Answer: Option (a) is the correct answer.
Explanation:
Combustion is defined as the reaction in which a hydrocarbon reacts with oxygen and leads to the formation of carbon dioxide and water.
During combustion reaction, both C-C and C-H bonds tend to break down. As a result, new H-O and C=O bonds are formed. So, the amount of energy released upon formation of H-O and C=O bonds is more than the amount of energy required for breaking the C-C and C-H bonds.
Hence, more is the number of C-C and C-H bonds more will be the release of heat of combustion.
Therefore, it means larger is the number of carbon and hydrogen atoms present in an alkane more will be its heat of combustion.
For example, heat of combustion of \(C_{6}H_{14}\) is -4163.0 kJ/mol.
- Heat of combustion of \(C_{6}H_{12}\) is -3919.5 kJ/mol.
- Heat of combustion of \(C_{2}H_{6}\) is -1559 kJ/mol.
- Heat of combustion of \(C_{3}H_{8}\) is -2219.2 kJ/mol.
Thus, we can conclude that \(C_{6}H_{12}\) is the hydrocarbon which will provide more heat when allowed to react with oxygen.
