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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.

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  • $\begingroup$ You are probably referring to this on the bottom of p169 "Whereas the partial molar Gibbs potentials characterize the equilibrium condition, the enthalpy finds its expression in the heat of reaction. This fact follows from the general significance of the enthalpy as a "potential for heat flux" at constant pressure (equation 6.29). That is, the flux of heat from the surroundings to the system, during the chemical reaction, is equal to the change in the enthalpy." but you should read further down, bottom of p170 $\endgroup$
    – hyportnex
    Jul 5, 2023 at 15:35
  • $\begingroup$ "The quantity $dH/d \tilde{N}$ is known as the heat of reaction; it is the heat absorbed per unit reaction in the vicinity of the equilibrium state. It is positive for endothermic reactions and negative for exothermic reactions. We have assumed that the reaction considered is not one that goes to completion. If the reaction does go to completion, the summation in equation 6.57 does not vanish in the equilibrium state, and this summation appears as an additional term in equation 6.58, etc." $\endgroup$
    – hyportnex
    Jul 5, 2023 at 15:36
  • $\begingroup$ I'm not sure if this is correct but I think this is because the number of atoms does not change in a chemical reaction, even if they're being re-arranged into different molecules and compounds, so dN, is probably not being changed anywhere. $\endgroup$
    – SK Dash
    Jul 5, 2023 at 16:26
  • $\begingroup$ To confirm, you are saying (which I agree with) that $dH=dQ$ if and only if we are at a point in configuration space whereat $\sum \mu_i \nu_i=0$ (i.e. in that particular case of chemical equilibrium)? @hyportnex $\endgroup$
    – EE18
    Jul 5, 2023 at 16:51
  • $\begingroup$ almost, per equation 6.58 if $dH/d\tilde N = -T \partial (\sum_j \nu_j \mu_j)/\partial T$ therefore if $\sum_j \nu_j \mu_j$ is independent of temperature then $dH/d\tilde N=0$. Not being a chemist I do not know what significance this may have in general but certainly the special case of $\sum_j \nu_j \mu_j = 0$ falls into that category with which I can at least "empathize". $\endgroup$
    – hyportnex
    Jul 5, 2023 at 18:43

4 Answers 4

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Writing dQ=TdS is the part of your analysis 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.

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  • $\begingroup$ 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 $dQ = TdS$? $\endgroup$
    – EE18
    Jul 7, 2023 at 18:46
  • $\begingroup$ If the system is in contact with a single reservoir at temperature T, then the Clausius inequality tells us that $Q=T\Delta S-T\sigma$ always. $\endgroup$ Jul 7, 2023 at 21:25
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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 $\Delta H =0$.

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  • $\begingroup$ How are we coming to the conclusion $\Delta H=Q$ here? The first law states $\Delta U = W + Q$. Are you then saying that $W = -P \Delta V$ so $Q = \Delta U + P \Delta V = \Delta H$? I buy that, but am confused as to why the derivation involving $dH = TdS -PdV + \sum_i \mu_i dN_i$ , given the identifications of $TdS = dQ$ and $-PdV = dW$, doesn't seem to lead to this conclusion? $\endgroup$
    – EE18
    Jul 5, 2023 at 20:42
  • $\begingroup$ If II can show you how this works out for a reaction of ideal gases, will that be sufficient to satisfy you? $\endgroup$ Jul 5, 2023 at 23:03
  • $\begingroup$ I'm not quite sure, but perhaps? My questions (in light of your initial answer here) are essentially 1) is the derivation of $\Delta H = Q$ which I supplied correct/are you saying $\Delta H = Q$ holds in general, whether or not we are at equilibrium and 2) how do we reconcile 1 with $dH = TdS -PdV + \sum_i \mu_i dN_i$ for $ \sum_i \mu_i dN_i \neq 0$, as is the case when we are not at equilibrium? $\endgroup$
    – EE18
    Jul 5, 2023 at 23:13
  • $\begingroup$ I'm so used to writing $dU$. I of course mean $dH=TdS+VdP + \sum_i\mu_idN_i$ in the above. $\endgroup$
    – EE18
    Jul 5, 2023 at 23:46
  • $\begingroup$ No. The two end states have to be thermodynamic equilibrium states. Your 2nd question is really 'How do I integrate this form of dH to get $\Delta H$ between the thermodynamic equilibrium end states?" Correct? $\endgroup$ Jul 6, 2023 at 0:41
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I already showed in my 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 $$\bar{H}_j=\mu_j+T\bar{S}_j=\bar{G}_j+T\bar{S}_j$$since the chemical potential is equal to the partial molar Gibbs free energy.

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 $$\Delta H=\left[\sum{\bar{H}_jn_j}\right]_{final}-\left[\sum{\bar{H}_jn_j}\right]_{initial}$$

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 $T_0$.

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You write $dU = TdS - P dV$, which as you say applies to constant $N_i$, but the most general form of the fundamental differential is $$ dU = T dS - P dV + \sum_i \mu_i dN_i $$ which can be mapped to the first law for open system under the associations $$ \begin{array}{lll} T dS & \to &\text{heat} \\ -P dV & \to &\text{mechanical work} \\ \mu_i dN_i & \to&\text{chemical work} \end{array} $$ For reversible process with chemical reaction $\sum_i \mu_i dN_i = 0$, even though $dN_i\neq 0$. Indeed the condition $\sum_i \mu_i dN_i = 0$ is how we obtain the equilibrium constant of the reaction.

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  • $\begingroup$ Hi Themis, thank you as always for your answer but in this case I'm afraid it doesn't answer my question. 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. $\endgroup$
    – EE18
    Jul 6, 2023 at 14:43
  • $\begingroup$ @EE18 I see now. Then the answer is that in a chemical reaction under equilibrium conditions, $\sum \mu_i dN_i=0$, even though $dN_i\neq 0$. The condition $\sum \mu_i dN_i=0$ is precisely how we derive the equilibrium constant of the reaction. $\endgroup$
    – Themis
    Jul 6, 2023 at 15:29
  • $\begingroup$ But I think per Chet’s answer, $\Delta H = Q$ holds for ANY quasi static process, not necessarily in equilibrium throughout. Indeed, it holds in taking us from an initial state to a final equilibrium state, right? $\endgroup$
    – EE18
    Jul 6, 2023 at 16:30
  • $\begingroup$ @EE18 I don't agree with Chet: $\Delta H=Q$ requires (1) constant pressure throughout the entire path and (2) reversible process. Remember that to derive this result we write $$Q = \Delta U - W$$ and to go from $W$ to $W=-P\Delta V$ we require both (1) and (2). $\endgroup$
    – Themis
    Jul 7, 2023 at 8:17
  • $\begingroup$ Yes, I think it may have gotten confused but Chet and I agree with you that those are the conditions. But to arrive at that, we are thinking of the system as a "black box" closed system. 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? $\endgroup$
    – EE18
    Jul 7, 2023 at 13:41

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