Question

I know that in a reversible isobaric process, the enthalpy change of the system is the change in heat for that system. What happens if the process is not reversible? I.e only the outside pressure is constant, for example atmospheric pressure?

For example, in a container with a movable piston on the top, a gas phase chemical reaction takes place with ΔH, in an uncontrolled manner so that the process is irreversible. The outside pressure is atmospheric. Is still ΔH=Δq for the gas (assume in the reaction the moles of gas are changed, increase?

Answer 1

Even for an irreversible process at constant external pressure, the enthalpy change (ΔH) can equal the heat exchanged (Δq), assuming that any work done is just expansion or contraction against this pressure, and there are changes in the gas moles.

Step-by-step explanation:

In thermodynamics, when a gas reacts in a container with a movable piston under a constant external pressure (like atmospheric pressure), and if the process is irreversible, the change in enthalpy (ΔH) may still be equal to the heat exchanged (Δq) if the system’s volume changes and the only work done is expansion or contraction work. This is because enthalpy is a state function related to heat under constant pressure conditions. However, the process differs from a reversible one, as you cannot simply reverse the direction of the change to return to the initial state due to the occurrence of internal irreversibilities like friction or turbulence that prevent the system from moving through equilibrium states.

Thus, for an irreversible isobaric process like an uncontrolled reaction within a container, if the system is open to the atmosphere and there’s a change in the amount of gas (moles), ΔH would still be equal to Δq because the atmospheric pressure acts as a constant external pressure and the only work done is the expansion or contraction against this pressure. This results in the enthalpy change being a measure of the heat exchanged during the process.