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

possibly, yes.

explanation:

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