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Exothermic reaction has negative enthalpy change because energy is flowing out of the system. Isolated system means no energy is flowing in and out of the system. If you have an exothermic reaction in an isolated system, is enthalpy change 0?

Also, assume

Substance 1: initial enthalpy of isolated system (i.e. reactants) is H1.

Substance 2: Enthalpy of isolated system after reaction (i.e products) is H2.

Substance 3: Enthalpy of products taken out of isolated system, and left to be cooled (i.e. products cooled down to room temperature) is H3

Is enthalpy change of reaction H2-H1? H3-H1? Or neither?

Correct me if I am wrong: If you have a sealed container with gas at 500K, the gas loses heat energy to its surroundings, then there is an enthalpy change. Since substance 3 is at a lower temp than substance 2, H3 is not equal to H2.

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  • $\begingroup$ Guys, If there something wrong with the question/my understanding, could you please comment. I'm asking this question precisely because I don't understand. $\endgroup$ Oct 13, 2022 at 14:42
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    $\begingroup$ Not downvoted by me, but if downvoted without a comment, a good guess can be the default downvote reasons displayed at mouse over the downvote triangle. $\endgroup$
    – Poutnik
    Oct 13, 2022 at 15:07

3 Answers 3

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Enthalpy is defined as $$H = U + pV$$

Isolated systems are implicitly of constant volume, otherwise they can exchange energy by volume mechanical work.

Isolated systems have therefore constant $U$ and $V$.

$H$ is then constant if and only if $p$ is constant.

Manipulation with content of isolated systems severely break their thermodynamic context.

Neither $H_2 - H_1$, neither $H_3 - H_1$. In contrary to a general enthalpy change, the reaction enthalpy is defined as thermal exchange at isothermal and isobaric conditions.

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An exothermic reaction is one in which you need to remove heat from the reaction mixture in order to hold the temperature constant. The amount of heat you need to remove to hold the temperature constant is minus the the enthalpy change for the reaction.

If you have an exothermic reaction of an ideal gas mixture in an isolated system, the internal energy change is zero, and the enthalpy change is $V\Delta P$, where V is the (constant) volume of the system and $\Delta P$ is the pressure change.

For the scenario you described, assuming no change in the number of moles in going from reactants to products, H2=H1 (really U2 = U1), and H3 < H1.

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  • $\begingroup$ Hm, even if the molar amount does not change, temperature and therefore also pressure raise. So even if U is constant, H is not. $\endgroup$
    – Poutnik
    Oct 13, 2022 at 16:08
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In totally isolated system the only thing that can happen is eventual equilibrium. this means everything at the same T, P, V, chemical potential energy, nuclear, gravitational and at maximum entropy. I cannot grasp what happens to PV work since return to an original state cannot happen without unisolating the system. Instinct says it is conserved but that is not a good reason.

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