Enthalpy Unleashed!

Let's examine the First Law of Thermodynamics yet another time. In a previous post entitled "Another Work Out," we discussed the different conditions for a system to have no PV work. Exploring these conditions (constant volume and pressure) further, we will define the effects upon the internal energy and introduce a new state function, the enthalpy of the system.

Process At Constant Volume 

We consider the system to only have PV work. 

This scenario is commonly encountered in a bomb calorimeter, which will be explored in an upcoming post. 

But, hey wait a minute, we know the change in volume to be zero, as the volume is constant. 

So that was pretty straight forward. 

This process is irreversible by the way. 

Process At Constant Pressure

Let's check out the system under constant pressure with only PV work. Just a reminder that the pressure under study is the external pressure. The internal energy is still defined under the First Law: 

We solve for the heat at constant pressure and expand the equation: 

The significance of this little manoeuvre is to recognize that the heat at constant pressure is made up state functions, and hence is a state function itself. 

Now, we define the heat at constant pressure as the enthalpy of the system: 

It is important to realize that although the internal energy equals the heat at constant volume and the enthalpy equals the heat at constant pressure, both internal energy and enthalpy exist at conditions where the pressure and/or volume aren't constants. 

Alternative Way of Defining Enthalpy

As with many series of equations you can approach the derivation backwards. I realize this is a trivial addition to this post, but because enthalpy is kind of abstract when you are first introduced to it, I feel it may add to a thorough approach to understanding the definition. As I like both of these derivations equally - maybe because they are equal *groan* - and found them both useful study tools, here goes! 

Expand the equation, considering the system to be irreversible, so useful work is negligible, and the external pressure is constant. 

We can cancel a bunch of terms. The "w" work, which is equal to the nonPV work in the system, is zero. The middle terms cancel, and the change in the pressure is also zero, as the pressure is constant. So the one lone heat at constant pressure term remains, showing it to be the enthalpy, just as we expected. 

Hee. :)

Enthalpy for Reactions Involved in Consumption or Production of an Ideal Gas

Okay, this is just a little calculation trick for problem solving. 

So, here is our old friend, the Ideal Gas Law: 

Hm, doesn't that PV look really similar to the PV in the enthalpy equation? 

But the reaction occurs at constant temperature, so 

Just a quick tip on how to calculate the change in the number of moles of gas. First you will probably have to write out your balanced chemical equation. So far, the problems I have commonly seen involve combustion reactions of a liquid hydrocarbon. The water produced in a combustion is also treated as a liquid. Omit the liquid and/or reactants and products, and calculate the difference between the gaseous products and reactants only. 

This equation can also be defined in terms of heat capacities, but I will return to it in an upcoming post. Until then... good night. :P