How does enthelpy change as the reaction proceeds?

[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?

 

[Borek]

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.

 

[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.

 

[sgstudent]

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.

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?

 

[Borek]

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.

 

[sgstudent]

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.

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?

 

[Borek]

the ΔH represent the difference between final and initial state while the dH represents the change in enthalpy for every infinitely small unit

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)?

 

[sgstudent]

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)?

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

 

[Borek]

Does it answer your original problem?

 

[sgstudent]

Does it answer your original problem?

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?

 

[Astudious]

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?

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 high school 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).

 

[sgstudent]

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 high school 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).

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 Gξ_eq - Gξ_initial 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?

 

[Chestermiller]

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

 

[DrDu]

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

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.

 

[sgstudent]

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.

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.

 

[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).

 

[sgstudent]

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).

But isn't ∆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.

 

[Chestermiller]

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.

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

 

[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.

 

[DrDu]

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.

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.

 

[Chestermiller]

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.

Didn't I say that for ideal solutions, the heat of mixing is zero?

Chet

 

[DrDu]

Of course you did :-)
But I wrote this before reading your answer in reply to sgstudents question why Delta G is so strongly dependent on concentrations.

 

[Chestermiller]

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.

You are confusing Heat of Reaction with the change in enthalpy in a chemical reactor, or the change in enthalpy for a stream passing through

Of course you did :-)
But I wrote this before reading your answer in reply to sgstudents question why Delta G is so strongly dependent on concentrations.

Where did I say anything about Delta G?

Chet

 

[DrDu]

Not you, but sgstudent in post #15

 

[Chestermiller]

Dr. Du,

The equation you wrote in post #14 is consistent with my definition for the case of a single reaction in an ideal solution. But, if it is an ideal solution, then your ΔH will not change with concentration. Also, if there are multiple reactions occurring, then the number of moles of species i can change as a result of multiple reactions, and the partial derivative of H with respect to species i will include the effects of all of these (so the equation will not give the correct result for the heat of reaction of any specific one of these).

Chet

 

[DrDu]

Dr. Du,

The equation you wrote in post #14 is consistent with my definition for the case of a single reaction in an ideal solution.

Agreed.

But, if it is an ideal solution, then your ΔH will not change with concentration. Also, if there are multiple reactions occurring, then the number of moles of species i can change as a result of multiple reactions, and the partial derivative of H with respect to species i will include the effects of all of these (so the equation will not give the correct result for the heat of reaction of any specific one of these).
Chet

I can't quite follow you here. For the formation of the partial derivative of H with respect to any n_i it is not relevant whether n_i changes due to one or several reactions.
So to get the enthalpy of reaction for reaction number j one has simply $\Delta H_j=\sum_i \nu_{ij} \frac{\partial H}{\partial n_i}$.
Specifically $dH=\sum_j d\xi_j \Delta H_j=\sum_i \frac{\partial H}{\partial n_i} \sum_j d\xi_j \nu_{ij}=\sum_i \frac{\partial H}{\partial n_i} dn_i$ as it should be. Here, $\sum_j d\xi_j \nu_{ij}=dn_i$.

 

[Chestermiller]

Agreed.

I can't quite follow you here. For the formation of the partial derivative of H with respect to any n_i it is not relevant whether n_i changes due to one or several reactions.
So to get the enthalpy of reaction for reaction number j one has simply $\Delta H_j=\sum_i \nu_{ij} \frac{\partial H}{\partial n_i}$.
Specifically $dH=\sum_j d\xi_j \Delta H_j=\sum_i \frac{\partial H}{\partial n_i} \sum_j d\xi_j \nu_{ij}=\sum_i \frac{\partial H}{\partial n_i} dn_i$ as it should be. Here, $\sum_j d\xi_j \nu_{ij}=dn_i$.

I'm not able to follow your math here, but I do know that H changes as a result of all the reactions, and not just the reaction under consideration, and dn_i is composed of the changes in n_i for all the reactions.

Chet

 

[DrDu]

I'm not able to follow your math here, but I do know that H changes as a result of all the reactions, and not just the reaction under consideration, and dn_i is composed of the changes in n_i for all the reactions.

Chet

I agree, that is the content of the last formula $dn_i=\sum_j \nu_{ij} d\xi_j$.

 

[Chestermiller]

Well, for whatever it's worth, this is how I learned it:

For a reacting mixture at constant pressure,
$$\frac{dH}{dt}=\sum_i {\left(H_i\frac{dn_i}{dt}+n_i{\frac{\partial H_i}{\partial T}}\frac{dT}{dt}\right)}$$
where H_i is the partial molar enthalpy of species i at temperature T. For an ideal solution, the partial molar enthalpy is equal to the enthalpy of the pure species at the same temperature and total pressure. The time derivatives of the species amounts are related to the rates of reaction by:
$$\frac{dn_i}{dt}=V\sum {\nu _{ji}r_j}$$
where V is the reactor volume and where forward and reverse reactions are treated separately.

The rate of change of partial molar enthalpy with respect to temperature is equal to the partial molar heat capacity C_pi. So,
$$\frac{dH}{dt}=V\sum_i \sum_j {\nu_{ij} H_i r_j}+\left(\sum_i {n_iC_{pi}}\right)\frac{dT}{dt}$$
But,
$$ΔH_j=\sum_i\nu_{ij} H_i$$
where $ΔH_j$ is the Heat of Reaction for reaction j at temperature T and pressure P.

Therefore,
$$\frac{dH}{dt}= V\sum_j {(ΔH_j) r_j}+\left(\sum_i {n_iC_{pi}}\right)\frac{dT}{dt}$$

For an adiabatic situation,

$$\left(\sum_i {n_iC_{pi}}\right)\frac{dT}{dt}=V\sum_j {(-ΔH_j) r_j}$$

Hope this helps.

Chet

 

[sgstudent]

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.

Hmm I get that the equation only has T as the variable and so H is only affected by temperature. But wouldn't the number of moles reaction also affect the amount of heat produced or absorbed which would be related to concentration? And so enthalpy seems like it should be affected by concentration. But I know this concept is wrong but I can't seem to justify why its wrong though.

And as Chestermiller and you have agreed ΔH is affected by both the forward and backwards reaction. So as the reaction goes on wouldn't the backwards reaction decrease the ΔH as well?

 

[DrDu]

Thermodynamically, it makes not much sense to distinguish between forward and backward reactions which are rather concepts from kinetics. All you see is a net reaction, i.e. a change of the concentration of the product with time. If this net rate is $r_i=dn_i/dt$, then $d\xi=r_i/\nu_idt$ and the heat produced per time step is $dH=\Delta H d\xi$. Of course if temperature is changing, too, then you have to take this into account as Chet did in his last post.

 

[sgstudent]

Thermodynamically, it makes not much sense to distinguish between forward and backward reactions which are rather concepts from kinetics. All you see is a net reaction, i.e. a change of the concentration of the product with time. If this net rate is $r_i=dn_i/dt$, then $d\xi=r_i/\nu_idt$ and the heat produced per time step is $dH=\Delta H d\xi$. Of course if temperature is changing, too, then you have to take this into account as Chet did in his last post.

Yes that makes sense. But isn't the rate of reaction not constant because of the mixture reaction equilibrium causing the concentrations of the reactants and products to change? And so dH would change too?

Something like the concentration of products increase over the extent of reaction and so the backwards reaction being more prominent causes the overall dH to decrease? Sorry if I'm bringing this into a loop I just can't seem to get why concentration isn't a factor in dH.

 

[Chestermiller]

And as Chestermiller and you have agreed ΔH is affected by both the forward and backwards reaction. So as the reaction goes on wouldn't the backwards reaction decrease the ΔH as well?

We seem to be talking about two different concepts here. One of these is the enthalpy variation of the reaction mixture, and the other is the molar "Heat of Reaction" of an individual reaction. The enthalpy of a reaction mixture will change as the concentrations of the species in the reactor changes, but the molar Heat of Reaction for each individual reaction of the mixture will not change as the concentration of the species in the reactor changes (if the mixture is an ideal gas mixture or an ideal fluid mixture, each of which exhibits a heat of mixing of zero). In my previous post, I expressed the molar Heat of each Reaction in terms of the partial molar enthalpies of the reactants and products comprising that particular reaction. For an ideal solution, the partial molar enthalpies of the reactants and products are equal to the molar enthalpies of the pure reactants and products at the same temperature and total pressure as the mixture. So the molar heat of an individual reaction is the same in the reaction mixture as it is for going from stoichiometric molar quantities of the pure reactants and ending up with corresponding stoichiometric molar quantities of the pure products.

If the molar Heat of Reaction for an individual reaction changed significantly with concentration even for ideal gas and liquid solutions, then there would be no point in preparing tables of heats of reaction and heats of formation, because they couldn't be used for anything. Their very existence tell us that, at least for ideal solutions, the molar Heats of Reaction do not change with concentration.

Chet

 

[Chestermiller]

Something like the concentration of products increase over the extent of reaction and so the backwards reaction being more prominent causes the overall dH to decrease? Sorry if I'm bringing this into a loop I just can't seem to get why concentration isn't a factor in dH.

From my post #28, for a single reversible reaction occurring in a reactor held at constant temperature, the equation for the variation of the mixture enthalpy H with time is given by:

$$\frac{dH}{dt}=V(r_fΔH_R-r_rΔH_R)=V(r_f-r_r)ΔH_R$$

where V is the reactor volume, r_f is the rate of the forward reaction, r_r is the rate of the reverse reaction and ΔH_R is the molar Heat of Reaction (which is independent of concentration for an ideal solution). So,while the enthalpy of the reaction mixture is changing with time, the molar Heat of Reaction is constant.

Chet

 

[DrDu]

From my post #28, for a single reversible reaction occurring in a reactor held at constant temperature, the equation for the variation of the mixture enthalpy H with time is given by:

$$\frac{dH}{dt}=V(r_fΔH_R-r_rΔH_R)=V(r_f-r_r)ΔH_R$$

where V is the reactor volume, r_f is the rate of the forward reaction, r_r is the rate of the reverse reaction and ΔH_R is the molar Heat of Reaction (which is independent of concentration for an ideal solution). So,while the enthalpy of the reaction mixture is changing with time, the molar Heat of Reaction is constant.

Chet

... and $r=r_f-r_r$ is the net rate of the reaction.