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Question: Answered & Verified by Expert
Average $\mathrm{C}-\mathrm{H}$ bond energy is $416 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following is correct?
ChemistryThermodynamics (C)AP EAMCETAP EAMCET 2004
Options:
  • A $\mathrm{CH}_4(g)+416 \mathrm{~kJ} \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)$
  • B $\mathrm{CH}_4(g) \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)+416 \mathrm{~kJ}$
  • C $\mathrm{CH}_4(g)+1664 \mathrm{~kJ} \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)$
  • D $\mathrm{CH}_4(g) \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)+1664 \mathrm{~kJ}$
Solution:
2895 Upvotes Verified Answer
The correct answer is: $\mathrm{CH}_4(g)+1664 \mathrm{~kJ} \longrightarrow \mathrm{C}(g)+4 \mathrm{H}(g)$
$\mathrm{CH}_4$ has four $\mathrm{C}-\mathrm{H}$ bonds. So, $4 \times 416=1664 \mathrm{~kJ}$ is required to break $\mathrm{CH}_4$ into $\mathrm{C}$ and $4 \mathrm{H}$.

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