Question

The approximate standard enthalpies of formation of methanol and octane are determined to be $-1.5kJ/mol$ and $-10.9kJ/mol$ respectively. The standard enthalpies of combustion of octane is denoted as $\Delta H$ (octane) and that of methanol as $\Delta H$ (methanol). The correct statement is:

• A. $\Delta H$ (octane) is more negative than $\Delta H$ (methanol)
• B. $\Delta H$ (octane) is less negative than $\Delta H$ (methanol)
• C. $\Delta H$ (octane) is equal to $\Delta H$ (methanol)
• D. $\Delta H$ (octane) +$\Delta H$ (methanol)=0

Answer 1

The correct option is A $\Delta H$ (octane) is more negative than $\Delta H$ (methanol)

$\Delta H_f^0$ (methanol) = -1.5 KJ/mol

$\Delta H_f^0$ (octane) = -10.9 KJ/mol

Heat of combustion $\alpha$ + Heat of formation.

$\therefore \Delta H$ (octane) is more negative than $\Delta H$ (methanol).