Use average bond energy data to calculate the value of deltaHfdegree for H_2O_2(
ID: 951946 • Letter: U
Question
Use average bond energy data to calculate the value of deltaHfdegree for H_2O_2(1) (liquid hydrogen peroxide). The enthalpy of vaporization of H_2O_2 is 51.47 kJ/mol. Compare your calculated value to the experimental value of-187.78 kJ/mol. Assuming one of the Kekule structures of benzene (3 C=C and 3 C-C bonds), calculate the enthalpy of combustion for benzene using bond energy data, assuming all species are in the gas phase for this reaction. Explain the discrepancy with the experimental value of-3225 kJ/mol. What values should be used for the carbon-carbon bond energies instead? Comment on your result.Explanation / Answer
(a) The formation equation for H2O2 will be :
H2(g) + O2 --> H2O2 (l)
Structure of H2O2 is H-O-O-H
The heat of formation = Heat of reaction = sum of bond energy of reactants - Sum of bond energies of reactants
Heat of formation = [Bond energy of H-H + Bond energy O=O] - [ 2X Bond energy of O-H + bond energy of O-O]
Heat of formation = [435 + 498 ] -[2X464 + 142] = 933 - 1070 = -137 KJ / mole
This is less than the experimental value
b) The combustion equation will be
C6H6 + (15/2)O2 --> 6CO2(g) + 3H2O (l)
Enthalpy of reaction = sum of bond energy of reactants - Sum of bond energies of reactants
Enthalpy of reaction = (6 X C-H + 3 X C=C + 3X C-C + 15/2 O=O) - ( 12 X C=O + 6 X O-H)
Enthalpy of reaction = (6X414 + 3x590 + 3X331 + 15/2 498) - ( 12 X 803 + 6 x 464)
Enthalpy of reaction = (2484 +1770 + 993 + 3725 ) - ( 9636 + 2784) = -3448 KJ / mol
so this is higher then the experimental value. The discrepancy can be understood by the fact that we have assumed that there are three complete C=C bonds and three C-C bonds, however in benzene it is observed that the bond distance is in between double and single bond for all the C -C linkage.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.