# Enthalpy in an irreversible isobaric process

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, increases)?

• If the initial pressure in the system is atmospheric and the final pressure at equilibrium of the system is atmospheric, then $q=\Delta H$ Commented Apr 24, 2023 at 10:34
• @ChetMiller unless there is non-PV work, right?
– Karsten
Commented Apr 26, 2023 at 0:42
• @Karsten Yes. I was trying to keep it simple and at the OP's level. Commented Apr 26, 2023 at 0:53

The reversibility of the reaction is irrelevant; any measured isobaric reaction will have ΔH = Δq. Keep in mind that simply knowing a process is reversible does not give any indication as to what equilibrium will actually look like for the process (meaning you can't generalize about how reversible vs. irreversible reactions will behave, as they are all simply reactions); you could easily have a reversible process that produces the same pressure change as an irreversible one. The reactions may have different mechanisms, activation energies, and enthalpies, but the underlying calculations will be the same. ΔH will always equal Δq if the process is in an isobaric environment. It does not matter how the reaction proceeds. The only thing that affects enthalpy in an isobaric environment is the energy changes within the reaction and how much heat is produced. Reversibility itself has no impact on this.

• Only if $P_i = P_f = P_{ext}$ as well. Otherwise you can have an "adiabatic" irreversible isobaric process where $P_i \ne P_f = P_{ext} =$ constant and $\Delta H = \gamma W \ne Q = 0$. Commented Feb 26 at 14:46

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?

I am going to extend a bit to clarify this simple but interesting case.

The key lies when you carry out the mathematical derivation, for a constant-pressure process, in a closed system and apply the 1st law \begin{align} dH &= dU + d(PV) \\ dH &= dQ + dW + PdV \\ dH &= dQ \underbrace{- PdV} + PdV \rightarrow \boxed{Q = \Delta H} \tag{1} \end{align} it is precisely where the underbraces appear, that you state that the work is $$dW=-PdV$$. This substitution is what we call a mechanically reversible process, which is equivalent to two propositions:

1. The system is uniform in pressure, i.e., $$P$$ is not a function of spatial coordinates.
2. The pressure of the system is only differentially displaced with respect to the pressure of the surroundings. Thus, we envision the system walking from $$P_1$$ to $$P_2$$ in an infinite number of steps, each one characterized by a mechanical equilibrium between the system and the surroundings.

We have three comments at hand now:

1. Regarding enthalpy: since it is a state function, if you can measure $$P_1$$, $$T_1$$, $$P_2$$, $$T_2$$, the enthalpy change can be computed directly $$\Delta H = H(P_2,T_2) - H(P_1,T_1) \tag{2}$$ independently if your process was highly irreversible or totally reversible.
2. Regarding heat transfer: if the process was not mechanically reversible $$Q \neq \Delta H \tag{3}$$ and we cannot calculate the heat transferred between the boundaries of the system/surroundings. I will emphasize: you cannot calculate it that way, because you cannot substitute $$W = -PdV$$ in Eq. (1).
3. Regarding reversibility: imagine the process is still mechanically reversible and thus $$W = -PdV$$, this doesn't mean that the process is reversible. Lets give a fast proof right now, and we will see that this isobaric process can be irreversible. For simplicity, we will consider $$\ce{Ar}$$, since it has a $$C_P \neq f(T) = 5R/2$$ and it behaves perfectly as an ideal gas. The compression is illustrated in the following image:

The calculations are straightforward: (1) the heat transfer to the system $$Q$$, the work done by the system $$W$$ to the surroundings, and the change of entropy of the universe yields \begin{align} Q &= \Delta H = \frac{5}{2}R(T_2 - T_1) \rightarrow Q = 2079 \; \pu{J/mol} \\ W &= -P(v_2 - v_1) = -(P_2v_2 - P_1v_1) = -R(T_2 - T_1) \rightarrow W = -831.4 \; \pu{J/mol} \\ \Delta S_\pu{uni} &= \Delta S_\pu{sys} + \Delta S_\pu{surr} \\ \Delta S_\pu{uni} &= c_P \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) + \frac{Q_\pu{surr}}{T'} \\ \Delta S_\pu{uni} &= \frac{5}{2}R \ln\left(\frac{T_2}{T_1}\right) + \frac{1}{T'} \left(-\frac{5}{2}R(T_2 - T_1)\right) = 3.646 \; \frac{\pu{J}}{\pu{mol·K}} > 0 \\ \end{align} The conclusion is: you can apply $$W = -PdV$$ if you want, but it doesn't mean that the process is reversible.

How to know $$Q$$ then? For an irreversible process there are many approaches, if you want to still apply thermodynamics. I will present you the simplest one. Consider that the system is mechanically reversible anyways and compute $$W^\pu{id} = -\int P \; dV$$ Say the process resulted in a compression (like the example). Therefore, the system has done work and $$W<0$$. In reality, we are being too optimistic, and the work in fact was less. Thereby we penalize it by an efficiency $$\eta < 1$$, and calculate the real work $$W = \eta W^\pu{id} \tag{4}$$ Finally, the heat is obtained by appling the 1st law. Like enthalpy, internal energy is independent of the path, and by Eq. (4) we have $$Q = \Delta U - W \rightarrow \boxed{Q = U(P_2,T_2) - U(P_1,T_1) - \eta W^\pu{id}} \tag{5}$$

To know the efficiency means to have valuable information of your system, and how it functions. Other approaches, as to calculate $$P$$ as a function of $$V$$, can also be applied. This one has again the same consequence, i.e., obtain information of how the system behaves when it expands or compress.

• Your diagram doesn't make sense -- if you have an isobaric compression, the temperature has to decrease per $PV = nRT$, so $T_2 < T_1$. Also your point #2 contradicts the other answer here which states "the reversibility of the reaction is irrelevant; any measured isobaric reaction will have $\Delta H = \Delta q$" as well as the comments by Chet Miller in the original question as well... Commented Jan 31 at 0:28
• Likewise since enthalpy $H$ is a state function it is thus path-independent and so in the absence of non-PV work, it has to follow that $\Delta H= \Delta q$ applies equally to all isobaric paths whether they are reversible or not. Commented Jan 31 at 1:05
• @ManRow Hello ManRow. I don't have much time to engage because I have to do some exams. You are not correct. I assure you that $Q = \Delta H$ is only true for isobaric and mechanically reversible processes. I attach to you the pages of the following book that explain this, in here. It is underlined in red. If you want, you can read its Chapter 2, and come back for next discussions. Commented Jan 31 at 1:33
• We can ask @ChetMiller. He is a chemical engineer as me, and has read Smith, Van Ness, and Abbott one million times. What is your answer Chet? If the process is isobaric but irreversible, can we say that $Q = \Delta H$? Commented Jan 31 at 1:35
• @ManRow Here you have another proof, in Chapter 2 of Atkins' book in Physical Chemistry. Pay attention to the underlined lines. If the process is irreversible, can you say $\mathrm{d}w = -p\,\mathrm{d}V$? No. If you can't say that, you can't replace that expression and reach $Q = \Delta H$. I await your answer to this discussion. Otherwise, I will flag these comments for words such as 'doesn't make sense' and not backing up your claims. Commented Jan 31 at 1:49