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In thermodynamics, we always quote a fixed temperature, whenever we mention enthalpy of a reaction. For instance, one can determine the enthalpy of combustion of methane at 25 °C. Now almost all reactions involve a change in temperature, i.e. they are either endothermic or exothermic.

How should we physically interpret the meaning of enthalpy change at a fixed temperature? I recall from one textbook on experimental physical chemistry (perhaps it was Shoemaker's Experiments in Physical Chemistry stating that these state functions; still how can physically interpret enthalpy change at a given arbitrary temperature? I have not seen any textbook explicitly addressing this point.

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    $\begingroup$ Well, you run your reaction, then you let the products cool down back to the temperature where you started, and that is your final state. $\endgroup$ Commented Feb 12, 2019 at 4:55
  • $\begingroup$ Interesting, but I never heard of this approach before. You mean T_initial is the same as T_final, eventually? Have you seen this written somewhere? I am in the field of chem for a while and searched scores of books but never saw this explanation. $\endgroup$
    – ACR
    Commented Feb 12, 2019 at 5:08
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    $\begingroup$ Not really. See, thermodynamics is full of imaginary processes. We talk about the things like enthalpy of formation all the time, even though the corresponding reaction (that of formation of our compound from the elements) is often impossible. Same thing here. Whether the process can be actually performed or not is irrelevant. With that in mind, yes, in some cases it can be done, and that's what they do in DSC. $\endgroup$ Commented Feb 12, 2019 at 5:41
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    $\begingroup$ This is Hess law really elementary stuff. $\endgroup$
    – Mithoron
    Commented Feb 12, 2019 at 17:19
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    $\begingroup$ @Mithoron, that is the same thing as saying that enthalpy is a state function, which was already stated in the original post. As Sherlock Holmes would have said, "it is elementary my dear Watson". $\endgroup$
    – ACR
    Commented Feb 12, 2019 at 20:06

2 Answers 2

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The enthalpy change at a fixed temperature is equal to the amount of heat you need to add in order for the final temperature to be equal to the initial temperature. If the enthalpy change is negative, this means that the reaction is exothermic, and you need to remove heat for the final temperature to equal the initial temperature.

If the reaction is carried out adiabatically (no heat addition or removal), the temperature of the products will be either higher or lower than the temperature of the reactants, depending on whether the reaction is exothermic or endothermic, respectively. To determine the temperature change, you can follow a two-step procedure: 1. allow the reaction to take place isothermally, by adding or removing the heat of reaction as necessary; then, 2. reverse the heat addition or removal that you did in step 1, by applying it to the products of the reaction from step 1. The net effect will be the adiabatic temperature rise (or fall) produced by the chemical reaction.

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  • $\begingroup$ Thanks. I got hold of Shoemaker's Experimental Physical Chemistry (reading it after 10 years again), and this is how he describes it as well. $\endgroup$
    – ACR
    Commented Feb 12, 2019 at 20:20
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First, in order to compute state functions we often devise idealized paths to get from initial to final states, since we know that the value of a state function at each extreme is independent of path. The path we pick is a reversible one usually, and where the intensive variables describing the system can be related functionally in some simple way that allows us to compute the difference in the state functions, using for instance the ideal gas law. These idealized paths have nothing to do with the actual course of the experiment. Still, as long as the initial and final points in the calculation and experiment match, the computation should (within the approximations of the model) be an accurate descriptor of what happened experimentally to the state functions.

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