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NEET CHEMISTRYMedium

Given the following bond energies: H-H bond energy =431.37 kJ mol−1= 431.37 \text{ kJ mol}^{-1} C=C bond energy =606.10 kJ mol−1= 606.10 \text{ kJ mol}^{-1} C-C bond energy =336.49 kJ mol−1= 336.49 \text{ kJ mol}^{-1} C-H bond energy =410.50 kJ mol−1= 410.50 \text{ kJ mol}^{-1} Based on the data given above, enthalpy change for the reaction C2H4(g)+H2(g)→C2H6(g)C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g) will be:

A

1523.6 kJ mol−11523.6 \text{ kJ mol}^{-1}

B

−243.6 kJ mol−1-243.6 \text{ kJ mol}^{-1}

C

−120.0 kJ mol−1-120.0 \text{ kJ mol}^{-1}

D

553.0 kJ mol−1553.0 \text{ kJ mol}^{-1}

Step-by-Step Solution

The given reaction is the hydrogenation of ethene: C2H4(g)+H2(g)→C2H6(g)C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g). The standard enthalpy of reaction can be calculated using bond enthalpies : ΔrH∘=∑BE(reactants)−∑BE(products)\Delta_r H^\circ = \sum BE(\text{reactants}) - \sum BE(\text{products}) ΔrH∘=[BE(C=C)+4×BE(C−H)+BE(H−H)]−[BE(C−C)+6×BE(C−H)]\Delta_r H^\circ = [BE(C=C) + 4 \times BE(C-H) + BE(H-H)] - [BE(C-C) + 6 \times BE(C-H)] ΔrH∘=BE(C=C)+BE(H−H)−BE(C−C)−2×BE(C−H)\Delta_r H^\circ = BE(C=C) + BE(H-H) - BE(C-C) - 2 \times BE(C-H) Substituting the given values: ΔrH∘=606.10+431.37−336.49−(2×410.50)\Delta_r H^\circ = 606.10 + 431.37 - 336.49 - (2 \times 410.50) ΔrH∘=1037.47−336.49−821.00=−120.02 kJ mol−1\Delta_r H^\circ = 1037.47 - 336.49 - 821.00 = -120.02 \text{ kJ mol}^{-1}.

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