The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2(g) and H2O (1) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation, ?f H? of benzene. Standard enthalpies of formation of CO2(g) and 2 H O(l)are –393.5 kJ mol–1 and – 285.83 kJ mol–1 respectively. ?

To calculate the standard enthalpy of formation (ΔfH) of benzene (C6H6), we can use the information provided about its combustion reaction. The combustion of benzene can be represented by the following balanced chemical equation:

C6H6 + 7.5 O2 → 6 CO2 + 3 H2O

In this reaction, one mole of benzene reacts with oxygen to produce six moles of carbon dioxide and three moles of water, while releasing 3267.0 kJ of heat. To find the standard enthalpy of formation of benzene, we can apply Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

Step-by-Step Calculation

1. **Identify the enthalpy changes**: We know the standard enthalpy of formation for the products:

• ΔfH for CO2(g) = –393.5 kJ/mol
• ΔfH for H2O(l) = –285.83 kJ/mol

2. **Calculate the total enthalpy of the products**: Since the combustion produces 6 moles of CO2 and 3 moles of H2O, we can calculate the total enthalpy change for the products:

ΔH(products) = [6 moles CO2 × (–393.5 kJ/mol)] + [3 moles H2O × (–285.83 kJ/mol)]

Calculating this gives:

• ΔH(products) = 6 × (–393.5) + 3 × (–285.83)
• ΔH(products) = –2361.0 kJ + –857.49 kJ
• ΔH(products) = –3218.49 kJ

3. **Apply Hess's Law**: The overall reaction can be expressed in terms of the enthalpy of formation of benzene:

ΔH(reaction) = ΔH(products) - ΔfH(reactants)

Since we are considering the formation of benzene from its elements, the enthalpy of formation of the reactants (elements in their standard states) is zero. Thus, we can simplify this to:

ΔH(reaction) = ΔH(products) - ΔfH(C6H6)

4. **Substituting values**: We know that the heat released (ΔH(reaction)) is –3267.0 kJ. Therefore:

–3267.0 kJ = –3218.49 kJ - ΔfH(C6H6)

5. **Rearranging to find ΔfH(C6H6)**:

ΔfH(C6H6) = –3218.49 kJ + 3267.0 kJ

ΔfH(C6H6) = 48.51 kJ/mol

Final Result

The standard enthalpy of formation of benzene (C6H6) is approximately 48.51 kJ/mol. This positive value indicates that the formation of benzene from its elements is an endothermic process, meaning it requires energy input.

This calculation illustrates how we can use combustion data and Hess's Law to derive the enthalpy of formation for a compound, providing valuable insights into its thermodynamic properties.