What is the standard enthalpy change for the combustion of one mole of benzene (C6H6) based on the given thermochemical equation: 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(l) ΔHrxn° = -6534 kJ?

• A. -13068 kJ/mol
• B. -6534 kJ/mol
• C. -2178 kJ/mol
• D. -3267 kJ/mol

Answer: (D)

Verified step by step guidance

1. Understand the given thermochemical equation: 2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(l) with ΔHrxn° = -6534 kJ. This represents the combustion of 2 moles of benzene.
2. Identify that the standard enthalpy change provided is for the combustion of 2 moles of benzene. To find the enthalpy change for 1 mole, you need to divide the given ΔHrxn° by 2.
3. Use the formula for calculating the enthalpy change per mole: ΔH per mole = ΔHrxn° / number of moles of benzene. Here, ΔHrxn° = -6534 kJ and the number of moles = 2.
4. Perform the division: (-6534 kJ) / 2 to find the enthalpy change for the combustion of one mole of benzene.
5. Interpret the result: The calculated value represents the standard enthalpy change for the combustion of one mole of benzene, which is the answer to the problem.