Given the following bond energies:
H-H bond energy $= 431.37 ext{ kJ mol}^{-1}$
C=C bond energy $= 606.10

Question

Given the following bond energies:
H-H bond energy $= 431.37 	ext{ kJ mol}^{-1}$
C=C bond energy $= 606.10 	ext{ kJ mol}^{-1}$
C-C bond energy $= 336.49 	ext{ kJ mol}^{-1}$
C-H bond energy $= 410.50 	ext{ kJ mol}^{-1}$
Based on the data given above, enthalpy change for the reaction $C_2H_4(g) + H_2(g) ightarrow C_2H_6(g)$ will be:

Accepted Answer

The given reaction is the hydrogenation of ethene: $C_2H_4(g) + H_2(g) ightarrow C_2H_6(g)$.
The standard enthalpy of reaction can be calculated using bond enthalpies :
$\Delta_r H^\circ = \sum BE(	ext{reactants}) - \sum BE(	ext{products})$
$\Delta_r H^\circ = [BE(C=C) + 4 	imes BE(C-H) + BE(H-H)] - [BE(C-C) + 6 	imes BE(C-H)]$
$\Delta_r H^\circ = BE(C=C) + BE(H-H) - BE(C-C) - 2 	imes BE(C-H)$
Substituting the given values:
$\Delta_r H^\circ = 606.10 + 431.37 - 336.49 - (2 	imes 410.50)$
$\Delta_r H^\circ = 1037.47 - 336.49 - 821.00 = -120.02 	ext{ kJ mol}^{-1}$.

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