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Given the two compounds propane and pentanone, the one with the higher boiling point is

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Pentanone, because it will boil at around 100 to 102 degree C, about the same as water, while propane boils at about -42 degree C, a temperature way below both pentanone and water's freezing point.

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Answer is: 28 kJ.
Chemical reaction: A₂ + B₂ ⇄ 2AB.
Ea(forward) = 105 kJ/mol.
Ea(reverse) = 77 kJ/mol.
ΔH(reaction) = ?
<span>The enthalpy change of reaction is the change in the energy of the reactants to the products.
</span>ΔH(reaction) = Ea(forward) - Ea(reverse).
ΔH(reaction) = 105 kJ/mol - 77 kJ/mol.
ΔH(reaction) = 28 kJ/mol; this is endothermic reaction (ΔH <span>> 0).</span>

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