Final answer:
To compare the standard heat of combustion of 6 moles of methanol with 1-hexene, the enthalpy changes for the combustion reactions of both the compounds to CO₂ and H₂O should be considered and can be calculated using standard enthalpies of formation or from bond energies.
Step-by-step explanation:
The question asks to compare the standard heat of combustion at 25 ℃ of 6 moles of gaseous methanol (6CH₃OH(g)) with the standard heat of combustion at 25 ℃ of 1-hexene (C₆H₁₂(g)) when the reaction products are carbon dioxide (CO₂(g)) and water (H₂O(g)). To answer this, we need to consider the enthalpy change (ΔH°) for the combustion of both compounds and compare them.
For methanol, the combustion reaction is typically written as:
CH₃OH(l) + 1.5O₂(g) → CO₂(g) + 2H₂O(g),
and for 1-hexene:
C₆H₁₂(g) + 9O₂(g) → 6CO₂(g) + 6H₂O(g).
The heat of combustion for each compound can be found in tables of standard enthalpies of formation or it can be calculated from bond energies if the enthalpies of formation are not readily available. Generally, for the same number of carbons, alkanes (like hexane) have a higher heat of combustion than alcohols (like methanol) because alkanes are more reduced (less oxidized) and can release more energy when completely combusted to CO₂ and H₂O.