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This question already has an answer here:

Considering all gases to be ideal, the enthapy change for the reaction $$\ce{I2 (g) + H2 (g) -> 2 HI (g)}$$ is $\pu{−106.78 kJ}$ at $\pu{350 K}$. To find the change in internal energy the formula we use is $$\Delta H = \Delta U + (\Delta n)RT \, .$$

To derive this equation we must have assumed the temperature during the reaction to be constant in the expression $$\Delta H = \Delta U + \Delta(nRT).$$ I know that $\Delta U$ depends on temperature and the amount of substance, but if we assumed the temperature to be constant then $\Delta U$ should be $0$, as the amount of substances of reactants and products are also the same, but it has a non-zero value. Is this possible?

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marked as duplicate by orthocresol, Freddy, M.A.R., Martin - マーチン Dec 17 '15 at 8:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Explain to me, why should $\Delta U$ be 0 for a constant-temperature reaction? $\endgroup$ – orthocresol Oct 2 '15 at 16:59
  • $\begingroup$ If the change in temperature during a reaction is 0 then ∆U=nCv∆T and ∆T is 0 so ∆U should be 0. $\endgroup$ – user2215860 Oct 2 '15 at 17:01
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    $\begingroup$ Under what conditions does that equation apply? This is the danger with thermodynamics. If you don't remember what equations apply when you will end up using them in entirely inappropriate situations. $\endgroup$ – orthocresol Oct 2 '15 at 17:02
  • $\begingroup$ Doesnt the change in internal energy is equal to nCv∆T always? $\endgroup$ – user2215860 Oct 2 '15 at 17:04
  • $\begingroup$ Nope. Go and reread your notes or whatever material you have. $\endgroup$ – orthocresol Oct 2 '15 at 17:08
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It is often said that $U = 5nRT/2$ for a diatomic gas (c.f. $U = 3nRT/2$ for a monatomic gas). According to this formula, the internal energy should not change in the reaction you considered, provided that the temperature is held constant. The reason is that in this reaction, one always has the same number of diatomic molecules in the system, regardless of how far the reaction has proceeded.

However, $5nRT/2$ is not the entire internal energy of the gas! This value reflects only the (translational and rotational) kinetic energy associated with the overall motion of each molecule. The total internal energy should also include the energy associated with the relative motion of atoms and electrons within each molecule. The energy of the relative motion manifests itself in the bonding energy. In this particular case, the total bonding energy of two HI molecules is different from that of an H2 molecule and an I2 molecule, so the internal energy should change in the reaction.

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The internal energy of an ideal gas mixture is determined not only by the temperature, but also by the chemical composition of the mixture, and, in the case of a chemical reaction, the composition changes. For a real gas, the internal energy also becomes pressure-dependent at higher pressures.

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  • $\begingroup$ What if the gas is ideal and temperature and the no of moles of reactant and product remains the same as in case of the reaction in my new question? $\endgroup$ – user2215860 Oct 18 '15 at 14:42

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