C10H8+12O2(g) GIVES 10CO2(g)+4H2O(l)

change in H=S+ngRT
Change in H=-5133-2×8.314×298
Change in H=-10088.144 KJ/MOL

Naphthalene (C₁₀H₈) is an organic compound commonly found in mothballs. The heat of combustion of naphthalene refers to the energy released when one mole of naphthalene undergoes complete combustion. In this case, the heat of combustion was measured at constant volume, which means that no work is done on or by the system during the reaction.



The combustion of naphthalene can be represented by the following balanced chemical equation:

C₁₀H₈ + 12O₂ → 10CO₂ + 4H₂O



From the balanced equation, we can see that for every mole of naphthalene combusted, 10 moles of carbon dioxide (CO₂) and 4 moles of water (H₂O) are produced. Therefore, the enthalpy change for the combustion of naphthalene can be calculated by using the enthalpy change for the formation of CO₂ and H₂O.



The enthalpy change for the combustion of naphthalene is equal to the sum of the enthalpy changes for the formation of CO₂ and H₂O:

ΔH = 10ΔH_f(CO₂) + 4ΔH_f(H₂O)

The enthalpy change for the formation of CO₂ and H₂O can be found in standard enthalpy of formation tables. The standard enthalpy of formation for CO₂ is -393.5 kJ/mol, and for H₂O is -285.8 kJ/mol.

Plugging in the values:

ΔH = 10(-393.5 kJ/mol) + 4(-285.8 kJ/mol)
= -3935 kJ/mol - 1143.2 kJ/mol
= -5078.2 kJ/mol

Therefore, the enthalpy change for the combustion of naphthalene is -5078.2 kJ/mol.



The enthalpy change for the combustion of naphthalene was calculated to be -5078.2 kJ/mol. This value represents the energy released when one mole of naphthalene undergoes complete combustion at constant volume. The calculation was based on the balanced chemical equation for the combustion reaction and the standard enthalpy of formation values for CO₂ and H₂