#73 a. Use bond enthalpies to estimate the reaction enthalpy for:
3C2H2(g)-->C6H6(g)
Could someone explain why in this problem the solution is to simply form 6 C-C bonds? Since there is already one C-C bond in C2H2, why does this bond have to be broken and reformed to produce the carbon ring? Also if this C-C bond is broken and reformed in the math, then why are the hydrogen bonds not broken and reformed as well?
Sorry if this question is somewhat convoluted....
Using bond enthalpies to find reaction enthalpy
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Re: Using bond enthalpies to find reaction enthalpy
Postby Shelby Yee 1J » Wed Jan 20, 2016 11:35 am
The Carbon atoms in the C2H2 molecule are bonded together by a triple bond, so this must be broken and they must be rearranged into a ring in which the Carbons are bonded by only single bonds. It takes more energy to break a triple bond than a single bond, so this might be the reason behind the discrepancy. And since the Hydrogens are already bonded to the carbon atoms with a single bond, it does not contribute to any energy given off or used in the making/forming of bonds since there is essentially no change.
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