>> Here, I'm going to briefly describe how we use the NIST WebBook.
This is a really valuable resource with a lot of data for thermodynamics.
Here I'm going to look at calculating enthalpy and entropy as a function of temperature.
So, let's go ahead and look at the WebBook by clicking this link.
And Chemistry WebBook, what we want to do now, we're going to look for a search by name,
and I'm going to put in here methane.
And I hit return, and it brings up a lot of data.
What we're interested in is the gas phase thermochemical data.
So, you'll see heats of formation.
And for multiple sources, but the values are pretty similar.
And likewise, heats a combustion, entropy, and what we're interested in now is, let's go down,
this heat capacity equation, and then I'll talk more about this,
but it's in terms of temperature divided by a thousand.
But it's the equation for heat capacity, equation for enthalpy, and equation for entropy.
It also then gives the values of the parameters.
So, the heat capacity over one temperature range, 298 to 1,300 has certain parameters
from 1,300 to 6,000, there are different parameters.
And so we'll talk about how we would use these.
So, I've copied some of these values.
So, let's go back and look at those and look at how we use this information from the WebBook.
So, there's a couple important things to point out here.
The zero indicates standard state one bar.
We're always calculating a difference in enthalpy.
So, we calculate the difference between the condition temperature we're at,
and the enthalpy at 298.
So, when we do integration, we're still starting at 298.
So, the interesting thing about the WebBook is this gives you an equation for enthalpy.
And it shows A, B, C, D, E from integration.
And it has these two additional parameters.
These account for the fact that we don't have to substitute in the limits.
We don't have to evaluate at 298 limit, for example, because that's all done by the WebBook.
And it's incorporated into these parameters F and H.
And this certainly makes life a lot easier.
The other thing to point out is the temperature
in these equations is Kelvin divided by a thousand.
And since heat capacity is joules per mole, you would expect then when we have the same A, B, C,
D, and E for enthalpy, it would also be joules per mole.
But because the integration, and we integrate respect to this temperature,
we need to end up multiplying the answer by a thousand,
because we really should integrate respect to Kelvin temperature.
And so that's taken care of by the fact that enthalpies are in kilojoules per mole,
even though these heat capacity parameters have joules.
On the other hand, for entropy, the calculation is in joules per mole.
But likewise, there's this parameter G that accounts for these limits.
So, let's just briefly describe what those limits are.
But the idea is we can calculate enthalpy, independent of which temperature range,
just by substituting in the appropriate parameters.
So, if I'm at 800 Kelvin, then I can substitute 800 divided by a thousand
into this enthalpy equation, and I would use these parameters.
If I'm at 2,000 Kelvin, I would use these parameters.
And, again, the WebBook values for F and H, which you notice are different
for the two temperature ranges, accounts for this.
And I'll show you just briefly how it accounts for that.
So, if we're in the temperature range between 298 and 1,300,
then the enthalpy at some temperature standard conditions minus enthalpy of 298,
remember we don't have absolute values for enthalpy.
We always are looking at difference.
It's just this integration 298 to temperature, heat capacity DT, but what we're going
to write that, because we want to use these values, is we integrate respect
to this lower case t, so that .298, lower case t, the heat capacity that I show here,
and now dt, which means I should multiply this by a thousand.
And I would get enthalpy in the correct units of joules.
And then what we've done is that's incorporated into the calculations in the WebBook,
so if you use this equation, you'll get an answer
in kilojoules, because it's taken into account.
And then what this integration looks like, so you should note that I'm using A1, B1, C1,
et cetera, because that corresponds to the parameters in this first column of numbers.
Right? Because if we're at a temperature, we're assuming we're at a temperature between 298
and 1,300, this is how we do the calculation.
And then, of course, we'd substitute in, do the integration, substitute in the limits.
Okay, so I've done the integration, I've substituted in the two limits.
Now, what the WebBook does for you, it calculates this term.
And this term is equal to F1 minus H. So, this is F1 minus H1.
So, it does that for you.
It puts in the appropriate parameters.
And this certainly makes then the calculation much easier to get [inaudible]
if the temperature is above the, so let's look at the table again.
If the temperature is above 1300 Kelvin, so we have to use the second parameters, again,
we'd just substitute into this equation here.
And the F and H values, which are different in this temperature range,
account for the calculation of the limits,
and for the other integration, as I'll show you in a second.
So, let's make sure to point out here that calculation I showed you so far is
if the temperature is between 298 and 1300 Kelvin.
Now let's look at the case where the temperature is greater than 1300 Kelvin, again,
I'm going to pause and write down the equation.
So, because we're above 1300, we first have to integrate from 298 to 1300 using the parameters
from the first set of numbers; the A1, B1 that are only appropriate between 298 and 1300.
And then we integrate from 1300 to our temperature of interest using the parameters
from the second column of numbers in the heat capacity table A2, B2, et cetera.
And we can do this integration.
Of course we do both integration.
We effectively have done this integration for this first interval.
And let's just write down the values then for the second interval.
Now, you'll note that I listed the integrals, perhaps the reverse, when I did the integration
for this term, I first wrote the evaluation of 1300.
That's why the minus sign.
And then I wrote a plus sign for evaluation of T. And the reason I did that is
because now we're going to look at all these terms together.
And they're equal to F2 minus H2 in the table.
So, instead of having to do this calculation each time we wanted
to do a determination enthalpy, the WebBook has done that for you, it's listed in the table
to calculate, and that's where the F2 and the H2 values combine come from.
So, that means the calculation of enthalpy is certainly much easier,
because we're just substituting in this plus the F2 minus H2.
So, the calculations for entropy are very similar.
Keep in mind, final result for enthalpy kill joules per mole.
But entropy, it's joules per mole Kelvin.