# What is the relationship between enthalpy and internal energy? Where does q = DeltaH come from?

##### 1 Answer
Jan 11, 2016

IMPLICIT ASSUMPTIONS

Okay, then let's look at that formula. You say that:

${q}_{p} = \Delta E + P \Delta V = \Delta H$

Here's how your formula turns out:

${q}_{p} = \Delta E + P \Delta V = \Delta H$

${q}_{p} = q + w + P \Delta V = \Delta H$

${q}_{p} = q - \cancel{P \Delta V} + \cancel{P \Delta V} = \Delta H$

$\textcolor{g r e e n}{{q}_{p} = \Delta H}$

Here, you have implicitly assumed ahead of time that $\Delta P = 0$ and thus $V \Delta P = \Delta P \Delta V = 0$. In this case, $\Delta H = {q}_{p}$, so this formula is correct.

But, it obscures the assumption that you may not have realized you already made: that the pressure is constant.

ENTHALPY VS. HEAT FLOW

To see what that means, let's compare it to a formula that I know to be correct (and is where I would start an enthalpy derivation of this kind), as well as two related equations:

$\setminus m a t h b f \left(\Delta H = \Delta E + \Delta \left(P V\right)\right)$

$\Delta E = q + w$

$w = - P \Delta V$

The bolded equation shows the full relationship before making any assumptions at all, and is thus more helpful when determining relationships between internal energy and enthalpy in multiple different conditions.

Next, you should notice that $\Delta \left(P V\right)$ requires you to use the product rule (plus a bit extra), which means you involve the change in volume and the change in pressure like so:

$\Delta \left(P V\right) = P \Delta V + V \Delta P + \Delta P \Delta V$

Thus, the relationship for enthalpy vs. heat flow is:

$\Delta H = q + w + P \Delta V + V \Delta P + \Delta P \Delta V$

$= q - \cancel{P \Delta V} + \cancel{P \Delta V} + V \Delta P + \Delta P \Delta V$

$\textcolor{b l u e}{\Delta H = q + V \Delta P + \Delta P \Delta V}$

So enthalpy is heat flow plus any pressure changes at a specific initial volume, plus the effects from changes in both simultaneously.

At this point, it is obvious that a constant pressure yields $\Delta H = q = {q}_{p}$.

ENTHALPY VS. INTERNAL ENERGY

Now, we can make comparisons between enthalpy and internal energy (recall the first law of thermodynamics for $\Delta E$):

$\textcolor{b l u e}{\Delta H = q + V \Delta P + \Delta P \Delta V}$

$\textcolor{b l u e}{\Delta E = q - P \Delta V}$

From this, here's what we can figure out:

You should notice that enthalpy at a constant pressure is entirely dependent on the heat flow.

The other thing you should notice is that internal energy at a constant volume is entirely dependent on the heat flow.

Finally, based on the bolded equation, we should see that:

Enthalpy is internal energy plus any work required to change the pressure or volume without worrying about keeping the other constant.