Calculate for the combustion of octane (C8H18), a component of gasoline, by using average bond energies and then calculate it using enthalpies of formation from Appendix IIB . What is the percent difference between your results? Which result would you expect to be more accurate?
Solution 106 E Step 1 of 4 First, we have to consider what is the balanced chemical equation for the combustion of octane The reaction can be written by indicating the Lewis structures for the same reactants and products Thus we can see that at the end of this reaction 7 CC bonds, 18 CH bonds and 25/2 double bonds O=O are broken, while 18 OH bonds and 16 C=O bonds are formed. Step 2 of 4 We have to remember that when a reaction happens, we have to spend energy to break bonds (positive H ), while broken bonds release energy (negative H) Moreover For C-H= 414 kJ/mol For C-C= 347 kJ/mol For O=O = 498 kJ/mol Thus H broken bonds= 18 (H C-H)+7 (H C-C+ 25/2 (H O=O)= 16106 kJ/mol For O-H= -464 kJ/mol For C=O = -799 kJ/mol Thus H formed bonds= 18 (H O-H+16(H C=O= -21136 kJ/mol Since H RXN= (H broken bonds) + (H formed bonds) H RXN= (16106 kJ/mol) + (- 21136 kJ/mol)= -5030 kJ/mol Step 3 of 4 Alternatively we can even calculate H RXN by using this formula: H 0 = (n H products) + (nH reactants) 0 RXN p r Here,n ap n are r ichiometric factors of products, reactants 0 H p,represent standard enthalpy of formation of every product and reactant In the table below we have the standard enthalpy of formation for products, reactants in the reaction: Reactant or product 0 0 0 0 Thus, substituting in H RXN = (n p products) + (nH reactrts) H 0 = (8 H 0 ) +(9 H 0 ) - [(1 H 0 ) +25/2 H )=[8(-393 kJ/mol) + 9 (- 241.8 RXN CO2 H2O C8H8 O2 kJ/mol)] - [1(-250.1 kJ/mol)+ 25/2 (0.0 kJ/mol)]= - 5074.1 kJ/mol Step 4 of 4 We not that the value of H obtained in this two methods (Step 2 and Step 3) are not the same ( - 5030 kJ/mol vs. -5074 kJ/mol) We can calculate % difference between the two results There’s a difference of 0.87%, but the value of H obtained in Step 4 is more accurate because 0 H of each compound has been obtained experimentally, while the values we used in