Question

Calculate the enthalpy of formation of ethene $ {C_2}{H_4} $ only magnitude in nearest integer (in $ {\text{kJ/mol}} $ )

Answer 1

Hint:Enthalpy of formation is also known as heat of formation or standard heat of formation. Enthalpy can be said as the amount of heat that is absorbed or liberated during a reaction in layman terms. If a certain amount of heat is absorbed during the course of the reaction, the enthalpy of the reaction is said to be positive. If a certain amount of heat is released during the course of the reaction, the enthalpy of the reaction is said to be negative.

Complete step by step answer:
To find the enthalpy of formation of a hydrocarbon, we use the combustion reaction of the hydrocarbons. This is because hydrocarbons on combustion produce carbon dioxide and water. The enthalpy of a reaction can be determined by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products.
The chemical reaction for the combustion of ethene can be written as $ {C_2}{H_4} + 3{O_2}(g) \to 2C{O_2} + 2{H_2}O $
The enthalpy of the reaction is known to be $ - 1393{\text{kJ/mol}} $
The enthalpies of formations of carbon dioxide and water are $ - 394{\text{kJ/mol}} $ and $ - 286{\text{kJ/mol}} $ respectively. The enthalpy of oxygen can be taken as zero.
Now we know that the enthalpy of a reaction is given by the formula, $ {\Delta _r}H = {\left( {{\Delta _f}H} \right)_{{\text{products}}}} - {\left( {{\Delta _f}H} \right)_{{\text{reactants}}}} $
Substituting the values we can find the enthalpy of formation of ethene molecule:
   $ - 1393 = 2 \times \left( { - 394} \right) + 2 \times \left( { - 286} \right) - {\Delta _f}{H_{{C_2}{H_4}}} $
Thus, we can write, $ {\Delta _f}{H_{{C_2}{H_4}}} = 1393 - 2\left( {394} \right) - 2\left( {286} \right) $
   $ {\Delta _f}{H_{{C_2}{H_4}}} = 33{\text{kJ/mol}} $

Thus, the enthalpy of formation of ethene is $ 33{\text{kJ/mol}} $ .
Note:
The enthalpy of formation of any element present in its standard state is always zero. In this reaction oxygen is present as a dioxygen molecule in the gaseous state. Hence, its enthalpy of formation is zero.