Chemistry multiple choice question HELP! Watch

Yatayyat
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What would I need to do to answer this?

https://imgur.com/vkJd7XC

I know that for an enthalpy to be most exothermic, it must release the most amount of energy from all 4 given options.

But I can't see how I would figure out which one to be the most exothermic?

Any help would be great. Thanks!
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BobbJo
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(Original post by Yatayyat)
What would I need to do to answer this?

https://imgur.com/vkJd7XC

I know that for an enthalpy to be most exothermic, it must release the most amount of energy from all 4 given options.

But I can't see how I would figure out which one to be the most exothermic?

Any help would be great. Thanks!
Use your understanding of the bonding in benzene. Alternating single and double bonds would form a resonance structure whereby the pi electrons are delocalised. The enthalpy of hydrogenation would be less exothermic as the bonds broken are stronger.
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Yatayyat
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(Original post by BobbJo)
Use your understanding of the bonding in benzene. Alternating single and double bonds would form a resonance structure whereby the pi electrons are delocalised. The enthalpy of hydrogenation would be less exothermic as the bonds broken are stronger.
Why do you exactly mean when you mention 'resonance structure' I have never heard of that term before in chemistry?
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BobbJo
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(Original post by Yatayyat)
Why do you exactly mean when you mention 'resonance structure' I have never heard of that term before in chemistry?
You must have encountered the structure of benzene.

Kekule first postulated a structure for the compound consisting of alternating single and double bonds. This was proven wrong by scientists later.

In fact the structure of benzene is a delocalised ring. The pi bonds are not static and can move. This gives rise to different structures, called resonance structures (or canonical forms, tautomers etc.). All six of the pi electrons in benzene are delocalized over all six atoms and are not uniquely shared by a given pair of two atoms.
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safa_k
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The answer would be A because the -1,3- double bonds in B are closer than the -1,4- double bonds in A. This means the p orbitals would overlap, hence more energy is required to break the bonds for B which is more stable.
Last edited by safa_k; 4 weeks ago
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BobbJo
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(Original post by safa_k)
The answer would be B because the -1,3- double bonds are closer than the -1,4- double bonds in A. This means the p orbitals would overlap, hence more energy is required to break the bonds.
But it is asking for the most exothermic enthalpy change
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Yatayyat
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(Original post by BobbJo)
Use your understanding of the bonding in benzene. Alternating single and double bonds would form a resonance structure whereby the pi electrons are delocalised. The enthalpy of hydrogenation would be less exothermic as the bonds broken are stronger.
Is it not that when bonds are broken it is an endothermic process, because you have to supply some energy to overcome the attractive forces between the atoms. So if the bonds are stronger then, wouldn't it be more exothermic since stronger bonds release more energy than weaker bonds where they break down? Sorry, but I just can't see how a resonance structure will help identify if the enthalpy of hydrogenation is more exothermic or not?
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safa_k
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(Original post by BobbJo)
But it is asking for the most exothermic enthalpy change
sorry yeah i edited my answer
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Yatayyat
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(Original post by safa_k)
The answer would be A because the -1,3- double bonds in B are closer than the -1,4- double bonds in A. This means the p orbitals would overlap, hence more energy is required to break the bonds for B which is more stable. A is less stable so it releases more energy??
Why would it be A in terms of the resonance structure?
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BobbJo
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(Original post by Yatayyat)
Is it not that when bonds are broken it is an endothermic process, because you have to supply some energy to overcome the attractive forces between the atoms. So if the bonds are stronger then, wouldn't it be more exothermic since stronger bonds release more energy than weaker bonds where they break down? Sorry, but I just can't see how a resonance structure will help identify if the enthalpy of hydrogenation is more exothermic or not?
Recall the following
Enthalpy change = Bonds broken - bonds made
If bonds broken are stronger, it will be less exothermic, not more

You say bond breaking is endothermic but go on to say that the enthalpy change would be more exothermic. There is a flaw in the reasoning

In each case, C-H bonds are formed. The same number of C-H bonds are formed. The determining factor is the strength of C=C bonds.

The stronger the bonds formed, the more exothermic
The stronger the bonds broken, the less exothermic

Enthalpy change = Bond energy of reactants - that of products
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BobbJo
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(Original post by Yatayyat)
Why would it be A in terms of the resonance structure?
A does not form a resonance structure

B,C,D contain alternating single and double bonds and there will be delocalisation of the pi electrons. The bonds would be stronger and require more energy to break.
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safa_k
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ok so think about it as delocalisation. The -1,3- double bonds in B C and D means there is greater delocalisation of the electrons (as the p orbitals overlap). This makes the structure more stable so more energy is required to break the bonds. For A, the bonds are -1,4- meaning the bonds are too far apart to show as much delocalisation so it is less stable and less energy is required to break the bonds. The same amount of energy is released when forming the C-H bonds in all the structures. This means A is more exothermic.
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BobbJo
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(Original post by safa_k)
too far apart to show as much delocalisation so it is less stable and less energy is required to break the bonds. The same amount of energy is required to form the C-H bonds in all the structures. This means A is more exothermic.
Energy is not required to form C-H bonds. Energy is released when C-H bonds form.
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safa_k
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(Original post by BobbJo)
Energy is not required to form C-H bonds. Energy is released when C-H bonds form.
Oops! Sorry! Thanks for correcting me
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Yatayyat
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(Original post by BobbJo)
A does not form a resonance structure

B,C,D contain alternating single and double bonds and there will be delocalisation of the pi electrons. The bonds would be stronger and require more energy to break.
Okay so I have to find the value that is most negative as this would be the most exothermic enthalpy change.

If a lot more of the energy used is done creating bonds then there is a lot less energy used breaking the bonds as this is the only way to get an exothermic reaction.

Does this mean that less energy is used up in breaking carbon double bonds? Why does it need to be C=C double bonds and not any other bond in the compound, can't it be single bonds instead?

Do I look for the compound that appears to be the least stable out of all 4 compounds then, as this need little energy to then re-stabilise?

I'm still confused to how resonance plays a big part in this?
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BobbJo
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(Original post by Yatayyat)
Okay so I have to find the value that is most negative as this would be the most exothermic enthalpy change.
https://www.chemguide.co.uk/basicorg.../benzene1.html

Reading the above will greatly help you think on the right track

If a lot more of the energy used is done creating bonds then there is a lot less energy used breaking the bonds as this is the only way to get an exothermic reaction.

Does this mean that less energy is used up in breaking carbon double bonds? Why does it need to be C=C double bonds are not any other bond in the compound, can't it be single bonds instead?

Do I look for the compound that appears to be the least stable out of all 4 compounds then, as this need little energy to then re-stabilise?

I'm still confused to how resonance plays a big part in this?
Enthalpy change = Bonds broken - bonds made
When hydrogenating, you break:
1. pi bonds in C=C
2. H-H bonds
You form:
1. C-H bonds

You don't use energy to create bonds. When bonds form, energy is released. Energy is used to break bonds. It is as if stretching an elastic band. You have to input the force to break it, but it will release heat when contracting back (i.e heat is released when bonds are formed).

Look at A,B,C,D. They all have 2 pi bonds. So number of C-H bonds formed will be the same. You have to consider the strength of C=C bonds or C-C (bond is partial single and partial double (intermediate bond order)).

You do not need to complicate it.

Look at benzene again. Recall that the question asks you to consider bonding in benzene.
Benzene has a delocalised pi ring. Kekule structure is alternating single and double bonds. The pi electrons in p orbitals are delocalised, forming a pi ring. Reading the above will tell you that the bonds are stronger in the ring than simply static single and double bonds. This is due to the hydrogenation enthalpy being more endothermic than expected.

Similarly,
B, C, D have carbon atoms with alternating single and double bonds. The pi electrons in p orbitals are delocalised, strengthening the bond. Additional energy ("delocalisation energy") has to be provided in bond breaking. So the enthalpy change is more endothermic.

A does not have alternating single and double bonds.
Hence answer is A.
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Yatayyat
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(Original post by BobbJo)
https://www.chemguide.co.uk/basicorg.../benzene1.html

Reading the above will greatly help you think on the right track

Enthalpy change = Bonds broken - bonds made
When hydrogenating, you break:
1. pi bonds in C=C
2. H-H bonds
You form:
1. C-H bonds

You don't use energy to create bonds. When bonds form, energy is released. Energy is used to break bonds. It is as if stretching an elastic band. You have to input the force to break it, but it will release heat when contracting back (i.e heat is released when bonds are formed).

Look at A,B,C,D. They all have 2 pi bonds. So number of C-H bonds formed will be the same. You have to consider the strength of C=C bonds or C-C (bond is partial single and partial double (intermediate bond order)).

You do not need to complicate it.

Look at benzene again. Recall that the question asks you to consider bonding in benzene.
Benzene has a delocalised pi ring. Kekule structure is alternating single and double bonds. The pi electrons in p orbitals are delocalised, forming a pi ring. Reading the above will tell you that the bonds are stronger in the ring than simply static single and double bonds. This is due to the hydrogenation enthalpy being more endothermic than expected.

Similarly,
B, C, D have carbon atoms with alternating single and double bonds. The pi electrons in p orbitals are delocalised, strengthening the bond. Additional energy ("delocalisation energy" has to be provided in bond breaking. So the enthalpy change is more endothermic.

A does not have alternating single and double bonds.
Hence answer is A.
Thanks a lot! That weblink as well as your explanation cleared up my doubt. Thanks again really!
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