Bond Enthalpy

Question

Using bond enthalpy data — C–H: 413 kJ/mol, C=C: 614 kJ/mol, C–C: 347 kJ/mol, H–H: 436 kJ/mol — calculate the enthalpy of hydrogenation of ethylene to ethane.

RankGuru · Accepted Answer

Bond enthalpy calculations use the principle that $\Delta H = \sum$(bonds broken) $-\sum$(bonds formed). The hydrogenation reaction is: $CH_2=CH_2 + H_2 \rightarrow CH_3{-}CH_3$. Bonds broken: one C=C (614 kJ) + one H–H (436 kJ) = 1050 kJ. Bonds formed: one C–C (347 kJ) + two C–H bonds ($2 \times 413 = 826$ kJ) = 1173 kJ. Note: we break the $\pi$ component of C=C and one H–H, then form one C–C and two C–H bonds in ethane compared to ethylene. $\Delta H = 1050 - 1173 = -123$ kJ/mol. Using rounded bond energies to the nearest standard JEE value gives approximately -137 kJ/mol when using slightly different standard values (C=C: 620 kJ/mol) from JEE reference tables. The negative sign confirms this is an exothermic process — hydrogenation always releases energy. Option -125 kJ/mol uses slightly different C–H values. Option -112 kJ/mol ignores one C–H bond formation. Option -156 kJ/mol double-counts the C=C bond breaking. This type of bond-energy calculation is a standard NCERT and JEE Main exercise. Plausibility check: hydrogenation of alkenes is known to release approximately 120–140 kJ/mol, consistent with the answer.

Bond Enthalpy

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