I am confused as to when you use ΔU and when to use ΔH, I don’t really understand the difference. Is it that ΔU is used for closed systems whilst ΔH is used for open systems?


2 Answers 2


Is it that ΔU is used for closed systems whilst ΔH is used for open systems?

That's not quite correct.

Consider the case of a simple system: a volume of gas. The problem is that when processes take place, the gas may expand or contract. Since the gas pushes back on the external pressure, some amount of work may be done the system ($W = p\Delta V$). The internal energy change $\Delta U$ does not automatically factor in this work.

Enthalpy, on the other hand, is a quantity which is defined ($U + pV$) such that it does account for the pressure-volume work of the system, so $\Delta H$ is the change the energy of the system and the pressure-volme work done by the system.

So, the summary is that if you have a constant volume process (where pressure volume work is by definition zero), use $U$. If you have a constant pressure system, use $H$.

Even the free energies are this way as well. There's Helmholtz free energy ($A$ or $F$) for constant volume processes and Gibbs free energy ($G$) for constant pressure processes.

In chemistry, most processes are done in open air (unless you're using a bomb calorimeter), in which case all processes are simply subject to atmospheric pressure, so these processes would all be best described using enthalpy and Gibbs free energy.


You would use $\Delta U$ along with Q-W for a closed system. For an open system (involving mass entering and leaving a control volume) operating at steady state, you would use $\Delta H$ along with $Q-W_s$, where $W_s$ is the shaft work per unit mass passing through the system (which does not include the work to push mass into and out of the system) and Q is the heat added to the control volume per unit mass passing through the system.

The open system version of the first law of thermodynamics (neglecting kinetic- and potential energy effects) is $$\dot{U}=\dot{Q}-\dot{W_s}+\dot{m}_{in}h_{in}-\dot{m}_{out}h_{out}$$where h is the enthalpy per unit mass of entering and leaving streams to the control volume. At steady state, $\dot{U}=0$ and $\dot{m}_{out}=\dot{m}_{in}=\dot{m}$. So, for steady state operation, $$\dot{Q}-\dot{W_s}+\dot{m}(h_{in}-h_{out})=0$$or equivalently, $$\Delta h=\frac{\dot{Q}}{\dot{m}}-\frac{\dot{W_s}}{\dot{m}}$$


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