The enthalpy change for the oxidation of naphthalene, $\mathrm{C}_{10} \mathrm{H}_{8},$ is measured by calorimetry.
$$
\begin{aligned}
\mathrm{C}_{10} \mathrm{H}_{g}(\mathrm{s})+12 \mathrm{O}_{2}(\mathrm{g}) \rightarrow & 10 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\ell) \\
\Delta_{i} H^{\circ} &=-5156.1 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}
\end{aligned}
$$
Use this value, along with the standard enthalpies of formation of $\mathrm{CO}_{2}(\mathrm{g})$ and $\mathrm{H}_{2} \mathrm{O}(\ell),$ to calculate the enthalpy of formation of naphthalene, in kJ/mol.