Can an exothermic (endothermic) process increase (decrease) internal energy? We classify exothermic and endothermic processes based on enthalpy change, $\Delta H &lt; 0$ and $\Delta H &gt; 0$ respectively.

I assume this translates to $\Delta U &lt; 0$ and $\Delta U &gt; 0$, respectively, for most processes. (where $\Delta H = \Delta U + \Delta (PV)$, in general).

But not all? In which case we might have processes where $\Delta H &lt; 0$ (exothermic) but $\Delta U &gt; 0$? Could you give an example?

Otherwise exo and endo could just be classified based on $\Delta U$ alone instead of $\Delta U + \Delta (PV)$.

Background: I'm just trying to find a motivation for using the concept of enthalpy. I know about the motivation due to additional $PV$ work needed for  a given process (hence $U$ not enough to account for all the energy involved). Just wondering if the exo/endo classification is another motivation for using enthalpy?

Side note: We know something like this for $\Delta H$ v $\Delta G$, for classifying spontaneous v non-spontaneous processes. In that case $\Delta H$ is not enough and we need $\Delta G$. Just wondering if a similar thing happens in case of $\Delta U$ v $\Delta H$ for exo/endothermic classification.
A primary reason for using enthalpy is the fact that heat transferred ($q$) is a fairly straightforward quantity to measure, whereas change in internal energy ($\Delta U$) is often not. 

Under conditions where volume is held constant and no form of work other than pressure-volume work is possible, $\Delta U = q$, so we can get direct information about the change in energy from the amount of heat transferred. 

But under the much more common laboratory condition of constant pressure, $q \neq \Delta U$. For this case, it is useful to equate $q$ with the change in a state function, so we have defined enthalpy such that $\Delta H = q$ when pressure is constant. Because enthalpy is a state function, we can make general conclusions about other paths between states based on our measurements of $q$, even though $q$ is path-dependent.