gowill
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Hi,

Would somebody kindly help me answer the following question please:

'Would you anticipate the activation energy of a reaction to be larger or smaller than a typical bond energy?' .

I imagine that the activation energy would be higher than the bond energy, but I am unsure of how to explain this and would appreciate an explanation so as to understand this.

Thanks for your help
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Big-Daddy
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(Original post by gowill)
Hi,

Would somebody kindly help me answer the following question please:

'Would you anticipate the activation energy of a reaction to be larger or smaller than a typical bond energy?' .

I imagine that the activation energy would be higher than the bond energy, but I am unsure of how to explain this and would appreciate an explanation so as to understand this.

Thanks for your help
A typical bond dissociation energy might be +400 kJ/mol. A typical activation energy would be between maybe 15 and 200 kJ/mol. I would expect the bond dissociation energy to be higher. And a bond formation energy, while lower than an activation energy, would be larger in magnitude.
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username913907
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(Original post by Big-Daddy)
A typical bond dissociation energy might be +400 kJ/mol. A typical activation energy would be between maybe 15 and 200 kJ/mol. I would expect the bond dissociation energy to be higher. And a bond formation energy, while lower than an activation energy, would be larger in magnitude.
you should take into account that the activation energy is the energy to the transition state so this included bond formation too
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Big-Daddy
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(Original post by JMaydom)
you should take into account that the activation energy is the energy to the transition state so this included bond formation too
Hmm I'm not sure what that adds to the question at hand ...?
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Ari Ben Canaan
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I hate these kinds of questions... I mean what is the point ? Its not adding anything to the learning process and just encourages rote memorisation of answers.
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username913907
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(Original post by Big-Daddy)
Hmm I'm not sure what that adds to the question at hand ...?
ehh..... It does add to the question thankyou. Maybe an explanation based upon theory rather than just taking an average of the bond enthalpies you know and comparing them to average activation energies would be answering the question!!!
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Big-Daddy
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(Original post by JMaydom)
ehh..... It does add to the question thankyou. Maybe an explanation based upon theory rather than just taking an average of the bond enthalpies you know and comparing them to average activation energies would be answering the question!!!
Of course that would be better. I am asking, for the OP's sake as well as my own, how the theory in your post adequately explains activation energies being much smaller than bond energies.

You did quote my post - now I'm looking for the answer!
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username913907
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(Original post by Big-Daddy)
Of course that would be better. I am asking, for the OP's sake as well as my own, how the theory in your post adequately explains activation energies being much smaller than bond energies.

You did quote my post - now I'm looking for the answer!
Lets use [OH]- + MeCl as an example. The bond enthalpy for the C-Cl is the energy required to break the bond, the activation energy is the energy to the transition state which is a state where the C-Cl bond is partially broken but the C-O bond is partially formed, so lower than the bond enthalpy.
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Big-Daddy
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(Original post by JMaydom)
Lets use [OH]- + MeCl as an example. The bond enthalpy for the C-Cl is the energy required to break the bond, the activation energy is the energy to the transition state which is a state where the C-Cl bond is partially broken but the C-O bond is partially formed, so lower than the bond enthalpy.
Ah I see

And what do you about the "transition state" for a reaction that involves a more stable intermediate, e.g. SN1?
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charco
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(Original post by Big-Daddy)
Ah I see

And what do you about the "transition state" for a reaction that involves a more stable intermediate, e.g. SN1?
There is still a transition state (energy barrier) required to achieve the stabilised intermediate...
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username913907
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(Original post by charco)
There is still a transition state (energy barrier) required to achieve the stabilised intermediate...

(Original post by Big-Daddy)
Ah I see

And what do you about the "transition state" for a reaction that involves a more stable intermediate, e.g. SN1?
In this (Sn1) solvation is VERY important to the energy changes. In calculations of bond enthalpy it will be for gas phase molecules, so no solvation.
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gowill
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Ok, I understand this now given the explanations.

Thank you all very much for your help and quick responses
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