Hello I was thinking about two thermodynamics problems and I wanted to get some insights into how to solve them.

The first problem was: Suppose we have one mole of ideal gas under constant external pressure (1 atm) conditions, and let it conduct a reaction (for example a photoinduced isomerization) where there is no net change in number of molecules. This reaction for example has an enthalpy change which is not known. We can measure the total heat Q evolved due to the reaction (like in a constant pressure calorimeter).

So because there is no change of molecules this can be treated as a formal isobaric change (can it?). As enthalpy change is equal to heat Q in isobaric processes, we can say that the measured heat when reaction goes to completion is equal to the reaction enthalpy change, and therefore we can calculate the temperature change.

My doubt is: Is the work the system performs on its surroundings included in ΔH?

How can we determine the temperature change then if we have an reaction of the type:

Also at constant external pressure? Here, the particle number changes.

  • $\begingroup$ Is the work the system performs on its surroundings included in ΔH? Remind yourself that the difference between enthalpy change and internal energy change at constant pressure is the expansion work that the system does. ΔH=ΔU+pΔV. $\endgroup$
    – Poutnik
    Commented Oct 5, 2023 at 9:08

1 Answer 1


When you measure the heat evolved in an isobaric calorimeter, it is assumed that the process takes place nearly isothermally (for the reaction mixture) and the heat involved is equal to the standard change in enthalpy $\Delta H^0$. Whether the total number of moles changes or not, $\Delta H^0$ includes the work done on the surroundings.

  • $\begingroup$ Yes I was aware that heat evolved will be equal to the change in standard enthalpy, at least in a mathematical sense, because for isobaric conditions ΔQ=ΔH, and ΔH=ΔU+pΔV. So in theory, we start recording ΔQ before and after reaction, and the difference in Q is then ΔH. But in the case of gaseous reactants, under isobaric conditions there is volume work. So during the reaction, the Q the reactants give off to our measurement apparatus gets smaller because of the volume work which removes energy from the reactants, so we need to measure Q and the volume work for determining ΔH or am I wrong? $\endgroup$
    – Mäßige
    Commented Oct 6, 2023 at 21:07
  • $\begingroup$ Yes, you’re wrong. The work is included in the $\Delta H$ and the Q. $\endgroup$ Commented Oct 6, 2023 at 21:44
  • $\begingroup$ Okay, I think I get it now. The ΔH=Q is bigger than ΔU by a factor of +pΔV. Suppose the Q is negative (for exothermic reaction). The consequence would be that the absolute value of ΔH is smaller than the absolute value of ΔU (as ΔU is negative, but +pΔV is positive). For example, we would get a value of -30 kJ/mol for ΔH, but ΔU would be -32 kJ/mol in reality right? $\endgroup$
    – Mäßige
    Commented Oct 7, 2023 at 16:51
  • $\begingroup$ It depends on whether the number of moles between reactants and products changes. $\endgroup$ Commented Oct 7, 2023 at 19:19
  • $\begingroup$ Let’s use my first example where it doesn’t. $\endgroup$
    – Mäßige
    Commented Oct 9, 2023 at 20:57

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