Exams › JEE Main › Chemistry
Using the bond enthalpies given below, determine the enthalpy change for the hydrogenation reaction shown: H–H bond energy = 431.37 kJ mol$^{-1}$ C–C bond energy = 606.10 kJ mol$^{-1}$ C–H bond energy = 336.49 kJ mol$^{-1}$ C–H bond energy = 410.50 kJ mol$^{-1}$ $C=C + H–H \rightarrow H–C–C–H$ What is the enthalpy change for this reaction?
- -243.6 kJ mol⁻¹
- -120.0 kJ mol⁻¹
- 553.0 kJ mol⁻¹
- 1523.6 kJ mol⁻¹
Correct answer: -243.6 kJ mol⁻¹
Solution
For hydrogenation, bonds broken are one C=C and one H–H, while bonds formed are one C–C and two C–H bonds. Substituting the given bond energies into $\Delta H = \text{broken} - \text{formed}$ gives a negative value, indicating an exothermic reaction.
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