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Hello I am currently struggling with enthalpy of reaction and its use cases.

Let's say I am conducting the following reaction in a piston as a closed system:

$$\ce{C2H4(g) + 2 O2(g) -> 2 CO(g) + 2 H2O(g)}$$

This reaction has an enthalpy of reaction of $\pu{-1560 kJ/mol}$. As far as I know, this refers to the heat released under constant pressure. So for example, if I let my reaction run in a piston with outside pressure of 1 bar, the reaction will heat the gases up by $\pu{-1560 kJ/mol}$, when the system is in pressure equilibrium with the outside pressure again (after heating).

But what about the work the gas does to its surroundings (in my case to the atmosphere)? Do I need to subtract the work done by the gas from the enthalpy of reaction to get the real value? Because for sure, the gas will expand against the outside pressure and lose some of the heat of reaction?

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  • $\begingroup$ Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of at least essential parts. // Optionally, here are formatting guides for texts and formulas/equations/expressions. $\endgroup$
    – Poutnik
    Aug 5, 2023 at 8:43
  • $\begingroup$ Ok will look into it $\endgroup$
    – Mäßige
    Aug 5, 2023 at 10:42
  • $\begingroup$ If you treat the mixture as a mixture of ideal gases, then the enthalpy is independent of pressure. But only at constant pressure is the heat added equal to the enthalpy change. $\endgroup$ Aug 5, 2023 at 10:49

2 Answers 2

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You appear to be conflating two distinct concepts: internal energy and enthalpy.

From my understanding, the term 'enthalpy' is applicable exclusively under constant pressure conditions. As such, its relevance extends only to reactions occurring at constant pressure. In cases like constant volume, enthalpy becomes a mere mathematical value devoid of any practical significance.

At constant pressure, enthalpy signifies the net heat energy exchanged between a system and its surroundings. Notably, this measurement excludes any involvement of pressure-volume work.

On the other hand, internal energy accounts for both heat and expansion work.

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To initiate, let's compute the expansion work carried out by the system. For this calculation, let's assume a constant temperature of 298K and a system perpetually in equilibrium with its surroundings, implying the reaction progresses in infinitesimal steps. This assumption enables us to replace the external pressure of the surroundings with the internal pressure of the system.

Expansion work done by gas Wexp = -∫Pext.dv = -∫Pgas.dv = -(∆ng)RT = (4-3)×8.314×298 ≈ -2477j ≈ -2.48kj

(∆ng) : change in no of moles of gaseous state in reaction

It's essential to note that the energy of approximately 2.48 kJ relinquished by the system due to expansion is unrelated to enthalpy changes.


But what about the work the gas does to its surroundings (in my case to the atmosphere)? Do I need to substract the work done by the gas from the enthalpy of reaction to get the real value? Because for sure, the gas will expand against the outside pressure and lose some of the heat of reaction?


Yes you must add or subtract, depending upon process, expansion work to get change in internal energy not in enthalpy.

Think of it this way: The system released more heat energy than was needed to break the bonds of the reactants. This resulted in a net release of 1560 kJ of heat energy. Additionally, the system lost 2.48 kJ of energy in the form of work. Therefore, the change in internal energy of the system is the sum of the heat released and the work done, which equals -1562.48 kJ.

I hope I have addressed your queries satisfactorily.

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    $\begingroup$ Enthalpy as a state function is applicable anytime. But only in isobaric cases it is really useful as it's change is equal to sum of exchanged heat and non expansion work. Expansion work is implicitly contained in Enthalpy definition here. $\endgroup$
    – Poutnik
    Aug 5, 2023 at 8:51
  • $\begingroup$ @Poutnik why are you not active in chat discussion chat.stackexchange.com/rooms/147631/… $\endgroup$ Aug 5, 2023 at 8:58
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    $\begingroup$ Partly because I was not notified as you have not used explicit @username syntax in chat, partly I am not going to get involved in extensive interactive education sessions. That would be resource demanding. There are more suited free study resources available to you. $\endgroup$
    – Poutnik
    Aug 5, 2023 at 9:10
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For isobaric systems, enthalpy change differs to internal energy change by the factor of the volume work done by the system,

This volume change related work mutually transforms with the extra heat provided to/obtained from isobaric system, compared to isochoric conditions.

$\Delta H = \Delta U + p \Delta V = Q_V - W_{pV} = Q_p \gt Q_V$

with the sign convention of positive work done on a system.

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  • $\begingroup$ So pV work is already included I guess? $\endgroup$
    – Mäßige
    Aug 5, 2023 at 10:44
  • $\begingroup$ So if I want to get internal energy change (and therefore the temperature change) I can just subtract the work ? $\endgroup$
    – Mäßige
    Aug 5, 2023 at 10:53
  • $\begingroup$ Questions should not be answered if the other side could get them on their own easily and quickly, as this way makes answers much more valuable. As the understanding records itself more permanently this way. $\endgroup$
    – Poutnik
    Aug 5, 2023 at 11:38
  • $\begingroup$ @Mäßige "So if I want to get internal energy change (and therefore the temperature change) ? " Note you can only find temperature change by just knowing internal energy change if there is only ideal gas which is not undergoing any reaction , by the expression ∆U = nCv∆T. Even for a real gas , not undergoing any reaction, this expression doesn't hold. Similarly for a reaction you can't calculate temperature change looking only at internal energy change. In my answer I assumed reaction occuring at constant temperature at 298 , but there is a change in internal energy by -1562.48 kj of energy. $\endgroup$ Aug 5, 2023 at 12:51
  • $\begingroup$ I think I got it; at constant pressure, the enthalpy of reaction already includes the internal energy change + work done by the gas. To only get the internal energy change, one needs to add the work done to the (negative) enthalpy of reaction. @Poutnik, However, in one comment you stated that “enthalpy is change is equal to sum of exchanged heat and non expansion work.” Why non expansion work? dH=dU+pdV $\endgroup$
    – Mäßige
    Aug 9, 2023 at 7:22

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