Question

From the reaction C10H8(s) + 12O2(g) ------> 10CO2 + 4H2O(l)            Delta(r) Ho = -5153.0kJ mol-1 and...

From the reaction

C10H8(s) + 12O2(g) ------> 10CO2 + 4H2O(l)            Delta(r) Ho = -5153.0kJ mol-1

and the enthalpies of formation of CO2 and H2O, calculate the enthalpy of formation of naphthalene (C10H8).

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Answer #1

we have:

deltaHo rxn = -5153.0 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Hof(CO2(g)) = -393.509 KJ/mol

Hof(H2O(l)) = -285.83 KJ/mol

we have the Balanced chemical equation as:

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

deltaHo rxn = 10*Hof(CO2(g)) + 4*Hof(H2O(l)) - 1*Hof( C10H8(s)) - 12*Hof(O2(g))

-5153 = 10*(-393.509) + 4*(-285.83) - 1*Hof(C10H8(s)) - 12*(0.0)

Hof(C10H8(s)) = 74.59 KJ/mol

Answer: 74.59 KJ/mol

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