Why is ##dQ = dH## still valid for chemical reaction (Callen)?

[EE18]

Consider a system which is characterized by the extensive variables $(U,V,N_1,...,N_m)$. For a quasistatic process which occurs in contact with some pressure reservoir and where the $N_i$ are constant, one has
$$dU = TdS -PdV \implies TdS = dQ,$$
where the implication follows from the First Law. Defining the relevant Legendre transform for constant pressure processes (the enthalpy, $H = U+PV$) leads to
$$dH = TdS +VdP = TdS = dQ,$$
where we have used $dP = 0$ for this quasistatic process occurring while in contact with a pressure reservoir.

Now I emphasize that this development has depended intimately on $dN_i = 0$ throughout. Nevertheless, in his Chapter 6.4, Callen uses $dH = dQ$ even for processes involving chemical reactions (but which are otherwise closed), wherein the $dN_i$ are certainly not constant in general. He provides no justification for using this though, so I am hoping someone can clear that up.

 

[Chestermiller]

If you consider the system a black box, then, irrespective of whether there is a chemical reaction within the system, if the process occurs with the system in contact with constant pressure environment equal to the initial pressure of the reacting system, the first law tells us that $$\Delta H=Q$$However, the left hand side of the equation will depend on the changes in the amounts of reactants and products and the initial and final temperatures. For example, for an ideal gas reaction, even if the initial and final temperatures is the same, $\Delta H$ and Q will not be zero. And, if Q = 0 (insulated system), $Delta T$ will not be zero even though $\Delta H =0$.

 

[EE18]

If you consider the system a black box, then, irrespective of whether there is a chemical reaction within the system, if the process occurs with the system in contact with constant pressure environment equal to the initial pressure of the reacting system, the first law tells us that $$\Delta H=Q$$However, the left hand side of the equation will depend on the changes in the amounts of reactants and products and the initial and final temperatures. For example, for an ideal gas reaction, even if the initial and final temperatures is the same, $\Delta H$ and Q will not be zero. And, if Q = 0 (insulated system), $Delta T$ will not be zero even though $\Delta H =0$.

Thank you for your reply as always! As discussed on the other site, I agree with this derivation but am curious as to why I can't obtain the result using the $dH$ fundamental relation:

Effectively, my concern is that Callen derived $dH = dQ$ by assuming that the $dN_i$ are constant ($dH=TdS+VdP + \sum_i\mu_idN_i = TdS = dQ$, after using $dP = 0$ when in contact with a pressure reservoir). However, in a chemical rxn, the $dN_i$ are not constant so that this derivation doesn't work, but $dH=TdS+VdP + \sum_i\mu_idN_i = TdS + \sum_i\mu_idN_i$ still holds. How do I use this latter expression to get to $dH = dQ$ is essentially the question.

 

[Chestermiller]

This is a really good question.

Suppose you had your closed system in contact with a constant temperature reservoir at temperature T (equal to the initial system temperature) during the chemical reaction process taking place at constant surroundings pressure P (equal to the initial system pressure)? Would you say that this process is reversible? Would you say that, for this process, $Q=T\Delta S$?

 

[EE18]

This is a really good question.

Suppose you had your closed system in contact with a constant temperature reservoir at temperature T (equal to the initial system temperature) during the chemical reaction process taking place at constant surroundings pressure P (equal to the initial system pressure)? Would you say that this process is reversible? Would you say that, for this process, $Q=T\Delta S$?

I am not sure! I think my confusion may lie in not appreciating/understanding the first law's form for processes wherein mole numbers change.

 

[Chestermiller]

I am not sure! I think my confusion may lie in not appreciating/understanding the first law's form for processes wherein mole numbers change.

Not really, Your issue is with the 2nd law.

This chemical reaction is occurring spontaneously, with no control over its rate in order to guarantee that it is proceeding quasi statically and reversibly. Also, if there is heat of reaction (exothermic or endothermic), the interior of the reaction mixture willl not be at the same temperature as the reservoir. So, in addition to the reaction proceeding spontaneously, the heat transfer is also proceeding with finite temperature gradients. Both these irreversibilities will prevent the integral of dQ/T from being equal to $\Delta S$ (as required for a reversible process). Instead, we will have $$Q=T\Delta S-T\sigma$$where $\sigma$ is the amount of entropy generated due to irreversibility. So we will also have $$Q=\Delta H=T\Delta S-T\sigma$$or$$\Delta G=-T\sigma<0$$$\Delta G$ being negative is likewise consistent with the process being irreversible.

 

[vanhees71]

Thank you for your reply as always! As discussed on the other site, I agree with this derivation but am curious as to why I can't obtain the result using the $dH$ fundamental relation:

Effectively, my concern is that Callen derived $dH = dQ$ by assuming that the $dN_i$ are constant ($dH=TdS+VdP + \sum_i\mu_idN_i = TdS = dQ$, after using $dP = 0$ when in contact with a pressure reservoir). However, in a chemical rxn, the $dN_i$ are not constant so that this derivation doesn't work, but $dH=TdS+VdP + \sum_i\mu_idN_i = TdS + \sum_i\mu_idN_i$ still holds. How do I use this latter expression to get to $dH = dQ$ is essentially the question.

One must not forget that you can introduce the $\mu_i$ only in connection with conserved quantities $N_i$ (in chemistry particle numbers for individual species of atoms/molecules). This implies that $\sum_i \mu_i \mathrm{d} N_i=0$. So the individual $\mathrm{d}N_i$ change due to the reaction but overall conservation of the various quantum numbers gurantees the vanishing of the sum.

 

[Chestermiller]

One must not forget that you can introduce the $\mu_i$ only in connection with conserved quantities $N_i$ (in chemistry particle numbers for individual species of atoms/molecules). This implies that $\sum_i \mu_i \mathrm{d} N_i=0$. So the individual $\mathrm{d}N_i$ change due to the reaction but overall conservation of the various quantum numbers gurantees the vanishing of the sum.

$\sum_i \mu_i \mathrm{d} N_i=0$ is the condition for chemical equilibrium, which is not the case in the present system. See my answer in post #6. This term is part of the entropy generation in this irreversible process.

 

[vanhees71]

Sure, but in the OP it was expclicitly said the process is understood to be quasistatic.

 

[Chestermiller]

Sure, but in the OP it was expclicitly said the process is understood to be quasistatic.

How do you make a chemical reaction proceed quasistatically?

 

[vanhees71]

I understood it such that it's an idealized situation, where the system stays in thermal and chemical equilibrium all the time. That's what's usually treated in the introductory physics texts on thermodynamics, and so in the quoted book by Callen (Sect. 6-4), i.e., only chemical equilibrium is considered (in the form $\sum_i \nu_i \mu_i=0$, where $\nu_i$ are the stoichometric coefficients). The non-ideal case is of course much more complicated.

 

[Chestermiller]

I understood it such that it's an idealized situation, where the system stays in thermal and chemical equilibrium all the time. That's what's usually treated in the introductory physics texts on thermodynamics, and so in the quoted book by Callen (Sect. 6-4), i.e., only chemical equilibrium is considered (in the form $\sum_i \nu_i \mu_i=0$, where $\nu_i$ are the stoichometric coefficients). The non-ideal case is of course much more complicated.

Again my question: In a closed system, how do you make a chemical reaction proceed quasi statically unless the reaction is at equilibrium to begin with? Under these circumstances, there would be no process, and the initial and final states would be the same state.

 

[vanhees71]

It's not a closed system. It's kept at constant pressure! Are you saying that all introductory textbooks on the subject are wrong?

 

[EE18]

One must not forget that you can introduce the $\mu_i$ only in connection with conserved quantities $N_i$ (in chemistry particle numbers for individual species of atoms/molecules). This implies that $\sum_i \mu_i \mathrm{d} N_i=0$. So the individual $\mathrm{d}N_i$ change due to the reaction but overall conservation of the various quantum numbers gurantees the vanishing of the sum.

I don't agree with this, whether quasistatic or not. I don't think quasistatic or not is the crux of this question. Whether or not we can actually arrange to have a chemical reaction proceed quasistatically, we can imagine it. Nevertheless, $\sum_i \mu_i \mathrm{d} N_i=0$ certainly does not hold unless we are right at chemical equilibrium. This is analogous to how we can have $\sum_i P_i \mathrm{d} V_i\neq0$ as we move towards mechanical equilibrium by slowly moving a piston.

Thus, my question is how we can reconcile thinking of the system as a "black box" closed system with which we can arrive at $\Delta H = Q$. However, we can also take the view of looking at the system and think about integrating $dH=TdS+VdP + \sum_i\mu_idN_i = TdS + \sum_i\mu_idN_i$, and yet it's not clear how $$\int dH = H = \int TdS + \int \sum_i\mu_idN_i = \int dQ + \int \sum_i\mu_idN_i = Q + \int \sum_i\mu_idN_i.$$ But why should $\int \sum_i\mu_idN_i=0$ in general?

 

[Chestermiller]

It's not a closed system. It's kept at constant pressure! Are you saying that all introductory textbooks on the subject are wrong?

In my terminology, closed means no exchange of mass with the surroundings. Heat transfer and work are okay.

 

[Chestermiller]

I don't agree with this, whether quasistatic or not. I don't think quasistatic or not is the crux of this question. Whether or not we can actually arrange to have a chemical reaction proceed quasistatically, we can imagine it. Nevertheless, $\sum_i \mu_i \mathrm{d} N_i=0$ certainly does not hold unless we are right at chemical equilibrium. This is analogous to how we can have $\sum_i P_i \mathrm{d} V_i\neq0$ as we move towards mechanical equilibrium by slowly moving a piston.

Thus, my question is how we can reconcile thinking of the system as a "black box" closed system with which we can arrive at $\Delta H = Q$. However, we can also take the view of looking at the system and think about integrating $dH=TdS+VdP + \sum_i\mu_idN_i = TdS + \sum_i\mu_idN_i$, and yet it's not clear how $$\int dH = H = \int TdS + \int \sum_i\mu_idN_i = \int dQ + \int \sum_i\mu_idN_i = Q + \int \sum_i\mu_idN_i.$$ But why should $\int \sum_i\mu_idN_i=0$ in general?

Like I said, dQ=TdS only for a reversible process, and this process is not reversible. That’s the fallacy in your reasoning.

 

[vanhees71]

Ok, but then you have this standard situation, discussed in all textbooks. The change of enthalpy in a reaction just gives the "production or consumption" of heat if this reaction takes place at finite pressure.

 

[EE18]

Ok, but then you have this standard situation, discussed in all textbooks. The change of enthalpy in a reaction just gives the "production or consumption" of heat if this reaction takes place at finite pressure.

My textbook, Callen, just asserts $H = \Delta Q$. As discussed above, I understand how to think of it from one perspective but not the other, and am looking for help in understanding that other perspective.

 

[Chestermiller]

My textbook, Callen, just asserts $H = \Delta Q$. As discussed above, I understand how to think of it from one perspective but not the other, and am looking for help in understanding that other perspective.

I thought I covered that in this https://physics.stackexchange.com/q...id-for-processes-involving-chemical-reactions

Here it is, reproduced for the present thread:
I already showed in a previous answer that, if the closed system is in contact with a constant pressure reservoir at the same pressure P as in the initial equilibrium state of the system, and, if the system pressure in the final equilibrium state is also P, then $$Q=\Delta H$$So the only remaining question now is how to determine $\Delta H$ from the equation $$dH=TdS+VdP+\sum{\mu_jdn_j}$$The first step is recognizing that, in this situation, for S we can write $S=S(T,P,n_1, n_2, ...)$such that $$dS=\frac{\partial S}{\partial T}dT+\frac{\partial S}{\partial P}dP+\sum{\bar{S}_jdn_j}$$where $\bar{S}_j$ is the partial molar entropy of species j in the reaction mixture: $$\bar{S}_j=\frac{\partial S}{\partial n_j}$$If we now substitute this into our equation for dH, we obtain:$$dH=T\frac{\partial S}{\partial T}dT+\left[V+T\frac{\partial S}{\partial P}\right]dP+\sum{[\mu_j+T\bar{S}_j]dn_j}$$
This is the same equation as $$dH=\frac{\partial H}{\partial T}dT+\frac{\partial H}{\partial P}dP+\sum{\bar{H}_jdn_j}$$with $$\frac{\partial H}{\partial T}=T\frac{\partial S}{\partial T}$$$$\frac{\partial H}{\partial P}=V+T\frac{\partial S}{\partial P}$$and, with the partial molar enthalpy of species j $\bar{H}_j$ given by $$\bar{H}_j=\mu_j+T\bar{S}_j=\bar{G}_j+T\bar{S}_j$$

In Chapter 11 of Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, it is shown that, if $\bar{M}_j$ is the partial molar value for species j of extensive property M of a mixture, then the following simple equation relates the overall mixture M to the partial molar values of M for the species in the mixture:$$M=\sum{n_j\bar{M}_j}$$Applying this powerful result to the enthalpy H of our reaction mixture gives$$H=\sum{n_j\bar{H}_j}=\sum{(\bar{G}_j+T\bar{S}_j)}$$
This equation is very general, and applies to all mixtures.

It follows from this that the change in enthalpy for a closed system experiencing a chemical reaction (or other process) is given by $$Q=\Delta H=\left[\sum{\bar{H}_jn_j}\right]_{final}-\left[\sum{\bar{H}_jn_j}\right]_{initial}$$

To apply these relationships, all one needs to know is how to determine mathematically the partial molar enthalpy of a species in the mixture. Let's consider the simplest case of an ideal gas mixture. According to "Gibbs theorem," the partial molar enthalpy of a species in an ideal gas mixture is the same as that of the pure species at the same temperature as the mixture: $$\bar{H}_j=H^F_j(T_0)+\int_{T_0}^T{C_pdT}$$where $H^F_j(T_0)$ is the heat of formation of species j at temperature $T_0$

 

[DrDu]

As long as there is no exchange of substances over the boundary, $$dU=\delta Q + \delta W$$, which defines heat.
If work is only equilibrium p-V work, $\delta W=-pdV$, so that for constant pressure
$$dH=\delta Q$$, irrespective of whether a chemical reaction is taking place near or far equilibrium.
However, then , $$TdS=\delta Q -\sum \mu_i dn_i$$.
So only in chemical equilibrium, where $$\sum \mu_i dn_i=0$$, $$ dS=\delta Q/T$$. This is clear as a chemical reaction far from equilibrium will go in hand with a change of entropy.

 

[DrDu]

Again my question: In a closed system, how do you make a chemical reaction proceed quasi statically unless the reaction is at equilibrium to begin with?

That's easy. Just short cut a battery with a high ohmic resistor. The system consists of both the battery and the resistor. So we can assume that it does not do other work than pV work with its surrounding.

 

[EE18]

I thought I covered that in this https://physics.stackexchange.com/q...id-for-processes-involving-chemical-reactions

Here it is, reproduced for the present thread:
I already showed in a previous answer that, if the closed system is in contact with a constant pressure reservoir at the same pressure P as in the initial equilibrium state of the system, and, if the system pressure in the final equilibrium state is also P, then $$Q=\Delta H$$So the only remaining question now is how to determine $\Delta H$ from the equation $$dH=TdS+VdP+\sum{\mu_jdn_j}$$The first step is recognizing that, in this situation, for S we can write $S=S(T,P,n_1, n_2, ...)$such that $$dS=\frac{\partial S}{\partial T}dT+\frac{\partial S}{\partial P}dP+\sum{\bar{S}_jdn_j}$$where $\bar{S}_j$ is the partial molar entropy of species j in the reaction mixture: $$\bar{S}_j=\frac{\partial S}{\partial n_j}$$If we now substitute this into our equation for dH, we obtain:$$dH=T\frac{\partial S}{\partial T}dT+\left[V+T\frac{\partial S}{\partial P}\right]dP+\sum{[\mu_j+T\bar{S}_j]dn_j}$$
This is the same equation as $$dH=\frac{\partial H}{\partial T}dT+\frac{\partial H}{\partial P}dP+\sum{\bar{H}_jdn_j}$$with $$\frac{\partial H}{\partial T}=T\frac{\partial S}{\partial T}$$$$\frac{\partial H}{\partial P}=V+T\frac{\partial S}{\partial P}$$and, with the partial molar enthalpy of species j $\bar{H}_j$ given by $$\bar{H}_j=\mu_j+T\bar{S}_j=\bar{G}_j+T\bar{S}_j$$

In Chapter 11 of Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, it is shown that, if $\bar{M}_j$ is the partial molar value for species j of extensive property M of a mixture, then the following simple equation relates the overall mixture M to the partial molar values of M for the species in the mixture:$$M=\sum{n_j\bar{M}_j}$$Applying this powerful result to the enthalpy H of our reaction mixture gives$$H=\sum{n_j\bar{H}_j}=\sum{(\bar{G}_j+T\bar{S}_j)}$$
This equation is very general, and applies to all mixtures.

It follows from this that the change in enthalpy for a closed system experiencing a chemical reaction (or other process) is given by $$Q=\Delta H=\left[\sum{\bar{H}_jn_j}\right]_{final}-\left[\sum{\bar{H}_jn_j}\right]_{initial}$$

To apply these relationships, all one needs to know is how to determine mathematically the partial molar enthalpy of a species in the mixture. Let's consider the simplest case of an ideal gas mixture. According to "Gibbs theorem," the partial molar enthalpy of a species in an ideal gas mixture is the same as that of the pure species at the same temperature as the mixture: $$\bar{H}_j=H^F_j(T_0)+\int_{T_0}^T{C_pdT}$$where $H^F_j(T_0)$ is the heat of formation of species j at temperature $T_0$

I read and appreciated your answer, but didn't understand directly how it explained how to reconcile the $dH$ differential with $dQ$. Perhaps I will need to read again to deal with this.

 

[EE18]

As long as there is no exchange of substances over the boundary, $$dU=\delta Q + \delta W$$, which defines heat.
If work is only equilibrium p-V work, $\delta W=-pdV$, so that for constant pressure
$$dH=\delta Q$$, irrespective of whether a chemical reaction is taking place near or far equilibrium.
However, then , $$TdS=\delta Q -\sum \mu_i dn_i$$.
So only in chemical equilibrium, where $$\sum \mu_i dn_i=0$$, $$ dS=\delta Q/T$$. This is clear as a chemical reaction far from equilibrium will go in hand with a change of entropy.

My question is how we can reconcile thinking of the system as a "black box" closed system with which we can arrive at $\Delta H = Q$ as you've wirrten, with the fact that we can also take the view of looking inside at the system and think about integrating $dH=TdS+VdP + \sum_i\mu_idN_i = TdS + \sum_i\mu_idN_i$, and yet it's not clear how $$\int dH = H = \int TdS + \int \sum_i\mu_idN_i = \int dQ + \int \sum_i\mu_idN_i = Q + \int \sum_i\mu_idN_i.$$ But why should $\int \sum_i\mu_idN_i=0$ in general? Are you saying that writing $dH=TdS+VdP + \sum_i\mu_idN_i$ is not even correct for a process with respect to a system outside of equilibrium, but converging to equilibrium?

 

[Chestermiller]

My question is how we can reconcile thinking of the system as a "black box" closed system with which we can arrive at $\Delta H = Q$ as you've wirrten, with the fact that we can also take the view of looking inside at the system and think about integrating $dH=TdS+VdP + \sum_i\mu_idN_i = TdS + \sum_i\mu_idN_i$, and yet it's not clear how $$\int dH = H = \int TdS + \int \sum_i\mu_idN_i = \int dQ + \int \sum_i\mu_idN_i = Q + \int \sum_i\mu_idN_i.$$ But why should $\int \sum_i\mu_idN_i=0$ in general? Are you saying that writing $dH=TdS+VdP + \sum_i\mu_idN_i$ is not even correct for a process with respect to a system outside of equilibrium, but converging to equilibrium?

No. Writing dQ=TdS is the part that is incorrect. It should read $$dQ=TdS-Td\sigma$$, where $d\sigma$ is the differential entropy generated due to irreversibility of the chemical reaction. So, $$Td\sigma=-\sum{\mu_j}dN_j$$which captures the entropy generation due to the spontaneous reaction.

 

[EE18]

No. Writing dQ=TdS is the part that is incorrect. It should read $$dQ=TdS-Td\sigma$$, where $d\sigma$ is the differential entropy generated due to irreversibility of the chemical reaction. So, $$Td\sigma=-\sum{\mu_j}dN_j$$which captures the entropy generation due to the spontaneous reaction.

Very interesting, thank you for pointing out precisely my mistake. I see now. If you're willing, if you can post even just these last two sentences as an answer on the other site I would be happy to accept it. Thank you again!

 

[EE18]

No. Writing dQ=TdS is the part that is incorrect. It should read $$dQ=TdS-Td\sigma$$, where $d\sigma$ is the differential entropy generated due to irreversibility of the chemical reaction. So, $$Td\sigma=-\sum{\mu_j}dN_j$$which captures the entropy generation due to the spontaneous reaction.

Last question for you: $\Delta H = \int dH = \int TdS + VdP = \int dQ + 0 = Q$ by considering the system as a black box and using $dQ = TdS$. But then we are also saying that we can't use $dQ = TdS$?

 

[EE18]

As long as there is no exchange of substances over the boundary, $$dU=\delta Q + \delta W$$, which defines heat.
If work is only equilibrium p-V work, $\delta W=-pdV$, so that for constant pressure
$$dH=\delta Q$$, irrespective of whether a chemical reaction is taking place near or far equilibrium.

Can you explain this in more detail? The first law says
$$dU=\delta Q + \delta W$$
which is no problem. If $dW = -PdV$ then $dQ = dU + PdV = dU + d(PV) = d(U+PV) = dH$. Is this how you're arguing?

 

[Chestermiller]

Last question for you: $\Delta H = \int dH = \int TdS + VdP = \int dQ + 0 = Q$ by considering the system as a black box and using $dQ = TdS$. But then we are also saying that we can't use $dQ = TdS$?

$$\Delta H=Q=T\Delta S-T\sigma$$So, what's the problem? Note that dH=TdS+VdP applies only to a reversible path between same two end states, and with not chemical reaction.

 

[DrDu]

Can you explain this in more detail? The first law says
$$dU=\delta Q + \delta W$$
which is no problem. If $dW = -PdV$ then $dQ = dU + PdV = dU + d(PV) = d(U+PV) = dH$. Is this how you're arguing?

Yes, exactly. And, as Chestermiller pointed out. If there is a chemical reaction, heat exchange is not the only source of entropy.

 

[DrDu]

That's easy. Just short cut a battery with a high ohmic resistor. The system consists of both the battery and the resistor. So we can assume that it does not do other work than pV work with its surrounding.

Just to give some other examples: You could run a reaction far from equilibrium quasistatically if it requires some homogeneous catalyst and you are using very small concentrations of it. In case of enzymatic reactions, reactions will run very slowly at lower temperatures.

 

[Chestermiller]

Just to give some other examples: You could run a reaction far from equilibrium quasistatically if it requires some homogeneous catalyst and you are using very small concentrations of it. In case of enzymatic reactions, reactions will run very slowly at lower temperatures.

You could also run a reaction reversibly using a van't Hoff equilibrium box.

 

[DrDu]

You could also run a reaction reversibly using a van't Hoff equilibrium box.

Yes, but I understood that the question was whether you can run a non-equilibrium reaction quasistatically, not reversibly.