Question

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)

Answer 1

To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states and use the standard enthalpies of combustion and formation.

Step-by-step explanation:

The standard enthalpy of formation of a compound can be calculated using the standard enthalpies of combustion and the equation:

∆Hf° = ∆Hc° + ∑∆Hf°(products) - ∑∆Hf°(reactants)

Given that the standard enthalpy of combustion of naphthalene is -5157 kJ/mol, we can use this value along with the standard enthalpies of formation of the products and reactants to calculate the standard enthalpy of formation of naphthalene.

Answer 2

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