Calculate enthalpy of reaction C2H4 + H2 gives C2H6.enthalpy of combustion of ethene, H2,and ethane are -1410,-286,-15.60kj/mol respectively
To calculate the enthalpy of the reaction C2H4 + H2 → C2H6, we can use the enthalpies of combustion for each compound involved.
First, we need to balance the chemical equation:
C2H4 + H2 → C2H6
The balanced equation is already given.
We can now calculate the enthalpy of the reaction by subtracting the sum of the enthalpies of combustion of the reactants from the sum of the enthalpies of combustion of the products.
Enthalpy of reaction = (∑ Enthalpy of products) - (∑ Enthalpy of reactants)
Enthalpy of reaction = (Enthalpy of C2H6) - [(Enthalpy of C2H4) + (Enthalpy of H2)]
Enthalpy of reaction = (-15.60 kJ/mol) - [(-1410 kJ/mol) + (-286 kJ/mol)]
Enthalpy of reaction = -15.60 kJ/mol + 1410 kJ/mol + 286 kJ/mol
Enthalpy of reaction = 1681.4 kJ/mol
To calculate the enthalpy of the reaction C2H4 + H2 -> C2H6, we need to use the enthalpy values of the substances involved.
The enthalpy of the reaction can be determined by considering the difference in enthalpy between the reactants and products.
The enthalpy change of the reaction can be calculated using the formula:
∆H = (enthalpy of products) - (enthalpy of reactants)
Let's assign the following variables:
∆H = Enthalpy change of the reaction
∆H1 = Enthalpy of combustion of C2H4
∆H2 = Enthalpy of combustion of H2
∆H3 = Enthalpy of combustion of C2H6
Given the following enthalpy values:
∆H1 = -1410 kJ/mol (enthalpy of combustion of ethene)
∆H2 = -286 kJ/mol (enthalpy of combustion of H2)
∆H3 = -15.60 kJ/mol (enthalpy of combustion of ethane)
Now, substituting the given values into the formula:
∆H = ∆H3 - (∆H1 + ∆H2)
∆H = -15.60 kJ/mol - (-1410 kJ/mol + -286 kJ/mol)
Simplifying:
∆H = -15.60 kJ/mol + 1410 kJ/mol + 286 kJ/mol
∆H = 1410 kJ/mol + 286 kJ/mol - 15.60 kJ/mol
∆H = 1681.40 kJ/mol
Therefore, the enthalpy change for the reaction C2H4 + H2 -> C2H6 is 1681.40 kJ/mol.