Chapter 15 Chemical Reactions Study Guide in PowerPoint to accompany Thermodynamics An Engineering Approach 8th edition by Yunus A engel and Michael A Boles 1 The combustion process is a chemical reaction whereby fuel is oxidized and energy is released Fuels are usually composed of some compound or mixture containing carbon C and hydrogen H2 Examples of hydrocarbon fuels are CH4 Methane C8H18 Octane Coal Mixture of C H2 S O2 N2 and non combustibles Initially we shall consider only those reactions that go to completion The components prior to the reaction are called reactants and the components after the reaction are called products 2 Reactants Products For complete or stoichiometric combustion all carbon is burned to carbon dioxide CO2 and all hydrogen is converted into water H2O O These two complete combustion reactions are as follows C O2 CO2 1 H2 O2 H2 O 2 Example 15 1 A complete combustion of octane in oxygen is represented by the balanced combustion equation The balanced combustion equation is obtained by making sure we have the same number of atoms of each element on both sides of the equation That is we make sure the mass is conserved conserved C8 H18 A O2 B CO2 D H2 O Note we often can balance the C and H for complete combustion by inspection C8 H18 A O2 8 CO2 9 H2 O 3 The amount of oxygen is found from the oxygen balance It is better to conserve species on a monatomic basis as shown for the oxygen balance O A 2 8 2 9 1 A 12 5 C8 H18 12 5 O2 8 CO2 9 H2 O Note Mole numbers are not conserved but we have conserved the mass on a total basis as well as a specie basis Also 12 5 kmol of O2 are used to form the CO2 and H2O O If 15 kmol of O2 were supplied to the process process 2 2 5 5 kmol of O2 would be excess oxygen in the products The complete combustion process is also called the stoichiometric combustion and all coefficients are called the stoichiometric coefficients 4 In most combustion processes oxygen is supplied in the form of air rather than pure oxygen Air is Ai i assumed d to t be b 21 percentt oxygen and d 79 percentt nitrogen it on a volume basis For ideal gas mixtures percent by volume is equal to percent by moles Thus for each mole of oxygen in air there exists 79 21 3 76 moles of nitrogen g Therefore complete p or theoretical combustion of octane with air can be written as C8 H18 12 5 O2 3 76 N 2 8 CO2 9 H2 O 47 N 2 5 Air Fuel Ratio Since the total moles of a mixture are equal to the sum of moles of each component there are 12 5 1 3 76 59 5 moles of air required for each mole of fuel for the complete combustion process Often complete combustion of the fuel will not occur unless there is an excess of air present greater than just the theoretical air required for complete combustion To determine the amount of excess air supplied for a combustion process let us define the air fuel ratio AF as kmol air AF kmol fuel Thus for the above example the theoretical air fuel ratio is 12 5 1 3 76 kmol air AFth 59 5 1 kmol fuel 6 On a mass basis the theoretical air fuel ratio is kg air kmol air kmol air AFth 59 5 kmol fuel 8 12 18 1 kg fuel kmol fuel kg air 1512 kgg ffuel 28 97 Percent Theoretical and Percent Excess Air IIn mostt cases more than th theoretical th ti l air i iis supplied li d tto ensure complete l t combustion b ti and d to reduce or eliminate carbon monoxide CO from the products of combustion The amount of excess air is usually expressed as percent theoretical air and percent excess air AFactual 100 AFth AFactual AFth Percent excess air 100 AFth Percent theoretical air 7 Show that these results may be expressed in terms of the moles of oxygen only as Percent theoretical air Percent excess air N O2 actual N O2 th 100 N O2 actual N O2 thh N O2 th 100 Example 15 2 Write the combustion equation for complete combustion of octane with 120 percent theoretical air 20 percent excess air C8 H18 12 12 5 O2 3 76 N 2 8 CO2 9 H2 O 0 2 12 5 O2 1 2 47 N 2 Note that 1 12 1 12 5 O 5 O2 is required for complete combustion to produce 8 kmol of carbon dioxide and 9 kmol of water therefore 0 2 12 5 O2 is found as excess oxygen in the products C8 H18 1 2 12 5 O2 3 76 N 2 8 CO2 9 H2 O 2 5 O2 12 47 N 2 8 Second method to balance the equation for excess air see the explanation of this technique in the text is C8 H18 12 Ath O2 3 76 N 2 8 CO2 9 H2 O 0 2 Ath O2 12 Ath 3 76 N 2 O 12 Ath 2 8 2 9 1 0 2 Ath 2 Ath 12 5 NOTE THIS METHOD WORKS ONLY FOR COMPLETE COMBUSTION WHEN PERCENT OR EXCESS THEROETICAL AIR IS SUPPLIED TO AN INCOMPLETE COMBUSTION PROCESS THE AMOUNT OF SUPPLIED O2 1 2ATH IN THIS EXAMPLE NOT USED FOR FORMING CO2 MUST BE ACCOUNTED FOR IN THE EXCESS O2 IN THE PODUCTS 9 Incomplete Combustion with Known Percent Theoretical Air Example 15 3 Consider the incomplete combustion of C8H18 with 120 theoretical air where 80 C in the fuel goes into CO2 and 20 of the carbon goes to CO Theoretical Combustion C8 H18 Ath O2 8 CO2 9 H 2O theoretical O Ath 2 8 2 9 1 Ath 12 5 12 5 The chemical reaction equation for incomplete Combustion with excess theoretical air and X amount of O2 in products is C8 H18 12 12 5 O2 3 76 N 2 0 8 8 CO2 0 2 8 CO 9 H2 O X O2 12 47 N 2 10 O balance gives O 12 12 5 2 0 8 8 2 0 2 8 1 9 1 X 2 X 3 3 Why is X 2 5 the amount of O2 in the previous complete combustion example Then the balanced equation is C8 H18 12 12 5 O2 3 76 N 2 6 4 CO2 16 CO 9 H2 O 3 3 O2 12 47 N 2 11 Combustion Equation When Product Gas Analysis Is Known Example 15 4 15 4 Propane gas C3H8 is reacted with air such that the dry product gases are 11 5 percent CO2 2 7 percent O2 and 0 7 percent CO by volume What percent theoretical air was supplied li d Wh Whatt iis th the d dew point i t ttemperature t off th the products d t if th the product d t pressure iis …