In case of few chemical reactions $\Delta n=0$,so according to the equation $\Delta H= \Delta U+\Delta nRT$ change in enthalpy equals change in internal energy. But what does this in essence actually mean? Like physically what does this indicate and do we get any additional information from that an equation? I got this query seeing the following equation of formation of hydrogen chloride.

$\ce{H2 +Cl2 =2HCl}$ with $\Delta n=0$.

Like can it be associated to why the reaction is zero order?

  • $\begingroup$ $\Delta n = 0$, not $\Delta T =0$. I guess you are taking isothermal condition $\endgroup$
    – Zenix
    Commented Apr 8, 2020 at 11:56

1 Answer 1


This equation holds for a reaction involving ideal gases at constant T. Under these conditions, $\Delta (pV) = \Delta n RT$. Assume now that a gas phase reaction occurs at constant pressure and temperature. Then $\Delta (pV) = p\Delta V$. This is equal to the negative of the pressure-volume work done by the system during the reaction since $w = -p\Delta V$. Therefore, at constant p and T, $\Delta H$ is equal to the change in energy of the system minus the pV work it did. This is consistent with the fact that we equate the enthalpy change for such a process at constant p and T with the heat exchanged, that is, $\Delta H = \Delta U -w = q_p$. Now if $\Delta n =0$ no expansion occurs at constant p and T, and so $w=0$ and $\Delta H = \Delta U = q$.

  • $\begingroup$ So if ∆n≠0 and so w≠0, in case of a chemical reaction like Haber's process of ammonia formation, on what is the work being done? Like is it used for breaking bonds? I just want to understand for what are we using the extra energy of ∆(PV) $\endgroup$ Commented Apr 9, 2020 at 11:29
  • $\begingroup$ Here the work is pV work (gas expanding or contracting against the surroundings). The making/breaking of bonds results in a change in thermal energy which either increases the T of the system or dissipates into the surroundings. If a reaction proceeds at constant T and p, and the reaction changes n, then the volume also has to change. That means work was done. $\endgroup$
    – Buck Thorn
    Commented Apr 9, 2020 at 13:28
  • $\begingroup$ Note that work doesn't have to be done. You might carry the reaction out in an isochoric bomb. In that case the likely outcome is a change in temperature and or pressure, in addition to the change in n. If the reaction is isothermal and isochoric, then if n changes you should see a change in p. $\endgroup$
    – Buck Thorn
    Commented Apr 9, 2020 at 13:30
  • $\begingroup$ Thank you very much bro. I understood it but there are still few doubts of mine regarding constant T,P and how can number of moles change with volume in reality ( not the equation). Could you please let me know where I can study this as I couldn't find anything when I looked up for this. $\endgroup$ Commented Apr 9, 2020 at 13:46
  • $\begingroup$ How do you think a internal combustion engine in a car works? It has a piston into which fuel is discharged and then ignited, expanding and thereby doing work which is used to propel the vehicle forward. $\endgroup$
    – Buck Thorn
    Commented Apr 9, 2020 at 18:38

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