 In this lecture what we're going to be doing is we're now going to be taking the heat of formation or enthalpy of formation that we looked at from the previous class and we're going to apply it to enabling us to use the first law to apply to reacting systems so we're finally going to be able to do something kind of interesting with all of the analysis that we've been doing with combustion. So what I'll do is I'll begin by reviewing the different forms of the first law for what we call now reacting systems. So these are systems whereby we have reaction taking place. Typically we usually look at combustion because that is our primary fuel source if you recall I said 83% of all the energy consumption within the U.S. I think that was in 2009 or 2010 comes from combustion. Consequently this is very important globally. People are looking at combustion and the associated side effects such as carbon dioxide generation of that and that will change the fuel mix over time but for current situation this is an important mechanism by which we get our energy generation. So if you recall what we had was the heat of formation and then we said that we had to correct that because the heat of formation is at 25 degrees C in one atmosphere so we have to have a way to correct the enthalpy for that with respect to whatever temperature we may be at. So the way that we correct it is with the enthalpy at whatever temperature our other product or reactant is at and then we subtract off the enthalpy and the zero reference state would denote the fact that that is at 298 Kelvin. So that's the way that we handle enthalpy for a reacting system. What we're going to do now is take a look at the different forms of the first law and then we'll work an example problem. It's going to take a couple of segments for this lecture. The example problem is kind of long but it will demonstrate how we can apply the first law to a reacting system. So let's begin by looking at the steady flow formulation for the first law. So there we have the steady flow formulation for the first law. We have heat transfer minus work. It's the same as what we saw before but you'll notice on the right hand side of the equation I'm neglecting kinetic and potential energy here. But what we're looking at is a change in enthalpy and we're doing this on a per kilomole basis given that we go through and do the stoichiometric balance. Whenever we go through reactions quite often we will be expressing things per kilomole and that's why we have the number of moles at the front. So in a way we're kind of like looking at the mole or flow rate I guess you could say when we're dealing with these problems. But there for the enthalpy it's just like above we have the heat of formation or enthalpy of formation corrected. For whatever temperature the product or reactant may be at and that's what the second and the third term are in each of the expressions. So that is the expression for steady flow. Let's continue on and take a look at fixed mass. And for fixed mass we will have two different formulations depending if we have constant pressure or constant volume. But for fixed mass to begin with what I'm going to do if you recall our definition for internal energy and its relationship to enthalpy was as follows. Now we don't want to come up with a new internal energy of formation like we did for enthalpy of formation. So we want a way to be able to kind of go from one to the other without having to come up with another set of tables. And the way that we're going to be able to do that is using this relationship here and we have specific volume on a per kilomole basis there. So if we look at the fixed mass form of the first law that's what we get. That should be reactants. Let me clean that up. So that's the form of the first law that we have and of course remember work. We said that that would be a combination of boundary work plus other work. And boundary work for fixed mass systems we defined as being integral pdv. So that is the form of the first law that we have. Now what we're going to do we're going to restrict this for either constant volume or constant pressure. First of all let's take a look at constant volume. And for constant volume what we have is the integral of pdv is equal to zero because dv is not changing. And for this that will simplify it a little bit. But what we'll do and we'll see this later in an example in the next lecture. We will rely on the ideal gas equation for a constant volume process. And what that enables us to do is it enables us to address this pv term that appears in the equation. And if you write out the ideal gas equation and what I will do is recast that now by dividing through by the number of moles. We're going to make a substitution of the pv for rut when we deal with this equation. And like I said you'll see that in the next lecture. We'll just put that off to the side for now. And the other process that we may have would be constant pressure. So let's take a look at what we would do with constant pressure. For constant pressure the boundary work is not equal to zero. But it would be equal to pv2 minus v1. That's assuming that we have a constant pressure process underway. And with that you'll recall when we looked at the first law that we said that we could make this substitution. And that was if we had a constant pressure process where we had boundary work. And with that we can rewrite our fixed mass form of the first law. So that's what we have for the first law. And what I've done here is I got rid of the pv term. And the way that I did that is by essentially moving the boundary work from this side of the equation over onto the other side of the equation. As well what we did is we made a little bit of a substitution using our delta u plus work boundary equals delta h. And so that has enabled us to recast the equation as you see it there. So that is the form that we would use of the first law for constant pressure. So we have constant volume. We have constant pressure for fixed mass. And then we have steady flow. So what I'm going to do in the next segment to this lecture is I'm going to go ahead and apply the first law for a steady flow process. And we will look at a way of being able to do the calculation for steady flow. And then in the next lecture what I'll do is I'll do a couple for the fixed mass be it constant volume or constant pressure. So that's where we're going and that concludes this segment of this lecture.