For C₂H₄ + H₂ → CH₃CH₃ with bond enthalpies C–H 416, C=C 612, C–C 348 and H–H 436 kJ, the enthalpy of hydrogenation is:
A+132 kJ
B+264 kJ
C−264 kJ
D−132 kJ
Answer & Solution
Correct answer: D. −132 kJ
1. Use $\Delta H=\sum(\text{bonds in reactants})-\sum(\text{bonds in products})$.
2. Reactants: one C=C at 612, four C–H at 416, one H–H at 436, giving $612+1664+436=2712$ kJ.
3. Products: one C–C at 348 and six C–H at 416, giving $348+2496=2844$ kJ.
4. So $\Delta H=2712-2844=-132$ kJ.
5. The negative sign means hydrogenation releases heat, so it is exothermic; reversing the subtraction gives $+132$ and the wrong direction.
_Source: NECTA ACSEE 2023 Chemistry 132/1, Question 6: Energetics_