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They say that enthalpy of reaction is the heat when at constant temperature and pressure. So if enthalpy is

$$\mathrm dH = \mathrm dU + p\,\mathrm dV = T\,\mathrm dS - p\,\mathrm dV + p\,\mathrm dV + μ\,\mathrm dN = T\,\mathrm dS + μ\,\mathrm dN$$

and heat is $T\,\mathrm dS,$ then what happens to the chemical potential term if enthalpy is heat?

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They say that enthalpy of reaction is the heat when at constant temperature and pressure.

They should not say that. At least not without saying in the absence of non-PV work. And in a closed system.

So if [...] heat is TdS [...]

You should not say that without specifying a reversible (near equilibrium) process in a closed system where dS refers to the surrounding:

$$q_\mathrm{rev} = T dS_\mathrm{sur}$$

The sign of $q_\mathrm{rev}$ is from the perspective of the system, i.e. if there is heat transfer from system to surrounding (positive value), the entropy of the surrounding will increase (we don't know what happens to the entropy of the system because there is more going on in the system whereas the only change in the surrounding is that it got heat transferred to it).

On the other hand, all the thermodynamic quantities in your formula refer to the system. By the way, the enthalpy and the entropy of a reaction are not linked, there is no general formula that lets you infer reaction enthalpy from reaction entropy without knowing other quantities.

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    $\begingroup$ Thanks, this makes some sense. So what you're saying is TdS is the reversible heat but is not the heat of reaction which is the enthalpy at constant T and P with no non-pv work and it being a closed system. $\endgroup$ – ChemEng Sep 28 at 15:09
  • $\begingroup$ Yes, and dS refers to the entropy change of the surroundings. $\endgroup$ – Karsten Theis Sep 28 at 18:54

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