I cannot understand the difference between thwe two although the internet suggests me there is a difference. Even asked my chemistry teacher but he could not give an explanation.


2 Answers 2


You may try your luck with the enthalpy article on Wikipedia, where

... $\mathrm{d}H$ refers to the amount of heat absorbed in a process at constant pressure...

I will try to alternatively put it down in more layman terms:

Enthalpy is defined as $$H = U + p \cdot V,$$


  • $U$ is system internal energy,
  • $p$ is system pressure,
  • $V$ is the system volume.

If we consider constant pressure ( isobaric ) scenario, the above equation in the difference form is $$\Delta H = \Delta U + p \cdot \Delta V = \Delta U + W,$$ where $W$ is the volume work done by the system on the surrounding.

If we add $\pu{110 J}$ of thermal energy to isobaric system ( $\Delta H$ ), which does $\pu{10 J}$ of volume work by expansion ( $W$ ), the change of its internal energy ( $\Delta U$ ) is $\pu{100 J}$. But those extra $\pu{10 J}$ is "stored" in the neighborhood and is released back if the reverse process is performed.

In an isobaric scenario, the change of enthalpy can be considered as the change of the system energy with the implied involved volume work. Formally, one can consider this work as as special way of energy storage of the system. Therefore, enthalpy is defined in such a way, that for isobaric systems, it is formally and effectively the energy of the system, as the reversible system work acts as a part of the system energy.

As majority of experiments are performed at isobaric, rather than isochoric conditions, thermodynamic data are collected in form of changes of enthalpy, rather then the internal energy.

  • $\begingroup$ That should be w, not $\Delta W$. Work is not a state function -- each process has associated with it a certain amount of heat and work. There's not a 'change in work.' And the idea that "formally...special way of energy storage" is (at least to my experience) unusual, and needs to be demonstrated in a sufficiently precise and formal manner to show that it is correct (it's not clear if it's correct based on what you've written). $\endgroup$
    – theorist
    Commented Jun 23, 2020 at 8:36
  • $\begingroup$ @theorist I agree there should be W - fixed. I also agree work is not generally a state function. But in a special case of isobaric scenario, it is. $W = p . \Delta V$, while in general case, like in thermal engines W obviously depends on path. About the rest, consider reversible isobaric thermal expansion of gases. I have also intentionally used wording close to the OP. $\endgroup$
    – Poutnik
    Commented Jun 23, 2020 at 8:56

The most simplest explanation I can give is this: Enthalpy is the measure of energy contained in a system plus the energy required to assemble it.

So, The measure of energy contained in a system is it's internal energy (U) and the energy to assemble is work. So,

$$ H= U + W$$


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