# what is the physical significance of ∆H=∆U for a chemical reaction?

In case of few chemical reactions $$\Delta n=0$$,so according to the equation $$\Delta H= \Delta U+\Delta nRT$$ change in enthalpy equals change in internal energy. But what does this in essence actually mean? Like physically what does this indicate and do we get any additional information from that an equation? I got this query seeing the following equation of formation of hydrogen chloride.

$$\ce{H2 +Cl2 =2HCl}$$ with $$\Delta n=0$$.

Like can it be associated to why the reaction is zero order?

• $\Delta n = 0$, not $\Delta T =0$. I guess you are taking isothermal condition – Zenix Apr 8 at 11:56

This equation holds for a reaction involving ideal gases at constant T. Under these conditions, $$\Delta (pV) = \Delta n RT$$. Assume now that a gas phase reaction occurs at constant pressure and temperature. Then $$\Delta (pV) = p\Delta V$$. This is equal to the negative of the pressure-volume work done by the system during the reaction since $$w = -p\Delta V$$. Therefore, at constant p and T, $$\Delta H$$ is equal to the change in energy of the system minus the pV work it did. This is consistent with the fact that we equate the enthalpy change for such a process at constant p and T with the heat exchanged, that is, $$\Delta H = \Delta U -w = q_p$$. Now if $$\Delta n =0$$ no expansion occurs at constant p and T, and so $$w=0$$ and $$\Delta H = \Delta U = q$$.