A question regarding bonds and their energy

[observer1]

Ok so I know that it takes energy to split a chemical bond and energy is released when a chemical bond is made
My question is:-
is the energy  required to break a bond = energy released when the Same bond is created

[studiot]

2 hours ago, observer1 said:

Ok so I know that it takes energy to split a chemical bond and energy is released when a chemical bond is made
My question is:-
is the energy  required to break a bond = energy released when the Same bond is created

Not necessarily because life if more complicated than you present.

 

Some bonds release energy when they break.

These are called exoergic   (also exothermic but that is strictly heat is released.

The reason they don't immediately break up by themelves and release energy is something called activation energy.

 

How far do you want to go with this topic ?

[MigL]

Systems tend to their lowest energy state.
Sometimes that's a bound state, sometimes it's a 'free' state.

Energy is released ( I use exothermic, but Studiot is correct ) to get to lower energy states.

[exchemist]

2 hours ago, observer1 said:

Ok so I know that it takes energy to split a chemical bond and energy is released when a chemical bond is made
My question is:-
is the energy  required to break a bond = energy released when the Same bond is created

Yes, it is identical. 

23 minutes ago, studiot said:

Not necessarily because life if more complicated than you present.

 

Some bonds release energy when they break.

These are called exoergic   (also exothermic but that is strictly heat is released.

The reason they don't immediately break up by themelves and release energy is something called activation energy.

 

How far do you want to go with this topic ?

I don't understand your answer here. As far as I know there is no bond that releases energy when it breaks. If that were to happen the "bonded" state would have a higher energy than the unbonded state, which would be an antibond rather than a bond, wouldn't it? So the "bond" would never form in the first place.  

But maybe I'm misunderstanding what you are getting at. Can you give an example, to help clarify? 

[swansont]

42 minutes ago, studiot said:

Some bonds release energy when they break.

Energy is released only if a stronger bond is formed. If all you’re doing is breaking a bond, it requires an input of energy.

[MigL]

My mistake also.
Didn't realize the OP specified chemical bonds.
I was also considering nuclear bonds in fission.

[exchemist]

57 minutes ago, swansont said:

Energy is released only if a stronger bond is formed. If all you’re doing is breaking a bond, it requires an input of energy.

The only rationalisation I can think of for @studiot's comment is that he is thinking of endothermic reactions, in which the internal energy of the products is greater than that of the reactants. That can happen if there is a sufficiently favourable entropy change, such that the free energy decreases even though the internal energy increases. ΔG= ΔU +PΔV -TΔS etc.

But all that means is that there is less energy released by the new bonds that form than is absorbed by those that break.    

[observer1]

the main reason why i asked this is because i wonder if the energy required to break a water bond and then when they are stored in tanks in rockets, the energy released when the combine, it the energy the same?

[exchemist]

30 minutes ago, observer1 said:

the main reason why i asked this is because i wonder if the energy required to break a water bond and then when they are stored in tanks in rockets, the energy released when the combine, it the energy the same?

Yes. When hydrogen is burnt with oxygen the reaction is 2 H2 + O2 -> 2H2O. You break 2 H-H bonds and an O=O double bond, which requires energy.  But the energy you get back from forming 4 H-O bonds, i.e. H-O-H + H-O-H is greater, by 286kJ/mol of water. If you run the reaction backwards, as you do in electrolysis of water to make hydrogen and oxygen, this is the energy you have to input to do it.   

[observer1]

what about nitrogen gas? since it takes A lot of energy to break its triplebond, is the energy released when the thripe bond bond is made the same as the energy input?

[exchemist]

29 minutes ago, observer1 said:

what about nitrogen gas? since it takes A lot of energy to break its triplebond, is the energy released when the thripe bond bond is made the same as the energy input?

Yes, in any chemical bond the energy to break it is what you get when it forms. There is an intrinsic energy reduction due to the bonding electrons entering a state of lower energy than in the isolated atoms. Because energy is conserved, if you reverse the process that is the energy needed to break the bond again. 

But be aware that, in industrial processes in practice, there are always inefficiencies. So for example in electrolysis, you don't get anything like 100% conversion of the actual electricity input into splitting water molecules, as there are a lot of energy losses along the way.  

 

[studiot]

6 hours ago, exchemist said:

I don't understand your answer here. As far as I know there is no bond that releases energy when it breaks. If that were to happen the "bonded" state would have a higher energy than the unbonded state, which would be an antibond rather than a bond, wouldn't it? So the "bond" would never form in the first place.  

But maybe I'm misunderstanding what you are getting at. Can you give an example, to help clarify? 

 

3 hours ago, exchemist said:

Yes. When hydrogen is burnt with oxygen the reaction is 2 H2 + O2 -> 2H2O. You break 2 H-H bonds and an O=O double bond, which requires energy.  But the energy you get back from forming 4 H-O bonds, i.e. H-O-H + H-O-H is greater, by 286kJ/mol of water. If you run the reaction backwards, as you do in electrolysis of water to make hydrogen and oxygen, this is the energy you have to input to do it.   

 

Thank you for asking what I meant rather than just taking an opposite view. +1

Is the solution of hydrogen flouride or sodium hydroxide not exothermic ?

 

4 hours ago, observer1 said:

the main reason why i asked this is because i wonder if the energy required to break a water bond and then when they are stored in tanks in rockets, the energy released when the combine, it the energy the same?

Thank you for the clarification.

I think the comments here show the value of a well posed question as opposed to a vague one, with respect, like yours.

 

Ok you have mentioned water.

So there are many types of bonds.

Some types of bonds have a very specific 'bond' energy.

Other types have a range of energies and a statistical average.

Water is subject to both of these types.

The intra molecular bonds exchemist refers to are of the fixed or specific type.

But the bonds which hold solid water together are of the statistical type.

So it is conceivable to create the bonds on the low side of the statistical average and replace them with new hydrogen bonds on the high side of the statistical average bond energy. And of course vice versa.

Of course this process cannot be continued indefinitely on account of conservation of energy.

[sethoflagos]

3 hours ago, exchemist said:

Yes. When hydrogen is burnt with oxygen the reaction is 2 H2 + O2 -> 2H2O. You break 2 H-H bonds and an O=O double bond, which requires energy.  But the energy you get back from forming 4 H-O bonds, i.e. H-O-H + H-O-H is greater, by 286kJ/mol of water. If you run the reaction backwards, as you do in electrolysis of water to make hydrogen and oxygen, this is the energy you have to input to do it.   

I think @studiot may have made a valid point.

As an example consider that your initial mixing of 2 moles of hydrogen with one mole of oxygen increased the entropy of the mixture by about 1.89 R. In the absence of reaction no energy change occurred and hence there is none to recover in the reverse reaction. However, mixing is an irreversible process (in the thermodynamic sense) and unmixing the products of the reverse reaction back into their pure elemental states will require quite a bit of unbudgeted additional work input.   

 

[studiot]

Just now, sethoflagos said:

I think @studiot may have made a valid point.

As an example consider that your initial mixing of 2 moles of hydrogen with one mole of oxygen increased the entropy of the mixture by about 1.89 R. In the absence of reaction no energy change occurred and hence there is none to recover in the reverse reaction. However, mixing is an irreversible process (in the thermodynamic sense) and unmixing the products of the reverse reaction back into their pure elemental states will require quite a bit of unbudgeted additional work input.   

 

Thanks +1

[exchemist]

1 hour ago, sethoflagos said:

I think @studiot may have made a valid point.

As an example consider that your initial mixing of 2 moles of hydrogen with one mole of oxygen increased the entropy of the mixture by about 1.89 R. In the absence of reaction no energy change occurred and hence there is none to recover in the reverse reaction. However, mixing is an irreversible process (in the thermodynamic sense) and unmixing the products of the reverse reaction back into their pure elemental states will require quite a bit of unbudgeted additional work input.   

 

That has nothing to do with bond energy, however.

1 hour ago, studiot said:

 

 

Thank you for asking what I meant rather than just taking an opposite view. +1

Is the solution of hydrogen flouride or sodium hydroxide not exothermic ?

 

 

Sure, dissolving solid NaOH in water is exothermic, like a great number of chemical reactions. That does not mean there is any bond that releases energy when it breaks, however. It simply means the new bonds that form release more energy than is absorbed by the bonds that are broken.  

[sethoflagos]

47 minutes ago, exchemist said:

That has nothing to do with bond energy, however.  

I take your point.

But in the reverse reaction you are breaking bonds by electrolysis, so aren't you steering hydrogen to the cathode and oxygen to the anode thereby unmixing them? I think the piper will still need to be paid his due.

Similar arguments can I think be made for all implicit irreversibilities in the process (in both directions).

However, the constant making and breaking of bonds in an equilibrium mixture of products and reactants (or similar system) has no irreversibilities. So here, the OP's premise must be true. The energy released by the forward reaction is exactly balanced by that consumed in the reverse.

[exchemist]

6 minutes ago, sethoflagos said:

I take your point.

But in the reverse reaction you are breaking bonds by electrolysis, so aren't you steering hydrogen to the cathode and oxygen to the anode thereby unmixing them? I think the piper will still need to be paid his due.

Similar arguments can I think be made for all implicit irreversibilities in the process (in both directions).

However, the constant making and breaking of bonds in an equilibrium mixture of products and reactants (or similar system) has no irreversibilities. So here, the OP's premise must be true. The energy released by the forward reaction is exactly balanced by that consumed in the reverse.

You are right about the overall thermodynamics of course, but you are talking of the entropy term in the free energy of reaction. The OP question was about bond energy, which is the internal energy term, the major part of the enthalpy of reaction. G= U+PV-TS.

You are talking about S while the OP is enquiring about U, isn’t it?

[sethoflagos]

3 minutes ago, exchemist said:

You are right about the overall thermodynamics of course, but you are talking of the entropy term in the free energy of reaction. The OP question was about bond energy, which is the internal energy term, the major part of the enthalpy of reaction. G= U+PV-TS.

You are talking about S while the OP is enquiring about U, isn’t it?

I'm sure you're better acquainted with the current technical definition of 'standard enthalpy of formation' than I am, and all the caveats that go along with it. And yes, if all those caveats are observed then delta H in the forward direction equals minus delta H in the reverse. But don't those caveats require that conditions at start and finish are globally unchanged? In other words, the equality of enthalpies is dependent on there being no overall change in entropy? 

That was my understanding but then I'm Chem Eng, not a Chemist. 

[MigL]

My understanding as a Physicist, not a Chemist or Chem Engineer 

13 hours ago, MigL said:

Systems tend to their lowest energy state.

For chemical bonds, which involve electron interactions, the bound state is always lower energy than the free state.

[studiot]

5 hours ago, MigL said:

For chemical bonds, which involve electron interactions, the bound state is always lower energy than the free state.

That would generally be true, all other things being equal.

But why do some molecules break up under the influence of catalysts, but not without the catalyst ?

We should be discussing which bonds, which energies and which states.

 

11 hours ago, exchemist said:

Sure, dissolving solid NaOH in water is exothermic, like a great number of chemical reactions. That does not mean there is any bond that releases energy when it breaks, however. It simply means the new bonds that form release more energy than is absorbed by the bonds that are broken.  

So exactly which new bonds are you thinking of, especially with reference to my hydrogen flouride example ?
The hydrogen flouride molecule breaks up but what new bonds do the hydrogen and flouride ions form ?

Edited August 18, 2022 by studiot

[exchemist]

1 hour ago, studiot said:

That would generally be true, all other things being equal.

But why do some molecules break up under the influence of catalysts, but not without the catalyst ?

We should be discussing which bonds, which energies and which states.

 

So exactly which new bonds are you thinking of, especially with reference to my hydrogen flouride example ?
The hydrogen flouride molecule breaks up but what new bonds do the hydrogen and flouride ions form ?

Electrostatic ion-solvent bonding. See solvation energy: https://en.wikipedia.org/wiki/Solvation

The process of an ionic solid dissolving in a polar solvent requires the ionic bonds in the crystal lattice to be broken. This requires energy input. The energy to do it comes from the greater energy released, when polar solvent molecules are electrostatically attracted to the cations and anions and settle into a solvent cage around them. (We can argue semantically about whether these ion-solvent interactions are to be called a formal "chemical bond" or not, but they are in terms of physics no different from an ionic "bond" in a crystal, viz. a reduction in electrostatic potential energy by the approach of unlike electric charges.)   

None of this is to dismiss the role that entropy also plays in determining whether a solid dissolves or not. Famously, dissolving ammonium nitrate in water leads to a cooling of the solution: it is an endothermic process. In this example, the entropy gain when it dissolves outweighs the enthalpy gain, so the reaction still occurs, even though it is "uphill" from the point of view of internal energy change alone. In terms of the classic chemist's formula ΔG =   ΔH - TΔS, even though   ΔH is +ve, ΔG can still be -ve, if the effect of the  TΔS term is big enough.

As for catalysis etc, this is to do with kinetics of reaction rather than thermodynamics. Catalysts lower the activation energy for the reaction to proceed and thereby increate the reaction rate. They allow a lower energy path as the reactant bonds break, reducing the energy of the transition state between reactants and products. But energy input is still needed, to get to even this lower energy transition state.   

 

Edited August 18, 2022 by exchemist

[studiot]

2 hours ago, exchemist said:

Electrostatic ion-solvent bonding. See solvation energy: https://en.wikipedia.org/wiki/Solvation

The process of an ionic solid dissolving in a polar solvent requires the ionic bonds in the crystal lattice to be broken. This requires energy input. The energy to do it comes from the greater energy released, when polar solvent molecules are electrostatically attracted to the cations and anions and settle into a solvent cage around them. (We can argue semantically about whether these ion-solvent interactions are to be called a formal "chemical bond" or not, but they are in terms of physics no different from an ionic "bond" in a crystal, viz. a reduction in electrostatic potential energy by the approach of unlike electric charges.)   

None of this is to dismiss the role that entropy also plays in determining whether a solid dissolves or not. Famously, dissolving ammonium nitrate in water leads to a cooling of the solution: it is an endothermic process. In this example, the entropy gain when it dissolves outweighs the enthalpy gain, so the reaction still occurs, even though it is "uphill" from the point of view of internal energy change alone. In terms of the classic chemist's formula ΔG =   ΔH - TΔS, even though   ΔH is +ve, ΔG can still be -ve, if the effect of the  TΔS term is big enough.

As for catalysis etc, this is to do with kinetics of reaction rather than thermodynamics. Catalysts lower the activation energy for the reaction to proceed and thereby increate the reaction rate. They allow a lower energy path as the reactant bonds break, reducing the energy of the transition state between reactants and products. But energy input is still needed, to get to even this lower energy transition state.   

 

I didn't say I wanted to argue semantics or that you do not make some very valid points.

What you are proving is that the situation is, as I said, more complicated than the OP thought so a simple answer will give the wrong impression.

[exchemist]

50 minutes ago, studiot said:

I didn't say I wanted to argue semantics or that you do not make some very valid points.

What you are proving is that the situation is, as I said, more complicated than the OP thought so a simple answer will give the wrong impression.

Not in the least. The OP asked the simplest of questions: whether the energy released when a bond forms is the same as the energy required to break it. And the answer to that is an unambiguous "yes". All the rest of the stuff we have been discussing since is merely all the other things that can influence the course of a chemical reaction. But, interesting thoughtit all may be,  that is not what was asked. 

[observer1]

so does entropy affect chemical energy released and needed?

[studiot]

18 minutes ago, observer1 said:

so does entropy affect chemical energy released and needed?

Have a look at Hess's law and Entropy.

 

https://en.wikipedia.org/wiki/Hess's_law

 

This is used as the basis for calculating many bond energies.

 

 

 

Edited August 18, 2022 by studiot