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Question: Answered & Verified by Expert
If at $298 \mathrm{~K}$ the bond energies of $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}$ and $\mathrm{H}-\mathrm{H}$ bonds are respectively $414,347,615$ and $435 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the value of enthalpy change for the reaction $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{H}_3 \mathrm{C}-\mathrm{CH}_3(\mathrm{~g})$ at $298 \mathrm{~K}$ will be
ChemistryThermodynamics (C)JEE MainJEE Main 2003
Options:
  • A
    $-250 \mathrm{~kJ}$
  • B
    $+125 \mathrm{~kJ}$
  • C
    $-125 \mathrm{~kJ}$
  • D
    $+250 \mathrm{~kJ}$
Solution:
1518 Upvotes Verified Answer
The correct answer is:
$-125 \mathrm{~kJ}$
$\mathrm{CH}_2=\mathrm{CH}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_3-\mathrm{CH}_3$
$\Delta H=1(\mathrm{C}=\mathrm{C})+4(\mathrm{C}-\mathrm{H})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-6(\mathrm{C}-\mathrm{H})=1(\mathrm{C}=\mathrm{C})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-2(\mathrm{C}-\mathrm{H})$
$=615+435-347-2 \times 414=1050-1175=-125 \mathrm{~kJ}$

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