The enthalpy of formation of ethane (C₂H₆) from ethylene by addition of hydrogen where the bondenergies of C – H, C – C, H – H are 414 kJ, 347 kJ, 615 kJ and 435 kJ respectively is - __________ kJ.

The correct answer is 125.

To find the enthalpy of the reaction for the formation of ethane (C₂H₆) from ethylene (C₂H₄) by addition of hydrogen, we can write down the balanced chemical equation:
C₂H₄(g) + H₂(g) → C₂H₆(g)

The enthalpy of reaction ($\mathrm{\Delta H}$) can be calculated using bond energies by subtracting the total bond energy of the products from the total bond energy of the reactants:
$\mathrm{\Delta H}=\sum \text{Bond Energies of Reactants}-\sum \text{Bond Energies of Products}$

Let us analyze the bonds involved in both the reactants and the products:

1. Bonds broken in Reactants:
- 1 mole of C=C double bonds in ethylene (C₂H₄)
- 4 moles of C-H single bonds in ethylene (C₂H₄)
- 1 mole of H-H single bonds in hydrogen gas (H₂)

Calculating the total energy required to break the reactant bonds:
$E_{\text{reactants}}={\mathrm{BE}}_{\text{C=C}}+4\times {\mathrm{BE}}_{\text{C-H}}+{\mathrm{BE}}_{\text{H-H}}$
Substituting the given values (BE of C=C = 615 kJ/mol, BE of C-H = 414 kJ/mol, BE of H-H = 435 kJ/mol):
$E_{\text{reactants}}=615+4\times 414+435$
$E_{\text{reactants}}=615+1656+435=2706\text{kJ}$

2. Bonds formed in Products (Ethane, C₂H₆):
- 1 mole of C-C single bonds
- 6 moles of C-H single bonds

Calculating the total energy released in forming product bonds:
$E_{\text{products}}={\mathrm{BE}}_{\text{C-C}}+6\times {\mathrm{BE}}_{\text{C-H}}$
Substituting the given values (BE of C-C = 347 kJ/mol, BE of C-H = 414 kJ/mol):
$E_{\text{products}}=347+6\times 414$
$E_{\text{products}}=347+2484=2831\text{kJ}$

3. Enthalpy of reaction ($\mathrm{\Delta H}$):
$\mathrm{\Delta H}=E_{\text{reactants}}-E_{\text{products}}$
$\mathrm{\Delta H}=2706-2831=-125\text{kJ}$

Therefore, the enthalpy of formation of ethane from ethylene by addition of hydrogen is -125 kJ.