Question

From the reaction C10H8(s) + 12O2(g) = 10CO2(g) + 4H2O(l) ΔrH° = –5153.0 kJ mol–1 and the enthalpies of formation of CO2 and H2O (–393.5 kJ mol–1 and –285.8 kJ mol–1, respectively) calculate the enthalpy of formation of naphthalene (C10H8).

Answer 1

Answer: We can use the enthalpy change of combustion formula: ΔrH° = ΣnΔrHf(products) - ΣnΔrHf(reactants), where n is the number of moles of each species. For this reaction, we have two reactants and three products, so n = 2 for the reactants and n = 3 for the products. The enthalpy of formation of carbon dioxide is given as -393.5 kJ mol-1, while the enthalpy of formation of water is -285.8 kJ mol-1.