While studying enthalpy $H=U+PV$ and its changes, I realized I am not clear on the following: chemical reactions happen with the external pressure being constant and equal to the atmospheric pressure $P_{atmos}$. But, in general, the system's internal pressure $P_{gas}$ would seem to be different from $P_{atmos}$ and variable during a reaction... The change in enthalpy $\Delta H= Q$, where $Q$ is thermal energy, only when the gas internal pressure matches the environment external pressure: $P_{gas}=P_{external}=constant$ Under those circumstances, the emitted or absorbed thermal energy $Q$ goes into change #H# while the gas expands/contracts. The work $PV$ done by the gas is perfectly matched by the $PV$ work done on it by the environment.

How common is for the gas internal pressure to be constant and equal to $P_{atmos}$ during chemical reactions to justify $\Delta H = Q$?

  • 1
    $\begingroup$ What are you talking about? Either reaction is conducted in open reactor - obviously atm. pressure, or closed one with pretty much whatever parameters. $\endgroup$
    – Mithoron
    Commented Jan 5, 2021 at 1:32
  • $\begingroup$ Your gas/diesel engine certainly does not... $\endgroup$
    – Jon Custer
    Commented Jan 5, 2021 at 1:48
  • $\begingroup$ If I understand Jon's comment correctly, the gas internal pressure is not constant during the reaction as it can happen in other reactions. Certainly, the external atmospheric pressure is constant if it happens in open air. $\endgroup$ Commented Jan 5, 2021 at 2:45

1 Answer 1


What happens during a reaction is important insofar as reversibility is concerned (a reversible path being a special way of performing a reaction in which equilibrium is sustained throughout), but not when computing $\Delta H$. The condition $\Delta H = q$ is derived from the basic law of conservation of energy and the definition of enthalpy. Because H is a state function, to compute the change in its value you need to know only the initial and final states of the system (and of course the values of H corresponding to those states). For $\Delta H = q$ to hold requires that (1) only expansion (pV) work be performed and that (2) the initial and final pressures of the system equal each other.

You can dream up scenarios in which you start at mechanical equilibrium (balanced pressure) between surroundings and system but end up at a constrained system volume (rigid). For instance, if an explosion moves a piston to a constrained setting within a cylinder. In that case it is not true that the enthalpy change and heat are equal. On the other hand, if the explosion is constrained such that the volume can expand until pressures balance (and external constraining pressure is constant), then enthalpy change and heat are equal.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.