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At 36:48 of this video lecture, The professor says that the reaction moving from reactants to products keeping temperature of both same and under standard condition is an isothermal constant volume process.

Some some four minutes later (41:57), he uses the sum of energy of whole cycle property and equates, the $U_{rx}$ to the other definition of other definition of heat change given by calorimeter using the law that sum of internal changes over a loop process is zero.

He gets:

$$ \Delta U_{rx} = - C_{v}^{calo} \Delta T$$

where $C_{v}^{calo}$ is energy change of calorimeter.

Now, what I find weird is that by the definition of the path-3 being isothermal, shouldn't $$ \Delta U_{rx}$$ be zero..?

My resolution for this was that maybe the gas not really ideal and that it may change energy loss by expanding but I'm not sure how I would justify this more authoritatively. So, my question is, when we talk about these chemistry reactions, do we treat gases as non-ideal?

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    $\begingroup$ When a chemical reaction occurs, internal energy is a function not only of temperature. It is also a function of chemical composition. $\endgroup$
    – Chet Miller
    Sep 7, 2020 at 3:23
  • $\begingroup$ I'm not quite sure what I'm searching for because the 'heat of formation' is regarding enthalpy not internal energy or is there some link between the two which makes what I said in correct? $\endgroup$
    – tryst with freedom
    Oct 7, 2020 at 16:13
  • $\begingroup$ Well, if $h^0$ is the molar heat of formation at the reference conditions of $T^0=298 K$ and $P^0=1\ bar$, then the molar internal energy at this reference state can be taken as $u^0=h^0-RT^0$, and the molar internal energy at temperature T can be expressed as $u=u^0+C_v(T-T^0)=h^0-RT^0+C_v(T-T^0)$ $\endgroup$
    – Chet Miller
    Oct 7, 2020 at 18:51
  • $\begingroup$ Hmm, that makes sense.. I think the internal energy change I found was the internal energy formation. I think if I divide that expression by moles of the quantity which was formed, then I'd get molar quantities. I'm not quite sure. If you can give me a reference on where I could find a discussion about this concept, then I'll try study that and eventually write an answer here. $\endgroup$
    – tryst with freedom
    Oct 7, 2020 at 18:57
  • $\begingroup$ See Moran et al, Fundamentals of Engineering Thermodynamics, Section 13.2. p823 has a subsection Analyzing Dlosed Systems that considers internal energies. Also see Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics, Section 4.4 onward. $\endgroup$
    – Chet Miller
    Oct 7, 2020 at 23:05

1 Answer 1

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Internal energy of any gas is given by $$U=U_\text{trans} + U_\text{rotational} + U_\text{vibrational} + U_\text{intermolecular} + U_\text{electronic} + U_\text{relativistic} + U_\text{bonds}$$ Last three aren't affected by ordinary heating. And $U_\text{intermolecular}$ for ideal gas is zero. That's why for ideal gas $U$ is only function of temperature. But during a chemical reaction (even if it involves an ideal gas) $U_\text{bonds}$ comes into play. The formation and destruction of bonds results in a decrease of potential energy.

Therefore, for a chemical reaction, even if the temperature remains constant the internal energy changes.

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