I struggle to see the difference between internal energy and enthalpy, or at least the need for differentiation between the two. I understand that enthalpy, supposedly unlike internal energy, accounts for the expansion work required for the system to exist, but it seems that that would only be necessarily stated if one had failed to account for the negative contribution to work in the first place.

Could enthalpy, as $$H = U + pV$$ and thus, $$H = q + w + pV$$ not be written as equally validly as, $$H = q + w_1 + w_2$$ i.e. $$H = U \ ?$$

It would seem that this definition of enthalpy implies that internal energy accounts only for work done to the system, disregarding work done by it.

  • $\begingroup$ In general, $H \neq U$ (most definitely). I don't quite follow your train of thought, could you please elaborate. $\endgroup$
    – getafix
    Sep 26, 2016 at 9:51
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    $\begingroup$ You have really bad misconceptions... Firstly $U \neq q + w$.. heat and work are ways of changing the internal energy, viz. $\Delta U = q + w$.. they are not the "two components of internal energy". Secondly, $w \neq pV$, but rather $w = -p\Delta V$... And lastly, just get this definition of enthalpy out of your head. It doesn't help at all to try and assign a meaning to the total enthalpy of a system. $\endgroup$ Sep 26, 2016 at 10:11
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    $\begingroup$ Please, go and look up a textbook, or ask a teacher. I recommend Levine's Physical Chemistry. The point of using enthalpy, is because enthalpy changes are equal to heat supplied at constant pressure. Therefore the change in enthalpy of a system is something that one can directly measure, e.g. using calorimetry: $q = mc\Delta T$. On the other hand, internal energy changes are equal to heat supplied at constant volume. The issue: most chemical rxns that we do are carried out at constant P, not V. Hence, $\Delta H$ more useful than $\Delta U$, not $H$ more useful than $U$! $\endgroup$ Sep 26, 2016 at 10:30
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    $\begingroup$ 1) That equation only holds at const-p. Not that it matters, I just wanted to get that out of the way. 2) So, you have $\Delta U = q + w = q - p\Delta V$ (definition of p-V work) which means $\Delta H = q - p\Delta V + p\Delta V = q$ That term $+p\Delta V$ is not a further component of work, in fact it cancels out the work term arising from $\Delta U$. As for the reasoning for why $H$ is defined in that way, it is not because of some magical significance of the total enthalpy (...) $\endgroup$ Sep 26, 2016 at 10:41
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    $\begingroup$ (...) but rather because defining it that way gives it certain special properties, of which the most important is $\Delta H = q_p$. Levine puts it this way. i.stack.imgur.com/BIXjn.png $\endgroup$ Sep 26, 2016 at 10:42

2 Answers 2


Trying to figure out a good intuitive physical interpretation of enthalpy has caused thermodynamics student over the last couple of centuries to waste huge amount of their valuable time (in my judgment). The really fundamental physical quantity is internal energy U, not enthalpy, and its physical interpretation is simple and straightforward. I regard enthalpy as just a convenient function to work with in solving many types of physical problems. If the function enthalpy had never been invented, we would still be able to solve these problems, but the results would not be as concise.


Your derivation is obviously flawed, and unfortunately you reasoning doesn't make much sense to me either.

You assume $U = q + w$ which is clearly not true; similarly $ \mathrm{d}w = -p \int \mathrm{d}V$ (and not $w = pV$ as your derivation seems to suggest).

Your reasoning/intuition about enthalpy seems ill founded as well, sorry. Perhaps, it's simplest to think of enthalpy as heat exchanged at constant pressure. Why? Because it turns out to be an extremely useful and convenient state function if we define it that way. Since most reactions, experiments take place at constant pressure (and calorimetric determination of heat exchanges is not particularly hard).

I did a derivation (of sorts) for enthalpy using Legendre Transforms, a while ago for another question. The question has since been closed, which is why I have reproduced the derivation below in case you are interested in it. Perhaps, it will help you understand why enthalpy is defined the way it is.

First a brief note on the mathematics of Legendre Transforms

Consider a function in two independent variables: $ f(x,y)$

We right the differential of this function as follows:

$df(x,y) = \Big( \frac{\partial f}{\partial x}\Big)_ydx + \Big( \frac{\partial f}{\partial y}\Big)_xdy$

Now, I shall define $u := \Big( \frac{\partial f}{\partial x}\Big) $ and $w := \Big( \frac{\partial f}{\partial y}\Big)$

so, $df = udx + wdy$ [equation 1]

Here, u and x (and similarly w and y) are called conjugate variables. (i.e $ux$ and $wy$ will have the same units as $f$)

[Think of it this way: In thermodynamics, PV, TS have the same units as H, U,G,A]

Now, our goal is transform our original function $f(x,y)$ to a function $g(x,w)$. We wish to replace y, with it's conjugate variable and the function $f $ and $g$ will have the same units.

so, we take the differential of the product $wy$ : $d(wy) = ydw + wdy$

We subtract this from [equation 1] and get $d(f-wy) = udx - ydw $ and can deduce that our Legendre Transformed function $ g(x,w) := f-wy$

Now, we can easily apply this method to Internal Energy (U) to obtain Enthalpy (H)

We have $U(S,V)$ and $dU(S,V) = \Big( \frac{\partial U}{\partial S}\Big)_VdS + \Big( \frac{\partial U}{\partial V}\Big)_SdV $

Now, here we seek $H(S,P) \equiv g$

a) $f \equiv U$

b) $S \equiv x$ is the variable we hold fixed

c) $V \equiv y$ is the variable we switch

d) $\Big( \frac{\partial U}{\partial S}\Big)_V := T \equiv u $(the conjugate of the variable we are holding in place

e) $\Big( \frac{\partial U}{\partial V}\Big)_S := -P \equiv w$ (the conjugate of the variable we change

$ g =f-wy $

so, $H(S,P) = U-(V)(-P)$

Or, $H(S,P) = U + PV$

From here, you can proceed to write down: $dH = dU + PdV + Vdp$

(or for macroscopic changes: $\Delta H = \Delta U + \Delta (PV) $)

We impose the constant pressure condition to get

$\Delta H = \Delta U + P\Delta V $


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