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We know, $\mathrm{d}H = nC_p\mathrm{d}T$ which implies $\mathrm{d}H = f(\mathrm{d}T)$ using $\mathrm{d}H = \mathrm{d}U + \mathrm{d}PV$ and $\mathrm{d}q = \mathrm{d}U + \mathrm{d}w$ we get $\mathrm{d}H=\mathrm{d}q$

In isothermal process, $\mathrm{d}T = 0$ implies $\mathrm{d}H=0$ since $\mathrm{d}H=\mathrm{d}q$, $\mathrm{d}q = 0$

for a reaction to be at constant temperature (unless it is a case of free expansion), there must be an exchange of heat, but $q = 0$ here. Same result can be obtained using $\mathrm{q} = ms(\mathrm{d}T)$

Where is the flaw in my reasoning?

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  • $\begingroup$ adiabatic means no exchange of heat. isothermal means no temperature change over time for the system. For a reaction involving a release of enthalpy, for isothermicity to be true, heat must be drained. Hence, the reaction cannot be adiabatic. Only isenthalpic reactions can concurrently be adiabatic and isothermal. That makes for a tiny selection of reactions. $\endgroup$ – Stian Yttervik Aug 25 '18 at 22:53
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The enthalpy change for reacting species is a function not just of temperature. In chemical reactions, bonds are broken and new bonds form. There is an enthalpy change associated with this as well. So the enthalpy of the reacting mixture depends both on the temperature and on the amounts of the various chemical species present. For a reaction occurring at constant pressure, the heat Q is still equal to the change in enthalpy of the reacting mixture, even if the temperature is also held constant.

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There is a very serious flaw in your reasoning.
You have written that $\mathrm{dH = dU + d(PV)}$, which is absolutely correct but, you should remember that $\mathrm{d(PV) = (PdV + VdP) \neq dW}$, because $\mathrm{dW}$ is only equal to $\mathrm{PdV}$.
So, your logic that $\mathrm{dH = dQ }$ is only true at constant pressure where $\mathrm{dP = 0}$. In general case, it is not at all true. In general, $\mathrm{dH = dQ + VdP}$.
So, in Isothermal process, surely $\mathrm{dH = 0}$, which means $\mathrm{dQ = -VdP}$, but in Isothermal processes, $\mathrm{dT=0 = d(PV)}$, which implies $\mathrm{PdV = -VdP}$. So, we have, $\mathrm{dQ = PdV}$ which doesn't say that it should be adiabetic always.

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  • $\begingroup$ Reactions on earth occur at const. pressure so all the reactions if isothermal are also adiabatic? But isn't this equation only valid for processes Not reactions? $\endgroup$ – quantised Aug 25 '18 at 16:13
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    $\begingroup$ The enthalpy change for reacting species is a function not just of temperature. It changes with the amounts of the various chemical species. For a reaction occurring isothermally, the heat Q is still equal to the change in enthalpy of the reacting mixture. $\endgroup$ – Chet Miller Aug 25 '18 at 17:10
  • $\begingroup$ @ChesterMiller I get it but what's wrong in Soumik Das's answer? $\endgroup$ – quantised Aug 26 '18 at 14:20
  • $\begingroup$ @quantised. It doesn't take into account chemical reactions, which was part of the original gist of your question. Even for an isothermal process, $\Delta H$ is not zero if chemical reaction occurs. $\endgroup$ – Chet Miller Aug 26 '18 at 16:34

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