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As far as I understand the internal energy of a system , $ U $, is the sum of all kinetic and potential energies of the particles constituting the system.

The first postulate states that $ U $ is a macroscopic equilibrium state function (i.e. described by a small number of macroscopic state variables) and that:

$$ \Delta U = Q + W + C $$

Any change in $U$ is due to heat , macroscopic mechanical work and chemical work.

Suppose we have a completely isolated system (isolated from the environment) , implying $Q,W = 0$.

Now, since it is completely isolated its total energy needs to be conserved:$$\Delta U = 0 $$

Suppose that there is a chemical reaction going inside the system - because $\Delta U = 0 $ it follows that $C = 0$. This makes sense because no matter what the particles do ( form bonds, break bonds, explode...) the total internal energy $U$ is always conserved. Online I have read various definitions of chemical energy, most of them say that chemical work is the energy "stored" in the molecules that gets released when they interact, and we have to take that into account when dealing with $\Delta U$, obviously for an isolated system there is nothing "stored" since $ U = const$.

The question is: Why do we include chemical energy in the First Postulate ? If any ongoing reaction in the system does not change the internal energy of the system ,why include it energy?

If the system is not isolated , then I guess the work from the chemical reaction of the system + environment and the influx of new particles from environment into the system ,make $ C \neq 0 $.

Maybe the better question to ask is: Is chemical work in an isolated system always 0 ?

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The C does not belong in the equation, as you have ascertained yourself. So the equation still remains $\Delta U=Q+W$ even with chemical reaction present. But, in the case of a chemical reaction U is a function not only of T and V (or T and P), but also a function of the amounts of the reactants and products in the final state vs the initial state. For for an adiabatic reaction at constant volume that goes to completion, for example, $$\Delta U=N\Delta U_R+n_PC_{Vp}(T-T_0)=Q+W=0$$Where $\Delta U_R$ is the internal energy "heat of reaction" (holding the temperature constant at constant volume), N is the number of moles of reaction executed, $n_P$ is the total number of moles of product present in the final state, and $C_{Vp}$ is the average molar heat capacity at constant volume of the products. This equation enables you to determine the change in temperature.

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  • $\begingroup$ suppose we have an isolated system with a fixed $N_1,N_2$ (say reactants before they react), fixed $T,P$ and other variable $x$. Since it is isolated $Q =0$, $W=0$, $C=0$. What ever happens to the system, its internal energy cannot change. If the reactants turn into $N_3$ , $ U $ still remains the same. $\endgroup$
    – lohe
    Commented Mar 15, 2022 at 22:45
  • $\begingroup$ is this correct ? I don't get how $ U$ can depend on the chemical reaction if the system is isolated... $\endgroup$
    – lohe
    Commented Mar 15, 2022 at 22:46
  • $\begingroup$ Any of the textbooks on thermodynamics by Sonntag and Van Wylen discuss in depth the use of the "heat of reaction" also called the "energy of formation" for chemical reactions, such as combustion. The "heat of reaction" for the specific chemical (or nuclear) reaction is the change in what you called "stored energy" in your question which we now know is associated with the change in the rest mass of the reactants compared to the products of the reaction. This "energy of formation" is equivalent to the change in stored energy, which is the change in rest mass. $\endgroup$
    – John Darby
    Commented Mar 15, 2022 at 23:19
  • $\begingroup$ Like I said, U doesn't only depend on the temperature and volume (or pressure); it also depends on the amount of each reactant and each product that is present in the mixture. So the contributions of T, V, and numbers of moles of the various species determine U. So, at constant U, if the amounts of chemical species changes adiabatically/isochoricly, the temperature must also change to hold U constant. $\endgroup$
    – Chet Miller
    Commented Mar 15, 2022 at 23:27
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This is a good question.

For a reaction (chemical or nuclear) within a closed thermodynamic system, the internal energy of the system does not change since there is no work, heat, or mass transfer between the system and its surroundings. However, the state of the system changes since the reaction causes a change in pressure and temperature, and that would lead you to think the internal energy of the system has changed. The change in the properties of the system (temperature, pressure, etc.) are a result of the change in the rest mass of the reactants relative to the products through $E = mc^2$, and considering the change in the rest mass there is no overall change in internal energy. (The change in rest mass is the change in "stored" energy that you refer to.)

Classical thermodynamics was developed before the change in rest mass was understood (it is not detectable in chemical reactions), so it was thought that the mass did not change. Classical thermodynamics uses an "enthalpy of formation" or an "internal energy of formation" to account for the change in the thermodynamic properties. For example, an exothermic reaction increases the pressure and temperature and therefore changes the state of the system; for the exothermic reaction we now know this happens because of a decrease in rest mass. (For an endothermic reaction, the rest mass increases.)

In classical thermodynamics, you include the "energy of formation" in the internal energy in the first law. You evaluate work, heat, and mass transfer for a non-isolated system using the classical first law with the addition of the "energy of formation".

See Can mass-energy equivalence be used to measure absolute internal energy? on this exchange for more information; specifically. how the overall internal energy does not change and how the "internal energy of formation" is used in classical thermodynamics.

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Only energy transfers between the system and surroundings in the form of heat and/or work belong on the right side of the first law equation for a closed system.

Therefore any heat or work as a consequence of a chemical reaction in the system would belong in the $W$ or $Q$ term on the right side, not as a separate $C$ term.

Hope this helps.

Hope this helps.

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  • $\begingroup$ well $C$ is the kinetic and potential energy change in the system (i.e. $ \Delta U$ of the system) caused by an exchange of matter between the system and the environment. $\endgroup$
    – lohe
    Commented Mar 15, 2022 at 16:47
  • $\begingroup$ I'm not sure what you mean by " $C$ is potential energy" and " it is a part of internal energy" . $C$ is never a part of internal energy , just a change of it $\endgroup$
    – lohe
    Commented Mar 15, 2022 at 16:49
  • $\begingroup$ I am referring to $C$ as being chemical energy. Chemical energy is the potential energy of the chemical bonds. That energy can be converted to kinetic energy during reactions. But the total energy internal energy is constant unless energy is obtained from or transferred to the surroundings during the reaction. $\endgroup$
    – Bob D
    Commented Mar 15, 2022 at 16:51
  • $\begingroup$ yes, in the context of chemical bonds. It has a different definition in thermodynamics $\endgroup$
    – lohe
    Commented Mar 15, 2022 at 16:54
  • $\begingroup$ The right side of the first law equation for a closed system only involves energy transfers in the form of heat or work. Those are not system properties (forms of energy), chemical or otherwise. $\endgroup$
    – Bob D
    Commented Mar 15, 2022 at 16:54

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