A question regarding bonds and their energy

[exchemist]

3 hours ago, observer1 said:

so does entropy affect chemical energy released and needed?

No it doesn't affect the energy released or needed but it does affect whether or how far the reaction proceeds. What determines whether or not a chemical reaction should proceed is whether the free energy of the products is lower than that of the reactants. We normally uses the Gibbs Free Energy, G. The relevant relationship is one I have mentioned in this thread, namely ΔG =   ΔH - TΔS, in which S is Entropy, T is absolute temperature and H is something called the Enthalpy, which is the internal energy in the chemical bonds, U, plus any PV work done during the reaction against the atmosphere (this can be important if there is a big volume change during the reaction for instance if 2 molecules of gas produce 3 molecules.)

You can see from this that whether the change in G is +ve or -ve can depend on the entropy change as well as the enthalpy change. In an earlier post I gave the example of a reaction (dissolving ammonium nitrate) that proceeds even though the products have a higher internal energy than the reactants. This is fun to do - the solution actually gets cold as the salt dissolves! Comparing the enthalpies of products and reactants tells you that if the  reaction proceeds energy input is required. However the reaction occurs spontaneously in spite of this, and it sucks the energy it needs from the environment - because of the effect of the entropy change. The system moves to a state of lower free energy but, in spite of this, an energy input occurs. 

    

[MigL]

Ahh ... Brings back memories of Thermodynamics classes, so many years ago.

Haven't heard 'Gibbs Free Energy' for about 40 years, and I work with Chemical Engineers, and a few Chemists, daily.

[exchemist]

1 hour ago, MigL said:

Ahh ... Brings back memories of Thermodynamics classes, so many years ago.

Haven't heard 'Gibbs Free Energy' for about 40 years, and I work with Chemical Engineers, and a few Chemists, daily.

Haha, yes. In my case we had a related formula dinned into us at school in the 1st yr 6th form: Log "how far" = -ΔH/RT +ΔS/R, (R being the gas constant) which we used to predict what reactions would go effectively to completion and which would not, based on looking up enthalpies and entropies of formation in something called the Rubber Book. This formula came (as we later learnt) from the relation between Gibbs Free Energy change and the equilibrium constant for the reaction:

ΔG =-RT lnK.

That would have been in 1970.....

 

Edited August 18, 2022 by exchemist

[observer1]

so if we take nitrogen gas (N2). the bond energy is 941kJ/mol. so if we add 2 nitrogen will 941kJ/mol energy be released?

can someone explain this

and

[exchemist]

35 minutes ago, observer1 said:

so if we take nitrogen gas (N2). the bond energy is 941kJ/mol. so if we add 2 nitrogen will 941kJ/mol energy be released?

can someone explain this

and

Yes, 941 (or is it 945?) kJ/mol will be released if 2 nitrogen atoms are allowed to form N2. 

Regarding the electrolysis question yes I agree it is confusing. I think it works like this, but I'm open to correction on it as it is a bit tricky to get right.

The lower number, 237.24kJ/mol is ΔG, the increase in Gibbs free energy involved in the dissociation and formation of hydrogen and oxygen from one mole of water, i.e. for the reaction H20 -> H2 + 1/2 O2. The higher number, 285.83 kJ/mol is ΔH, the change in enthalpy.

The change in entropy ΔS is 163J/K mol, which at 25C, i.e. 298K gives 163 x 298 = 48.57kJ/mol. You will see this is the difference between the two figures (apart from a minor rounding error) 

It is ΔH that is closer to the change in the bond energies. (But it is not exactly equal to the change in bond energies, as there is PV work done, due to one molecule generating 1 and 1/2 molecules, i.e. 2 molecules of water generate 3 molecules, 2 of hydrogen plus one of oxygen. This occupies a greater volume so, under atmospheric pressure, work is done against the atmosphere to make room for the extra gas molecules created. Remember H =  U + PV.)

The actual energy input required is ΔH, but the theoretical amount of electrical input needed is given by ΔG. The remainder of the enthalpy is taken in by the system from the environment, i.e. the system will get colder and will suck heat in from around it, like my example of ammonium nitrate. 

At least, I think that is how it works out. 

 

 

[sethoflagos]

9 hours ago, exchemist said:

The actual energy input required is ΔH, but the theoretical amount of electrical input needed is given by ΔG. The remainder of the enthalpy is taken in by the system from the environment, i.e. the system will get colder and will suck heat in from around it, like my example of ammonium nitrate. 

At least, I think that is how it works out. 

+1

Just to complete the circle, in the combustion cycle only a theoretical maximum of 237 kJ/mol of the enthalpy change is thermodynamically available for conversion to eg electricity for electrolysis (this is the meaning of Gibbs free energy). 

[exchemist]

24 minutes ago, sethoflagos said:

+1

Just to complete the circle, in the combustion cycle only a theoretical maximum of 237 kJ/mol of the enthalpy change is thermodynamically available for conversion to eg electricity for electrolysis (this is the meaning of Gibbs free energy). 

Phew, so it was right! Yup, the Gibbs free energy is the energy available to do work, and electrical energy counts as "work" in this context. I must admit I was hesitant about thinking it through for the backwards reaction, but I think it makes sense.  I suppose one can think of it as getting the reaction to go backwards using an 80% "push" from electrical input with the remaining 20% coming from the entropic "pull", due to 2 liquid phase molecules being replaced by 3 gas phase ones. This pull helps to force the bond formation to go in the "uphill direction, energetically speaking, so the balance of energy required will come from the environment.

Of course all this is a bit theoretical, since what with electrode inefficiency losses and the desire for a rapid reaction rate, I'm sure a higher electrical input is required in practice than the mere Gibbs free energy theoretical amount.  

 

[sethoflagos]

1 hour ago, exchemist said:

Of course all this is a bit theoretical, since what with electrode inefficiency losses and the desire for a rapid reaction rate, I'm sure a higher electrical input is required in practice than the mere Gibbs free energy theoretical amount.  

Hydrogen-Oxygen combustion in a fuel cell with full heat recovery is currently capable of around 85% thermodynamic efficiency, so we should be looking at a nett electrical output of a smidgen over 200 kJ/mol.

Conventional alkaline electrolysis has a thermodynamic efficiency of ~70%, but a lot of work is currently being done on Proton Exchange Membrane (PEM) technology and we should soon be able to achieve 85% on that too bringing the energy cost (electrical) down to about 280 kJ/mol. 

Edited August 19, 2022 by sethoflagos
small clarification