The standard enthalpy of combustion of naphthalene is - 5157 kj mol-1. calculate its standard enthalpy of formation. (use data in tables 2c. 4 and 2c. 5 in the resource section)

The combustion of naphthalene is given as:

C10H8 (s) + 12 O2 (g) - -> 10CO2 (g) + 4 H2O (l)

Enthalpy of combustion ΔH = - 5157 kJ/mol

Now, the enthalpy change of a reaction is given

ΔH = ∑nΔHf (products) - ∑nΔHf (reactants)

where n = number of moles

ΔHf = enthalpy of formation

Therefore,

ΔH = [10*ΔHf (CO2) + 4*ΔHf (H2O) ] - [1*ΔHf (C10H8) + 12*O2]

-5157 = [10 * (-393.5) + 4 * (285.83) ] - [ΔHf (C10H8) + 12 * (0) ]

ΔHf (C10H8) = 78.68 kJ/mol