# Does the temperature during this reaction remain constant? [duplicate]

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?

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

• Explain to me, why should $\Delta U$ be 0 for a constant-temperature reaction? – orthocresol Oct 2 '15 at 16:59
• If the change in temperature during a reaction is 0 then ∆U=nCv∆T and ∆T is 0 so ∆U should be 0. – user2215860 Oct 2 '15 at 17:01
• 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. – orthocresol Oct 2 '15 at 17:02
• Doesnt the change in internal energy is equal to nCv∆T always? – user2215860 Oct 2 '15 at 17:04
• Nope. Go and reread your notes or whatever material you have. – orthocresol Oct 2 '15 at 17:08

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.