Question

Naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) is a solid aromatic compound often sold as mothballs. The complete combustion of this substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(25^{\circ} \mathrm{C}\) yields \(5154 \mathrm{~kJ} / \mathrm{mol}\). (a) Write balanced equations for the formation of naphthalene from the elements and for its combustion. (b) Calculate the standard enthalpy of formation of naphthalene.

Short answer

The standard enthalpy of formation of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) is -3343.2 kJ/mol, which was calculated using balanced equations for its formation from elements and combustion, as well as applying Hess's law.

Step-by-step solution

Writing the balanced equation for the formation of naphthalene from its elements

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: \( 10\: \mathrm{C}(s) + 4\: \mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{10} \mathrm{H}_{8}(s) \)

Writing the balanced equation for the combustion of naphthalene

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: \( \mathrm{C}_{10} \mathrm{H}_{8}(s) + 12\: \mathrm{O}_{2}(g) \rightarrow 10\: \mathrm{CO}_{2}(g) + 4\: \mathrm{H}_{2} \mathrm{O}(l) \)

Calculating the standard enthalpy of formation of naphthalene

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.