# How do you use the equation for heat flow at constant temperature and pressure (latent heat of fusion or vaporization)? q = nDeltaH

Jun 9, 2015

The equation you're referring to is $q = n \cdot \Delta H$.

#### Explanation:

This equation is used to establish a relationship between how much heat is involved when a number of moles of a substance undergoes a phase change.

Let's say, for example, that you want to know how much heat is needed to melt 3 moles of ice at its melting temperature of ${0}^{\circ} \text{C}$.

The equation will look like this

$q = n \cdot \Delta {H}_{\text{fus}}$, where

$q$ - the amount of heat;
$n$ - the number of moles of a substance;
$\Delta {H}_{\text{fus}}$ - the molar enthlapy of fusion.

In water's case, the molar enthalpy of fusion is equal to 6.02 kJ/mol. This means that, in order to convert 1 mole of solid ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you need to impart 6.02 kJ of heat.

If you want to know how much heat would be needed to convert 3 moles of ice at its melting point to liquid water, you'd have

q = 3cancel("moles") * 6.02"kJ"/cancel("mol") = "18.06 kJ"

Here's another example. Let's say that you need to determine how much heat is released when 5 moles of water condense from steam at ${100}^{\circ} \text{C}$ to liquid at ${100}^{\circ} \text{C}$. This time, the equation would be

$q = - n \cdot \Delta {H}_{\text{vap}}$, where

$\Delta {H}_{\text{vap}}$ - the molar enthalpy of vaporization.

Notice that the equation has a negative sign this time. This is used to show that heat is released, not absorbed, when condensation occurs.

The molar enthalpy of vaporization will tell you how much energy is absorbed when 1 mole of liquid water is converted to 1 mole of steam. In water's case, $\Delta {H}_{\text{vap}}$ is equal to 40.7 kJ/mol.

This means that you release 40.7 kJ for every mole of water that goes from steam at ${100}^{\circ} \text{C}$ to liquid water at ${100}^{\circ} \text{C}$.

This time, you have

q = -5cancel("moles") * 40.7"kJ"/cancel("mol") = "-203.5 kJ"

Converting that many moles of water from steam to liquid would release that much heat.

As a conclusion, you need to remember that this equation uses number of moles, not grams, and that it can only be used for phase changes, since they take place at constant temperature.

Jul 17, 2015

Just a note about phase changes: they occur at a constant temperature, but the pressure can change (but it doesn't have to).

When you say $q = n \Delta H$, you perhaps mean the following:

$\textcolor{b l u e}{{q}_{\text{vap" = nDeltaH_"vap}}}$
$\textcolor{b l u e}{{q}_{\text{fus" = nDeltaH_"fus}}}$

...so that the units match. For example, molar enthalpy $\left(\overline{H}\right)$, molar enthalpy of phase changes ($\Delta {\overline{H}}_{\text{vap}}$, $\Delta {\overline{H}}_{\text{fus}}$), and standard enthalpy [i.e. of formation] $\left(\Delta {H}_{f}^{o}\right)$ have units of $\frac{k J}{m o l}$, but generic enthalpy $\left(\Delta H\right)$ is in just $k J$.

You may not have learned this explicitly yet, but a way of defining $\Delta H$ is:

$\Delta {H}_{\text{vap" = DeltaU_"vap}} + \Delta \left(P V\right)$
where $U$ is internal energy and $\Delta \left(P V\right)$ is what I might call "cumulative" work (meaning both work at a constant pressure and at a constant volume are included/embedded).

You can also see this here.

The result of working with this equation is as follows (dropping the subscripts):

$\Delta H = \Delta U + {\overbrace{P \Delta V + V \Delta P}}^{\text{Product Rule}}$

$= {\overbrace{q + \left(- P \Delta V\right)}}^{\text{1st Law of Thermodynamics}} + P \Delta V + V \Delta P$

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

Thus, if the pressure is constant (like a normal day, really)...

$= q + {\cancel{V \Delta P}}^{0}$

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

where ${q}_{p}$ is heat transfer at a constant pressure, such as during a phase change (but it doesn't have to be), and this enthalpy is generic. At a constant pressure, only certain events can occur, but we tend to walk into rooms of constant pressure anyways, so it's not like it's rare.

Basically, the idea of this is that when you are in General Chemistry, those teachers/professors purposefully do not tell you that you are working under constant pressure because it is supposed to be self-evident, usually, and it is also irrelevant to the problem because that's all you're working with at the time. You're working at a constant altitude, so the pressure tends to not change unless you induce a change.

You may have also encountered something in General Chemistry that is similar:

$q = n \Delta {H}^{o}$

More explicitly:
$\textcolor{\mathrm{da} r k red}{\Delta {H}_{r x n}^{o} \left(\frac{k J}{m o l}\right) = {q}_{r x n} / \left(\text{mol Limiting Reagent}\right) \left(\frac{k J}{m o l}\right)}$

where the standard enthalpy of reaction is basically a "normalization" of the heat transfer that occurs while using a certain number of $m o l s$ of limiting reagent. This "normalization" gives one $m o l$ of product at STP.