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Thermodynamics — Medium JEE chemistry MCQ

Using the bond enthalpies: C-H = $414$ kJ/mol, C-C = $347$ kJ/mol, C=C = $611$ kJ/mol, H-H = $435$ kJ/mol. The enthalpy change for the hydrogenation of ethene ($\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6$) is:
  1. A. $-125$ kJ/mol
  2. B. $+125$ kJ/mol
  3. C. $-250$ kJ/mol
  4. D. $+250$ kJ/mol

Solution

Bonds broken (reactants): - 1 C=C bond: $611$ kJ - 1 H-H bond: $435$ kJ - 4 C-H bonds in ethene: $4 \times 414 = 1656$ kJ Total energy absorbed = $611 + 435 + 1656 = 2702$ kJ Bonds formed (products): - 1 C-C bond: $347$ kJ - 6 C-H bonds in ethane: $6 \times 414 = 2484$ kJ Total energy released = $347 + 2484 = 2831$ kJ $\Delta H = \text{Energy absorbed} - \text{Energy released}$ $\Delta H = 2702 - 2831 = -129 \approx -125$ kJ/mol (The hydrogenation is exothermic)

CHEMISTRY

medium • PYQ Reworded
Question
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Using the bond enthalpies: C-H = 414414 kJ/mol, C-C = 347347 kJ/mol, C=C = 611611 kJ/mol, H-H = 435435 kJ/mol. The enthalpy change for the hydrogenation of ethene (C2H4+H2→C2H6\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6) is:
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Thermodynamics — Medium JEE Chemistry MCQ | MyGoalPrep