Back to Directory
NEET CHEMISTRYMedium

Given the following bond energies: H−H\text{H}-\text{H} bond energy: 431.37 kJ mol−1431.37 \text{ kJ mol}^{-1} C=C\text{C}=\text{C} bond energy: 606.10 kJ mol−1606.10 \text{ kJ mol}^{-1} C−C\text{C}-\text{C} bond energy: 336.49 kJ mol−1336.49 \text{ kJ mol}^{-1} C−H\text{C}-\text{H} bond energy: 410.50 kJ mol−1410.50 \text{ kJ mol}^{-1} Enthalpy for the reaction, CH2=CH2+H2→CH3−CH3\text{CH}_2=\text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3-\text{CH}_3 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 chemical reaction is the hydrogenation of ethene: CH2=CH2+H2→CH3−CH3\text{CH}_2=\text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3-\text{CH}_3 The enthalpy of reaction (ΔrH\Delta_r H) can be calculated using bond energies: ΔrH=∑B.E.(reactants)−∑B.E.(products)\Delta_r H = \sum \text{B.E.(reactants)} - \sum \text{B.E.(products)} In the reactants, there is one C=C\text{C}=\text{C} bond, four C−H\text{C}-\text{H} bonds, and one H−H\text{H}-\text{H} bond. In the products, there is one C−C\text{C}-\text{C} bond and six C−H\text{C}-\text{H} bonds. ΔrH=[B.E.(C=C)+4×B.E.(C−H)+B.E.(H−H)]−[B.E.(C−C)+6×B.E.(C−H)]\Delta_r H = [\text{B.E.(C}=\text{C)} + 4 \times \text{B.E.(C}-\text{H)} + \text{B.E.(H}-\text{H)}] - [\text{B.E.(C}-\text{C)} + 6 \times \text{B.E.(C}-\text{H)}] This simplifies to: ΔrH=[B.E.(C=C)+B.E.(H−H)]−[B.E.(C−C)+2×B.E.(C−H)]\Delta_r H = [\text{B.E.(C}=\text{C)} + \text{B.E.(H}-\text{H)}] - [\text{B.E.(C}-\text{C)} + 2 \times \text{B.E.(C}-\text{H)}] Substituting the given values: ΔrH=[606.10+431.37]−[336.49+2(410.50)]\Delta_r H = [606.10 + 431.37] - [336.49 + 2(410.50)] ΔrH=1037.47−[336.49+821.00]\Delta_r H = 1037.47 - [336.49 + 821.00] ΔrH=1037.47−1157.49=−120.02 kJ mol−1\Delta_r H = 1037.47 - 1157.49 = -120.02 \text{ kJ mol}^{-1} Therefore, the enthalpy of the reaction is approximately −120.0 kJ mol−1-120.0 \text{ kJ mol}^{-1}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started