Can an exothermic (endothermic) process increase (decrease) internal energy?

Question

We classify exothermic and endothermic processes based on enthalpy change, $\Delta H < 0$ and $\Delta H > 0$ respectively.

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

But not all? In which case we might have processes where $\Delta H < 0$ (exothermic) but $\Delta U > 0$? 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.

Answer 1

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.

Answer 2

From my perspective, this is easiest to tackle from a straight forward energy balance.

The energy balance determines how energy, heat and work are distributed when you go from State 1 to State 2.

Thermodynamic States

Just to emphasize this, a state is a fixed point where all properties are set in stone. So state 1 can be an unreacted system, state 2 a reacted system... State 1 can be the inlet of a compressor at a certain T and P and state 2 can be the outlet at a different T and P. State 1 can be a bomb calorimeter before combustion (at a T and V) and state 2 can be the calorimeter after combustion (same V, different T and consequently, different P).

You can do an energy balance between any two states, and it will tell you directly what is happening to the energy, heat and work of your system. It is a beautiful thing actually.

Energy Balance

For an unreacting and open system at steady state

$$ \left[ U + \frac{P}{\rho} + KE + PE + W + Q \right]^{\rm State_1} = \left[ U + \frac{P}{\rho} + KE + PE + W + Q \right]^{\rm State_2} $$

Where $U$ is internal energy, $\frac{P}{\rho}$ is flow energy, $KE$ is kinetic energy, $PE$ is potential energy, $Q$ is heat and $W$ is work. I should say each variable is really the summation of all sources contributing. There could be several sources producing heat, same with work. There could be several streams entering and several exiting. Traditionally $State_1$ and $State_2$ represent IN and OUT respectively.

If there is flow, then $\frac{P}{\rho}$ is present, and we typically denote it as $Pv$. Thus on both sides we have $U + Pv$ which we write as $H$. It is typical to say changes in $KE$ and $PE$ are negligible. I will also put IN and OUT in place of States Our energy balance is now

$$ \left[ H + W + Q \right]^{\rm IN} = \left[ H + W + Q \right]^{\rm OUT} $$

At this point the actual problem naturally falls out. If you want to know what the change in enthalpy is rearrange the sides

$$ \Delta H = W_{IN} - W_{OUT} + Q_{IN} - Q_{OUT} $$

It is entirely possible for $\Delta H$ to be positive or negative for either positive $\Delta Q$ or $-\Delta Q$ so I don't see how you can draw any conclusions about $\Delta H$ with respect to what heat is doing. If there are no work interactions, this simplifies the equation $$ \Delta H = Q_{IN} - Q_{OUT} $$ With no work interactions the system would need to be exothermic i.e., lose heat for $\Delta H$ to be negative.

Let me know if I made a mistake in my derivations, It would not be the first time. But remember all of the assumptions that led us to this point!