I am trying to solve a question that involves studying the heat released of a combustion process at constant pressure versus at constant volume.

The question is

The heat of combustion of ethanol into carbon dioxide and water is $\pu{-327 kcal}$ at constant pressure. The heat evolved (in cal) at constant volume and $\pu{27^\circ C}$ (if all gases behave ideally) is?

The solution to this problem is given using the equation

$$ \Delta H = \Delta U + RT\Delta n (\mathrm{gas})$$

where $\Delta H$ is the heat released at constant pressure, and $\Delta U$ is stated as the heat that is released for this reaction at constant volume.

What I am unable to understand is how we can place the values of enthalpy and internal energy changes for the reaction taking place under different conditions in the same equation. Does this mean that the enthalpy change that would take place under constant volume would be the same as the one at constant pressure? Doesn't placing the value of $\Delta H$ as $\pu{-327 kcal}$ give us the internal energy change for the reaction at constant pressure?


1 Answer 1


You are first trying to find the internal energy change for the reaction at constant temperature and pressure, given the enthalpy change at the same temperature and pressure. So we are using $$\Delta H=\Delta U+\Delta (PV)$$ For a reaction of ideal gases at constant temperature and pressure, $$\Delta (PV)=(\Delta n)RT$$So, $$\Delta U=\Delta H-(\Delta nRT)$$for the reaction at constant temperature and pressure. Of course, this applies only where the ideal gas approximation is valid.

Next, we reduce the volume of the product mixture back down to the original volume. We know that the internal energy of a gas mixture is a function only of T, and independent of mixture volume. So there is no further change in internal energy in reducing the product mixture volume back down to the initial reactant mixture volume. So the $\Delta U$ of the reaction we calculated for constant temperature and pressure is also the $\Delta U$ of the reaction for constant temperature and volume.

  • $\begingroup$ Will $\Delta U$ not change as the reaction involves phase change too? For example the internal energy change of boiling water at 100 deegrees celsius is not 0 $\endgroup$
    – mark
    Nov 17, 2023 at 13:17
  • $\begingroup$ Futhermore, if we say that $U $ only depends on temperature for this reaction, then we obtain that $\Delta U$ is zero for both the reactions as it takes place under constant temperature, implying that the heat evolved at constant volume is also zero $\endgroup$
    – mark
    Nov 17, 2023 at 13:30
  • $\begingroup$ We are talking about ideal gases, so there is no phase change involved. $U$ depends on temperature and chemical composition (not just temperature) for a reacting system. However, for the final product mixture, the chemical composition is fixed, and its U depends only on its temperature. But, for the entire process, certainly the chemical composition of the reaction mixture is different from the product mixture. So going from the reaction mixture to the product mixture at constant volume will involve a $\Delta U$. For a process where $\Delta U=0$, temperature would have to change. $\endgroup$ Nov 17, 2023 at 14:04
  • $\begingroup$ If I understand correctly, the internal energy change associated with the change in chemical compsition can be represented by the energy changes involved in each step of the reaction mechanism, i.e breaking and forming bonds. Does this energy change not depend on the pressure/volume? $\endgroup$
    – mark
    Nov 17, 2023 at 14:19
  • $\begingroup$ Not for an ideal gas mixture. $\endgroup$ Nov 17, 2023 at 14:33

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