Question

Find out the enthalpy of reaction for hydrogenation of ethylene when \[\Delta {H_f}\] of ${C_2}{H_4}$ and ${C_2}{H_6}$ is x and y J/mol.

Answer 1

Hint: Enthalpy is a property of a thermodynamic system, and is defined as the sum of the system's internal energy and the product of its pressure and volume. The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states

Complete answer:
The heat of hydrogenation of ethyne is equal to the difference between enthalpy of formation of ethene from the enthalpy of formation of ethyne.
Explanation:
Enthalpy of formation of ethene:
\[2C + 2{H_2} \to {C_2}{H_{4,}}\Delta {H_f} = x\]
Enthalpy of formation of ethane:
\[2C + 3{H_2} \to {C_2}{H_{6,}}\Delta {H_f} = y\]
Hydrogenation of ethene will produce ethyne as hydrogenation involves the addition of hydrogen thus we can get this by adding ethene to hydrogen molecules to produce ethylene. In the question it is asked for the enthalpy change for hydrogenation reaction of ethyne, thus by subtracting the enthalpy of formation for ethene and ethyne we get:
On subtracting enthalpy of formation of ethane from enthalpy of formation of ethene, we get:
\[{C_2}{H_4} + {H_2} \to {C_2}{H_{6,}}\Delta {H_{hydrogenation}} = ?\]
\[\Delta {H_{hydrogenation}} = y - x\] J/mol
The heat of hydrogenation of ethyne is equal to the difference between enthalpy of formation of ethene from the enthalpy of formation of ethyne.

Note:
We must have to know that the enthalpy of formation for an element in its elemental state will always be zero because it takes no energy to form a naturally-occurring compound. When a substance is formed from the most stable form of its elements, a change in enthalpy takes place.