Chemists say something like "amount of heat consumed for a chemical reaction equals the change in enthalpy" but I cannot understand why this is the case.

Here is my argument: Since $H = U +PV$, we have $dH = T dS + V dP + \sum_i \mu_i dN_i$. If we assume that the heat flow is quasistatic so that we can use $dQ=TdS$, and assuming that $P$ is constant during the reaction so that $dP=0$, we have $dH = dQ + \sum_i \mu_i dN_i$.

Apparently we have an additional term $\sum_i \mu_i dN_i$, so that $dH \neq dQ$.

Where am I wrong?


1 Answer 1


For a closed system (no mass transfer into or out of system) at constant pressure, $$\Delta U=Q-P\Delta V$$This equation applies irrespective of whether a chemical reaction is occurring within the system. So, $$\Delta H=\Delta U+P\Delta V=Q$$The heat of reaction is also defined such that T does not change between the initial and final states of the system.

Also, how can it be quasi static if there is a chemical reaction occurring, presumably at finite rate?

  • $\begingroup$ So Dr. Miller, are you saying that the fault in my reasoning is the assumption of quasi-staticity? BTW, thanks for your derivation for $\Delta H = Q$ ! $\endgroup$
    – Moca Aoba
    Dec 12, 2020 at 15:47
  • $\begingroup$ I would say yes. It takes some doing to make a reaction proceed quasisstatcly. Google van't Hoff equilibrium box. $\endgroup$ Dec 12, 2020 at 15:57

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