Search any question & find its solution
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
Options:
Solution:
1518 Upvotes
Verified Answer
The correct answer is:
$-125 \mathrm{~kJ}$
$-125 \mathrm{~kJ}$
$\mathrm{CH}_2=\mathrm{CH}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{CH}_3-\mathrm{CH}_3$
$\Delta H=1(\mathrm{C}=\mathrm{C})+4(\mathrm{C}-\mathrm{H})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-6(\mathrm{C}-\mathrm{H})=1(\mathrm{C}=\mathrm{C})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-2(\mathrm{C}-\mathrm{H})$
$=615+435-347-2 \times 414=1050-1175=-125 \mathrm{~kJ}$
$\Delta H=1(\mathrm{C}=\mathrm{C})+4(\mathrm{C}-\mathrm{H})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-6(\mathrm{C}-\mathrm{H})=1(\mathrm{C}=\mathrm{C})+1(\mathrm{H}-\mathrm{H})-1(\mathrm{C}-\mathrm{C})-2(\mathrm{C}-\mathrm{H})$
$=615+435-347-2 \times 414=1050-1175=-125 \mathrm{~kJ}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.