# How does enthalpy change as the reaction proceeds?

1. Mar 28, 2015

### sgstudent

My concept of enthalpy change is that the change is based on limits that approach zero. Something like the heat evolved/absorbed per second (but instead of second its a infinitely small).

A+B→C ΔH=100J/mole of A (standard enthalpy change)

And as such when the reaction first proceeds the initial ΔH is the enthalpy change of reaction for the reactants undergoing reaction(100J/mole of A reacted). However, at the next instance some product is formed and as such some of the backwards reaction would occur and so there is a negative enthalpy change (-100J/mole of A formed). And the backwards reaction keeps increasing until the forward reaction = backwards reaction and so the ΔH drops to zero.

Is this concept right? If so how do we determine the total energy released in that reaction?

Last edited: Mar 28, 2015
2. Mar 28, 2015

### Staff: Mentor

I have serious problems deciphering what you wrote.

Aren't you talking about kinetics - how fast the heat is produced? (I am using "heat" loosely here, somehow the idea of "producing enthalpy" doesn't sound right to me). If so, what is happening is not different from the way we can treat changes in the amount of substance produced (concentration changes), it is exactly the same math.

3. Mar 28, 2015

### epenguin

I hesitated to reply earlier because of the same discomfort as Borek.
To put it another way, any chemical reaction produces a change in some physical parameter (which is the only way we know about it). The chemical change A →B produces a certain amount of heat per mole converted. It may produce also a certain change in optical absorbance at a particular wavelength, or fluorescence per mole and many other properties can be used. In all cases the reaction is accompanied by some change in the physical parameter per mole, and measuring that parameter one can follow the reaction in time. Calorimetry is not the most used, but its use has increased and very sensitive calorimeters with continuous recording are available. The actual kinetics of the change must be the same whatever instrument you use.

4. Mar 29, 2015

### sgstudent

Perhaps this post on reddit might have been better in asking this question (more specifically the answer by the user smugbug23) http://www.reddit.com/r/explainlikeimfive/comments/30ksvi/eli5_how_does_enthalpy_change_change/

I was thinking that the enthalpy change of reaction changes as the reaction proceeds because during the course of the reaction some of the reactants are converted to products and because of that some of the products would undergo the backwards reaction reducing the enthalpy change of the overall reaction. And so at equilibrium ΔH=0.

However that seemed to be debunked by smugbug23 as the ΔH is defined in the terms of the reaction in the direction it was written. And so despite the backwards reaction occurring it does not affect the enthalpy change of the forward reaction. He said that ΔH is the energy change per infinitely small progression of the reaction where the progression is measured in an infinitely small change in concentration. Is what he said true?

5. Mar 29, 2015

### Staff: Mentor

Define precisely what you mean by ΔH - especially what is the "delta" here.

And reread epenguin post, as IMHO he precisely addresses what is important here.

6. Mar 29, 2015

### sgstudent

The ΔH should be a dH i believe? Because the ΔH represent the difference between final and initial state while the dH represents the change in enthalpy for every infinitely small unit right? Similarly at equilibrium we should say that dG=0 rather than ΔG=0 because again ΔG represents final-initial?

7. Mar 29, 2015

### Staff: Mentor

Looks OK to me.

Now, do you agree that dH=αdn - where n is number of moles of the substance produced (consumed), and α is just some proportionality constant/conversion factor (that already includes stoichiometric conversion if it is needed)?

8. Mar 29, 2015

### sgstudent

Yes that makes sense. The number of moles of substance produced or reactant consumed will be proportional to dH i believe.

9. Mar 29, 2015

### Staff: Mentor

10. Mar 29, 2015

### sgstudent

Yes it does. To confirm that I have it right, would it be correct to say that ΔH is used for the final and initial states of the chemical equation while dH is used to determine the actual energy change during the reaction?

If so, then can we tie this together with Gibbs Free Energy? Because I've seen in notes and books that ΔG=0 but then again since ΔH represent the different in final and initial states ΔG should also be so. So it shouldn't be dG=0 instead?

11. Mar 29, 2015

### Astudious

You will run into serious problems here because of ambiguity in the definition of ΔG typically used.

dG/dξ = 0 at equilibrium, yes. But ΔG = 0 too.

This is because of the (non-trivial) fact that dG/dξ evaluated at a certain extent ξ1 (which itself is defined as coming from your set initial state) is equivalent to ΔG, that is change in G between the initial state (ξ=0) and that extent ξ1 being considered. This should require detailed proof for you and it is available on Wikipedia. http://en.wikipedia.org/wiki/Chemical_equilibrium

Because dG/dξ = ΔG for all ξ, in any process, and dG/dξ = 0 by definition at equilibrium (since we have a minimum in G at equilibrium), ΔG = 0 (as ξ for equilibrium is just one particular value of extent for a given process). This is a subtle truth. Meanwhile, dH/dξ does not equal ΔH in general. So while dH/dξ = 0 at the equilibrium extent, ΔH does not (rather, it represents what you throughout highschool chemistry etc. knew as "the enthalpy change of the reaction" - the change in H that occurs when the system goes from initial state to equilibrium "completion" - and that is all we can say about it).

12. Mar 29, 2015

### sgstudent

Thanks for the insightful reply I have read the proof from the wiki page but I can't really understand it. They ended with dG/dξ = ΔG = 0 But how can this be so?

From the diagram they gave us shouldn't ΔG be Gξeq - Gξinitial giving us a negative value instead of zero?

Also in this page http://en.wikipedia.org/wiki/Extent_of_reaction they stated that dH/dξ=ΔH so that's confusing as well. Is there a difference in the 2 terms?

13. Mar 29, 2015

### Staff: Mentor

As a reaction procedes, the enthalpy change does not drop to zero, even as equilibrium is approached. The rate at which the reaction mixture generates heat is R(-ΔH), where R is the net rate of the reaction and where ΔH is negative for an exothermic reaction. So, as the rate of the reaction approaches zero, the rate of heat generation drops to zero.

The heat of reaction ΔH represents the amount of heat that has to be added at a system between the following two thermodynamic equilibrium states:

State 1: Pure reactants in stoichiometric proportions at temperature T and at a pressure of 1 atm.

State 2: Pure products in corresponding stoichiometric proportions at temperature T and at a pressure of 1 atm.

So the heat of reaction is a function of state, and has nothing to do with how fast a reaction is occurring.

Chet

Last edited: Mar 29, 2015
14. Mar 30, 2015

### DrDu

Does it really? I'd rather have said that $\Delta H := \sum_i \nu_i \frac{\partial H(T, p, \{n_i\})}{\partial n_i}|_{T,p, \{n_{j\neq i}\}}=\frac{\partial H}{\partial \xi}$, where $\nu_i$ are the stochiometric coefficients of the reaction. So it will change during the reaction, too, although not as much as does $\Delta G$.
To put it differently, $\Delta H$ is the difference of the momentary molar enthalpies of the reactants and products, and the molar enthalpies depend somewhat on the changing concentrations of all reaction partners.

Last edited: Mar 30, 2015
15. Mar 30, 2015

### sgstudent

Why does it change less compared to ∆G? I had the impression the 2 were similar so they should both decrease as the reaction reaches equilibrium.

16. Mar 30, 2015

### DrDu

Delta G varies so strongly with concentration as the entropy is strongly concentration dependent, even for ideal mixtures (for which Delta H would be independent of concentration).

17. Mar 30, 2015

### sgstudent

But isnt ∆H also dependent on the concentration of the reactants and products? As the reaction proceeds the concentration of reactions decrease and products increase so wouldn't that affect the ∆H too? But I'm not too sure if the backwards reaction will contribute to the ∆H that we are talking about here.

18. Mar 30, 2015

### Staff: Mentor

Well, maybe you would say that, but that's not consistent with how the quantity called Heat of Reaction is defined in the literature and it is also not consistent with how it is used in practice. I know this because I have analyzed actual real world reactors involving reaction kinetics (featuring dozens of reactions), material balances, and heat balances during my career as a practicing professional chemical engineer.

Heat of reaction is a precisely defined quantity, obtained according to the definition I presented in post #13, and is available in published tables or is calculated from heats of formation in published tables:

Introduction to Chemical Engineering Thermodynamics, Smith and Van Ness
Chemical Process Principles, Volume 1, Hougen, Watson, Ragatz
Elements of Chemical Reaction Engineering, Fogler
Chemical Reaction Engineering, Levenspeil

Heat of Reaction does not change with concentration because, in the definition, the reactants and products start and end in their pure states, respectively. In doing a heat balance on a real reactor, the effects of concentration come into play when you have to include the heats of mixing which, for an ideal solution, are zero.

Chet

19. Mar 30, 2015

### DrDu

Not for ideal mixtures. E.g. for gasses, the internal energy is only a function of T, but not of p and/ or V. Same holds for H, as H=U+pV=U+nRT. In mixtures, p must be replaced by the partial pressure $p_i$ so that $p_i V=n_i RT$. So in the end, for ideal gasses, the molar enthalpy does only depend on T, but not on concentration.
This is not true for entropy. Upon mixing two gasses, the entropy changes by $-R n_1 \ln X_1 -R n_2 \ln X2$ where $X_i =n_i/(n_1+n_2)$, i.e. the fraction of moles.

20. Mar 30, 2015

### DrDu

For me, that sounds more like the definition of the standard enthalpy of reaction. You are right in that the standard enthalpy of reaction and the actual enthalpy of reaction differ by the heat of mixing, which can often be ignored in practice.