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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 are 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

Last updated date: 13th Jun 2024
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Hint: Combustion of an organic compound is the conversion of all the organic species into carbon dioxide and water. Hence, If the amount of carbon and hydrogen content is greater in the organic species, the amount of carbon dioxide formed will be greater and hence more energy will be released.

Complete step by step answer:
In the question given to us, we have been given the information regarding enthalpy of formation of methanol as well as octane.
The Question asks us to determine the value for enthalpy of combustion:
The reaction for combustion of methanol will be:
$C{H_3}OH + \dfrac{3}{2}O_2\rightarrow{{}}C{O_2} + 2{H_2}O$
The enthalpy of formation for methanol will be: $ - 1.5kJ/mol$
The reaction for Combustion of octane will:
${C_8}{H_{18}} + \dfrac{{25}}{2}O_2\rightarrow 8C{O_2} + 9{H_2}O$
The enthalpy of formation of octane will be: $ - 10.9kJ/mol$
Combustion of organic compounds in the presence of oxygen will always give carbon dioxide and water as products.
According to the law of conservation of matter, the amount of matter in the reactants, will be the same as the amount of matter in the product.
Hence if the carbon and hydrogen content is more in the reactant, then the product will have more carbon dioxide and water molecules formed.
Compare the above two reactions for octane and methanol, burning of methanol gives one molecule of carbon dioxide whereas burning of octane gives eight carbon dioxides.

Hydrocarbons with the longer chain will give more energy on burning since they have more number of bonds and each bond has a potential energy.
This is why octane will give more energy on burning than methanol, since octane has a bigger chain.
The value of enthalpy of combustion will be more negative for octane than methanol.

So, the correct answer is Option A.

Note: Enthalpy of combustion can be determined mathematically by using Hess’s law. It states that if a process occurs in stages then the enthalpy for the final step will be equal to the sum of the enthalpy of the individual steps. It means that enthalpy change is not path dependent.