Are chemical treated as ideal gases when trying to figure out heat of formation using calorimetry? 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?
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.