91Ó°ÊÓ

To form naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) from its elements, we combine carbon (C) and hydrogen (H) elements. The balanced equation for this reaction can be given as: \( 10\: \mathrm{C}(s) + 4\: \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{10} \mathrm{H}_{8}(s) \)

Naphthalene reacts with oxygen gas (\(\mathrm{O}_{2}\)) during combustion and forms carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\: \mathrm{O}\)). The balanced equation can be written as: \( \mathrm{C}_{10} \mathrm{H}_{8}(s) + 12\: \mathrm{O}_{2}(g) \rightarrow 10\: \mathrm{CO}_{2}(g) + 4\: \mathrm{H}_{2} \mathrm{O}(l) \)

We are given that the enthalpy change, ΔH, for the combustion of naphthalene is -5154 kJ/mol. We can use the combustion equation and Hess's law to calculate the standard enthalpy of formation of naphthalene. Let ΔHf represent the enthalpy of formation of naphthalene, then: Formation of naphthalene: \( 10\: \mathrm{C}(s) + 4\: \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{10} \mathrm{H}_{8}(s), \Delta H = \Delta H_{f}\) Combustion of naphthalene: \( \mathrm{C}_{10} \mathrm{H}_{8}(s) + 12\: \mathrm{O}_{2}(g)\rightarrow 10\: \mathrm{CO}_{2}(g) + 4\: \mathrm{H}_{2} \mathrm{O}(l), \Delta H = -5154 \: \mathrm{kJ/mol} \) Connecting these two reactions using Hess's law: \( 10\: \mathrm{C}(s) + 4\: \mathrm{H}_{2}(g) + 12\: \mathrm{O}_{2}(g) \rightarrow 10\: \mathrm{CO}_{2}(g) + 4\: \mathrm{H}_{2} \mathrm{O}(l) \) We can find the enthalpy changes for the reaction of the carbon and hydrogen elements with oxygen: Formation of \(\mathrm{CO_{2}}\): \( \mathrm{C}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g), \Delta H = -393.5 \: \mathrm{kJ/mol} \) Formation of \(\mathrm{H_{2}O}\): \( \mathrm{H}_{2}(g) + \frac{1}{2} \: \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l), \Delta H = -285.8 \: \mathrm{kJ/mol} \) Applying Hess's Law: \( \Delta H_{f} + (-5154) = 10 \times (-393.5) + 4 \times (-285.8) \) Solve for ΔHf: \( \Delta H_{f} = 10 \times (-393.5) + 4 \times (-285.8) + 5154 \) \( \Delta H_{f} = -3343.2\: \mathrm{kJ/mol}\) Thus, the standard enthalpy of formation of naphthalene is -3343.2 kJ/mol.