The enthalpy of formation of ethane (C₂ H₆) from ethylene by addition of hydrogen where the bond-energies of C-H, C-C, C=C, H-H are 414 kJ, 347 kJ, 615 kJ and 435 kJ respectively is - __________ kJ

Solution

Chemistry ma’am · 4 steps

Only the bonds that change matter: break C=C and H–H, then form C–C and two new C–H.

1. Write the reaction. CH₂=CH₂ + H₂ → CH₃-CH₃. The four C–H bonds of ethylene survive unchanged, so they cancel out of the calculation.
2. Bonds broken. One C=C and one H–H: 615 + 435 = 1050 kJ. Breaking bonds costs energy, so this term is positive.
3. Bonds formed. One C–C and two new C–H: 347 + 2 × 414 = 347 + 828 = 1175 kJ. Forming bonds releases this energy.

   Δ H = Σ BE_(broken) - Σ BE_(formed) = 1050 - 1175 = -125 kJ

4. So. The enthalpy change is -125 kJ. The minus sign is already printed in the blank, so the answer is 125.

The idea to keep. With bond energies, it is always bonds broken minus bonds formed. Draw both structures and count only the bonds that actually change.

Answer from the NTA final answer key. Solution by PrepWiser.