Questions: Calculate the combustion energy (Delta E) for benzene, mathrmC6 mathrmH6 in mathrmkJ / mathrmmol. Express your answer in kilojoules per mole to three significant figures. Delta E= mathrmkJ / mathrmmol

Solution Steps

The balanced chemical equation for the combustion of benzene (\(\mathrm{C}_6\mathrm{H}_6\)) is:

\[ 2 \mathrm{C}_6\mathrm{H}_6(l) + 15 \mathrm{O}_2(g) \rightarrow 12 \mathrm{CO}_2(g) + 6 \mathrm{H}_2\mathrm{O}(l) \]

To calculate the combustion energy, we need the standard enthalpies of formation (\(\Delta H_f^\circ\)) for the reactants and products. These values are typically found in thermodynamic tables:

• \(\Delta H_f^\circ\) for \(\mathrm{C}_6\mathrm{H}_6(l)\) = 49.0 kJ/mol
• \(\Delta H_f^\circ\) for \(\mathrm{O}_2(g)\) = 0 kJ/mol (element in its standard state)
• \(\Delta H_f^\circ\) for \(\mathrm{CO}_2(g)\) = -393.5 kJ/mol
• \(\Delta H_f^\circ\) for \(\mathrm{H}_2\mathrm{O}(l)\) = -285.8 kJ/mol

The combustion energy \(\Delta E\) (or \(\Delta H\) for constant pressure) is calculated using the formula:

\[ \Delta E = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]

Substitute the values:

\[ \Delta E = [12(-393.5) + 6(-285.8)] - [2(49.0) + 15(0)] \]

\[ \Delta E = [-4722.0 - 1714.8] - [98.0] \]

\[ \Delta E = -6436.8 - 98.0 \]

\[ \Delta E = -6534.8 \, \text{kJ for 2 moles of } \mathrm{C}_6\mathrm{H}_6 \]

Since the reaction is for 2 moles of benzene, divide the total energy by 2 to find the energy per mole:

\[ \Delta E = \frac{-6534.8}{2} = -3267.4 \, \text{kJ/mol} \]

Final Answer