2 m [Note, these are relative masses and volumes, the actual amounts taking part in the reaction can be any so long as they are in the same proportion as above.] Teesside University Open Learning © Teesside University 2011 (Engineering)8 Let's do a second example of balancing a combustion process, this time for benzene, formula C H , a liquid fuel, being completely burnt.When only 6 6 carbon and hydrogen are present and the combustion is complete, the products are always carbon dioxide and water so we start from a base equation CH+? O CO + H O 6 6 (l) 2 (g) 2 (g) 2 (g) We can see that the carbon, hydrogen and oxygen are all unbalanced.Let's start with the carbon (6 on left, 1 on right) and balance that.Because we cannot change benzene we canonly balance it by producing 6 lots of carbon dioxide soCH+? O 6CO + H O 6 6 (l) 2 (g) 2 (g) 2 (g) Now the hydrogen (6 on left, 2 on right).To balance this we will need to have 3 lots of water formed, so the equation now becomesCH+? O 63 CO + H O 6 6 (l) 2 (g) 2 (g) 2 (g) Finally, the oxygen (2 on left and 15 on right). We need to use 7.5 times the amount of oxygen to enable the balance.We therefore have CH+? 75 . O 6CO+ 3H O 6 6 (l) 2 (g) 2 (g) 2 (g) [Balanced equations are often referred to as having stoichiometric quantities of materials reacting together and stoichiometric quantities being produced.] Teesside University Open Learning © Teesside University 2011 (Engineering)"