# Shouldn't all isothermal processes be adiabatic?

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?

• 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. – Stian Yttervik Aug 25 '18 at 22:53

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.
• @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. – Chet Miller Aug 26 '18 at 16:34