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Topic: Bond Dissociation Energy for Double/Triple Bonds  (Read 5727 times)

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Offline galpinj

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Bond Dissociation Energy for Double/Triple Bonds
« on: July 25, 2016, 01:12:57 PM »
Hi everyone,

Quick question about BDE; normally BDE for double and triple bonds is higher for single bonds. My question, which may be dumb, is if enough BDE is given to break a double bond, does this mean that only the double bond has been broken and the single bond is still present or that both bonds have been broken?

For example, C=C has a BDE of 144 kcal/mol. If 144 kcal/mol were added to break the bond, would the product be C. + .C or would the result be C-C (double bond broken)

Thank you!

Offline AWK

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #1 on: July 25, 2016, 01:34:59 PM »
Data show the whole energy needed to disconect atoms.
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Offline j.k.11

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #2 on: July 25, 2016, 02:43:46 PM »
Bond dissociation energy gives the energy required for homolytic bond cleavage.

For a single bond that's sigma bond cleavage; for double bonds sigma + pi; for triple bonds sigma + 2 pi.

Offline Enthalpy

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #3 on: July 25, 2016, 04:25:14 PM »
As a sidenote, I suggest that you use joules instead of calories for your own use. But you must know calories too because useful docs use them.

Offline galpinj

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #4 on: July 25, 2016, 05:48:34 PM »
Really, eh? I was pretty confident the answer would be that it only breaks one bond - go figure! So, in a weird way, it's more efficient to break a C=C double bond than it would be to break two C-C single bonds. That seems a little off to me, but I guess that's the way it is!

Offline Mitch

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #5 on: July 25, 2016, 05:52:59 PM »
Really, eh? I was pretty confident the answer would be that it only breaks one bond - go figure! So, in a weird way, it's more efficient to break a C=C double bond than it would be to break two C-C single bonds. That seems a little off to me, but I guess that's the way it is!

A single bond uses stronger sigma bonding. A double bond uses a weaker pi bond plus a strong sigma bond. So, yes a double bond does not simply mean two times the strength of a single bond.
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Offline galpinj

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #6 on: July 25, 2016, 06:34:42 PM »
Really, eh? I was pretty confident the answer would be that it only breaks one bond - go figure! So, in a weird way, it's more efficient to break a C=C double bond than it would be to break two C-C single bonds. That seems a little off to me, but I guess that's the way it is!

A single bond uses stronger sigma bonding. A double bond uses a weaker pi bond plus a strong sigma bond. So, yes a double bond does not simply mean two times the strength of a single bond.

Thanks Mitch! Any reason why the pi bond would not be as strong as a sigma bond? Is it because sigma bonds contain more "s" character?

Appreciate the *delete me*

Offline Mitch

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #7 on: July 25, 2016, 07:29:51 PM »
Because sigma orbitals have more overlap, bonding, than pi orbitals. More overlap, the stronger the bond.
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Offline mjc123

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #8 on: July 26, 2016, 10:00:40 AM »
But consider
O-O 146 kJ/mol; O=O 495
N-N 163; N=N 418; N≡N 941
Attributed to the effects of lone pair repulsion

Offline galpinj

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Re: Bond Dissociation Energy for Double/Triple Bonds
« Reply #9 on: August 09, 2016, 09:33:32 PM »
But consider
O-O 146 kJ/mol; O=O 495
N-N 163; N=N 418; N≡N 941
Attributed to the effects of lone pair repulsion

How does lone-pair repulsion influence our BDE in your examples? I can't see how the lone pairs are influencing the BDE between two atoms

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