What's the unit of enthalpy?

Dec 25, 2015

$\text{Joule}$ = $\frac{k g \cdot {m}^{2}}{s} ^ 2$ = $N \cdot m$

Dec 26, 2015

Enthalpy is a measure of heat flow at a constant pressure.

In other words:

$\setminus m a t h b f \left(\Delta H = {q}_{p}\right)$

You should have used this relationship before in General Chemistry. You just may not have known it at the time. For instance, dropping a piece of metal in water? That's at a constant surrounding atmospheric pressure.

We can show this by starting from the following equations:

$\setminus m a t h b f \left(\Delta H = \Delta U + \Delta \left(P V\right)\right)$ $\text{ } \boldsymbol{\left(1\right)}$

$\Delta \left(P V\right) = P \Delta V + V \Delta P + \Delta P \Delta V$ $\text{ } \boldsymbol{\left(1.1\right)}$
(where we have simply included very combination for how pressure and volume could change)

$\setminus m a t h b f \left(\Delta U = q + w\right)$ $\text{ } \boldsymbol{\left(2\right)}$

$w = - P \Delta V$ $\text{ } \boldsymbol{\left(2.1\right)}$

where $H$ is enthalpy, $U$ is internal energy, $q$ is heat flow, $w$ is work, $P$ is pressure, and $V$ is volume.

Thus:

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

but the pressure is assumed to be constant, so:

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

As a result, we define a new relation; at a constant pressure:

$q = {q}_{p}$

So:

$\textcolor{b l u e}{\Delta H = {q}_{p}}$

Why does that help us? Well, we should know that heat flow is going to be in $\text{J}$, joules, from General Chemistry.

So naturally, enthalpy must then be in $\setminus m a t h b f \left(\text{J}\right)$.

Of course, enthalpy can have other units. It's not to say it can't be converted to other units, but the basic unit for it is $\text{J}$, and more often, $\text{kJ}$.

Some other forms of enthalpy with their common units:

Molar enthalpy $\Delta \overline{H} : \text{ kJ/mol}$
Standard enthalpy of reaction $\Delta {H}_{\text{rxn"^@: " kJ/mol}}$
Standard enthalpy of formation $\Delta {H}_{f}^{\circ} : \text{ kJ/mol}$