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PROFESSOR NELSON: Well, so last time we did a little bit
more work on thermodynamic cycles, and basically went
around a cycle and looked at some state functions, delta u
and delta H. Saw that around a closed cycle those zero
because they're state functions.
So if you start and end at the same place, they've got end at
the same place that they started.
There's no change in them, and then we also looked at some at
non-state functions, work and heat, and saw that those
aren't zero going around a cycle.
Of course you can do work in a cyclic process, and heat can
be exchanged with the environment at the same time.
So we calculated those values too.
And we also looked at this other funny function that's
our special function.
We looked at this integral dq over T. It's so special that
we called this derivative dS, and saw that at least for the
cycle we looked at, it also behaved as a
state function behaves.
That is, going around the cycle, it had no net change,
and we'll see this come back later on.
Then we went on to look at thermochemistry, and that's
what I want to continue today.
So really, the main result that we saw last time is that
we can describe the enthalpy of reaction, so if we have
reactants going to products, and we'd like to know our heat
of reaction or enthalpy of reaction, remember it's a
constant pressure as we're considering it, so it's
equivalent to heat of reaction, than the way we can
do this is construct a cycle where we're going to express
these in terms of their constituent elements.
We're going to all the way back to the atoms.
Right, so if we have the elements in their standard
states, that is their most stable forms at room
temperature and pressure, then what we're really doing is
we're saying let's take the reactants and let's pull them
all apart into their separate elements,
they're separate atoms.
And then put them back together this way, and we'll
calculate the enthalpy of this process, the
enthalpy of this process.
It's a cycle, so that'll give us the enthalpy of
reaction, all right?
So this delta H is our sum for delta H of formation for the
reactants, right.
And this is our sum of the quantity for
the products, right.
And in this way, then we can express our delta H of
reaction is just going to be given a sum of the heats of
formation or the difference between the heats of formation
of the products and the reactants, right?
Now, of course, we have to weight this by the number of
moles of each reactant to each product
involved in the reaction.
So to write this a little more carefully, it's, you want to
write the sum over i of nu i delta H naught of formation of
i summed over the products.
This is per mole minus sum over i, nu i here over the
reactants in the same quantity.
Delta H per mole f i, right?
So in other words, were taking each of the products and we're
going to weight them by their stoichiomeric coefficients,
and we're going to have in this delta H of reaction their
heats of formation, and then we're going to have the
negative heats of formation of the reactants, because that's
being subtracted, okay?
So, just to go through a basic example, let's just look at
the burning of methane.
So we've got methane gas, combining with oxygen gas, so
you have carbon dioxide gas and liquid water, and we'll
consider the whole thing at room temperature
and pressure, right.
So let's just undertake this set of processes.
Let's decompose these to the elements.
So in other words, our methane gas, we're going to write the
chemical equilibrium between this and the solid carbon.
That is to say let's specify graphite, it's not diamond,
right, graphite is the stable form of carbon at room
temperature and pressure.
And then, hydrogen gas two moles of hydrogen gas and
we're going to have a delta H of formation.
This is for CH4 in the gas phase, right?
And then over here, we have our oxygen but 2 O2 gas going
to, well 2 O2 gas -- the point is oxygen already is in it's
stable form at room temperature
and pressure, right?
That is the elemental form of oxygen, of course, is as
oxygen molecules in the gas so there's no delta H formation
for oxygen.
The gas is zero, right.
And then second phase of this, now we're going to take the
elements that we've produced, and we're going to put them
back to make products, right?
So there we've got carbon as graphite, and a solid plus an
oxygen gas goes to make CO2 gas, right.
And we've got our positive delta H of formation
associated with that.
And finally we've got hydrogen gas plus oxygen gas giving us
two water molecules and this is in the liquid phase, and
there's some delta heat of formation associated with
that, okay.
So there is our set of individual processes that's
going to constitute our cycle, right.
If we take the combination of all those, which is to say we
just take both steps in this cycle, then we'll have our net
reaction, which is to say we'll have this, right.
So what that says is our heat of reaction is our heat of
formation per mole of CO2 gas, right.
Plus 2 times the heat of reaction or heat of formation
per mole of liquid water minus heat of formation per mole of
methane, right.
And things we can easily look up.
So in a simple way, we can determine, we can calculate
what the heat of reaction is for
something like this, right.
And you'll see in the back of your book, you'll see
moderately extensive tables of heats of formation, and if you
go online you'll see, you can find extremely extensive
tabulated values of heats of formation, that have been
measured for an enormous number of compounds.
And from that, then you can look at enthalpies of reaction
for countless numbers of reactions, right.
So by using the tabulated data, we can really determined
heats of formation for most reactions that you might
contemplate, OK?
Of course, the cyclic steps that we've taken to do this
apply not just for breaking down reactants and product
into the elements in their standards states, but of
course we could also look at whole sets of reactions and
write cycles as well, right.
So for example, if we wanted to look at the reaction of
something like carbon graphite plus oxygen gas to make carbon
monoxide gas, well normally this isn't the way the
reaction would work.
Normally if you just tried to make this happen what you
would wind up with the CO2, not CO, right.
But one way to approach this is you could look at the
individual reactions of going, you could go to CO2 and then
you can go with CO2 and oxygen and combine, or combine CO and
oxygen to get CO2, right?
So normally, CO2 is formed, but you could calculate the
heat of reaction.
Of course you could go back to the elements, but you can also
say well, let's just take a known heat of reaction for, or
heat of formation for CO2, and then also combine that with
the heat of reaction for CO plus oxygen forming CO2 and
determine this that way, all right, OK, so once you have
done this, once we've got the heat of reaction then there
are a few simple results that are useful.
One is we know immediately whether when we run a reaction
heat is going to be released in the process out to the
environment, or whether heat is going to
be taken in, right.
So, and of course if you ever used a hot pack or a chemical
hot pack or cold pack, you've seen examples of the two of
those, where the constituents are selected to give you
either heat released or heat taken up, right.
So in particular delta H of reaction less than zero, that
means a negative amount of heat, which means that heat is
released, right.
Yes, that is it's exothermic.
Positive enthalpy of reaction, which is to say a heat of
reaction is positive.
Remember the way we define heat is positive it means that
there is heat that is coming from the environment into the
system, right.
In other words heat is absorbed or taken up by the
reacting system.
That's endothermic, okay.
All right, one more important aspect of this that we need to
be able to deal with, and that is the heats of formation that
you'll find tabulated typically are going to be
given for room temperature and pressure.
And it's very common that you might want to know the
thermodynamic condition for reactions, especially at other
temperatures, very common, of course.
And it's not hard to see how the heat of reaction at room
temperature can be related to they heat of reaction at other
temperatures.
So let's just look at that.
So remember your fundamental relation, dH, if you write
that as a function of temperature and pressure,
partial of H with respect to T, constant pressure, dT, plus
partial of H with respect to p at constant temperature, dp.
But if we're working at constant
pressure then dp is zero.
So we just got the derivative with respect to temperature.
That's our heat capacity, right, Cp.
And we may have tabulated values of Cp for an enormous
number of materials.
So now, if we look at the temperature dependence of
delta H for reaction, really what we're going to need to do
is look at how that heat capacity changes going from
reactants to products, right?
So in other words, d delta H of reaction, with respect to
temperature.
That's what we want to find out, right?
So how does delta H change if we change the temperature?
Well, it's just given by Cp dT for the reactants and the
products, right?
For the products minus the reaction, that is it's delta
Cp, we need the sum over the products.
Nu i, Cp i minus same sum over reactants,
i, nu i Cp i, right.
Similar to what we saw before.
Now we need to integrate, right.
So if we integrate between some pair of temperatures, T1
and T2, we do this and we have this quantity, d delta H of
reaction with respect to temperature, dT, all this is
at constant pressure, right.
Let's just be careful to specify that.
Well, so of course, this is then just delta H of reaction
at T2 minus delta H of reaction at T1.
This is looking good.
This is what we want to get, is our heat of reaction at
some new temperature T2.
If we already know it at some initial temperature, usually
room temperature, T1.
And it's just given by the integral from T1 to T2 of
delta Cp dT.
Now in general, this is as far as we can go, but at least if
the temperature change isn't too big, for most materials
that heat capacities doesn't depend very strongly on
temperature, right.
If you take some material, and you measure its heat capacity,
right, how much heat do I have to put into it in order to
change its temperature by a degree, right.
And I make that measurement at room temperature.
And then maybe I raise the temperature to whatever, room
temperature, maybe 20 degrees hotter than room temperature.
And I again say OK, now how much heat do I need to raise
this thing's temperature by 1 degree?
The amount of heat I need to put in to do that is not very
different from what it was at room temperature right.
In other words the heat capacity
didn't change very much.
So, the result of that is that at least for modest
temperature excursions, it may be the case that we can
simplify this and simply say it's just the difference in
heat capacities times the temperature change.
Again, we can't always be assured of that, but it's
often the case.
Any questions on any of this so far?
OK, now what I want to do is just describe a little bit of
how do you measure all this stuff, right?
So what we've done so far is just lay out in principle how
we should describe heats of reaction, and of course
assuming that we've got everything in tables we can
always look them up, but sometimes you want to make new
compounds, right.
Sometimes you'd like to know what the energetics are
involved there, right, and so it's useful to be able to
actually measure something.
So how do we do It?
This is what's called calorimetry.
So we're going to make measurements to determine heat
of reaction values.
Actually, calorimetry is used for all sorts of things.
It can be used to determine, for example, the energetics of
phase transitions.
Maybe you're not looking at a reaction, but you've got some
new compound, and you're looking at it go from liquid
to solid or to gas.
And you can see what the heat involved in a process like
that is as well.
So what happens?
Well, the objective in the case of a reaction, we have
reactants at some temperature going to products at that
temperature, and that is our heat of reaction at that
temperature.
That's how it's defined, right, at constant
temperature.
OK?
What do we actually measure?
Well, what we can really do is we can put the reactants in
some container, put the whole thing in an insulated place,
you know, hold things in a big insulated box.
Run the reaction.
Maybe it produces heat.
If it does, then the whole thing will heat up, right?
The stuff that's inside the reacting volume, and
whatever's right around it there inside the big box, it's
all going to heat up.
So I'll start at some initial temperature, T1.
I won't end up at the same temperature.
It'll be hotter.
How much hotter?
Well, that depends on how much heat was produced.
So in principle, if I measure how much hotter, I can
determine how much heat was produced, and from that, I
should be able to calculate delta H at T1.
So let's think about how that's really working.
I've really got reactants at T1, plus a calorimeter at T1,
and I'm going to end up with products -- well let's let me
just write that's over.
In fact, let's start by just putting a little sketch of the
whole thing up.
So here's my reacting stuff, maybe it's a liquid.
It's going to take place in there.
It's going to be a constant pressure, it might be open to
the air, or even if it isn't, there might be plenty of room,
and it's a liquid anyway, so the pressure isn't going to
change significantly.
I want to measure the temperature.
And the whole thing is insulated right.
I don't want to have to deal with heat escaping to the
outside environment in a way that might be difficult or
complicated to measure or calculate.
OK, so this, what I've sketched here would be a
constant pressure calorimeter.
There's a reaction.
Normally this is used for a reaction in the condensed
phases and liquid usually.
So this is a constant pressure calorimeter.
It's set up to be
well-insulated so it's adiabatic.
That's the set up.
We're going to run the reactants, the reaction.
The reactants are going to turn into products.
Let's look at what happens.
So what happens is my reactants at T1 plus
calorimeter at T1 turn into products at T2 plus my
calorimeter at T2, right.
That's my process.
That's what really happens.
Now that, the enthalpy of that process isn't what I want,
right, because the temperature has changed, also the
calorimeter is heated up.
So I need to relate this to what I do want.
Let's label this one.
It is adiabatic.
It is taking place inside this thing, and it's a constant
pressure, and we'll do it reversibly, right.
So that's what we've got.
Now, we can always take these things the products and the
calorimeter at temperature T2 and cool them or warm them to
get to products plus calorimeter at T1, right.
Let's label that two.
It's still at constant pressure.
It's reversible.
It's a temperature change.
Now to make that happen, it's not adiabatic, right.
If I wanted to do that, I'd need a heating element or
something to cool, so I could make that temperature change
happen, right.
Well, so now I can complete the cycle.
I've got reactants and calorimeter at T1.
Fewer products and calorimeter at T1, right.
Calorimeter doesn't change in this process.
So here I've got some delta H associated with changing the
temperature.
This delta H though, this is what I want.
This is delta H of reaction at T1, right?
It's isothermal, constant pressure, reversible.
Just what I want.
That's how I've defined delta H of reaction.
This is what I'm after.
All right, so now let's see how we execute it, and do the
calculations that allow us to calculate this.
So, one, what's delta H in step one?
Yes, exactly, it's adiabatic, right constant pressure.
it's zero.
Delta H1 is zero, right.
What we're really going to do in practice is we're going to
measure, we're going to use our thermometer and say great,
how much did the temperature change, right.
So now, then we're going to use what we're going to learn
from step two in order to calculate this part, what we
could call step three.
OK, so two, that's the crucial part.
It's constant pressure.
It's just given by the corresponding heat, and it's
just a temperature change, right.
So we know how to do this, integral from well it's T1 to
T2, depending on which way we go of Cp for the
whole system dT.
OK?
It's just how much heat is involved when we change the
temperature.
Now, the products have some heat capacity associated with
them right, it takes a certain amount of heat if we make
their temperature change, to either put it in or take it
away, depending on which direction the
temperature is changing.
Same with the calorimeter, OK.
In actual practice though, the calorimeter, is going to be
this big hunk of metal, and really what's going to,
there's going to be some stuff in here, right, they'll be a
fluid, it will usually be some sort of oil.
There's a pretty big thermal mass.
Now we run the reaction and it produces heat.
Most of the heat is going to warm up or cool off all that
oil and stuff.
There's a moderate amount of material
in the actual reaction.
You want to design the calorimeter to fulfill that
condition,
Of course, you can't make it so enormous that even for,
that for any ordinary reaction there's not even a measurable
temperature change, right, because if you have your
enough oil to fill up this room, it would take a huge
amount of heat to change it by even a tiny, it's temperature
but even a tiny amount.
So the calorimeter is designed sort of to scale to match the
maybe the ordinary volume of reactants that you'll put in
there, such that, pretty much all the heat's just going to
heat up the calorimeter, and there's only a small amount
that's going toward heating up the products, right.
So in general, or at least usually it'll be the case,
that this is approximately equal to integral from T1 to
T2 Cp of just the calorimeter dT.
And that generally is just given by the heat capacity the
calorimeter times delta T, right..
Because the heat capacity of the calorimeter just like this
thing is not strongly temperature dependent, OK.
So the point is we can calculate delta H associated
with this process pretty readily, OK.
That leaves three, right.
But delta H for step three must just be the opposite of
this delta H because this was zero and we know that were
going around a cycle, right.
And that's our heat of reaction.
So we're going to be able to do this, right?
OK, now this is one way to do calorimetry, and it's
practical as long as all the materials are in condensed
phases, solids and liquids.
If gases are involved it can work, but it can be tricky to
keep the pressure constant, right, because that means that
now the moles of gas might be changing, and that means in
some way the volume has to be adjustable.
You'd need some sort of balloon or membrane that will
allow that to happen.
Could be done, but easier is to just do the whole thing at
constant volume, right, and just run the reaction that way
and redo the calculation to be a constant volume rather than
constant pressure calorimeter, right.
And it's not hard to do that.
So let's just look at what happens in that case.
It's almost the same.
So let's see, just to finish the job here though, all this
says is delta H of reaction is just negative Cp for the
calorimeter times delta T, all right. so that's what we think
we know in constant pressure calorimetry.
Yes, and if we have gases involved, it's pretty similar,
but now what will have is something like this.
We'll have a reaction vessel that's sealed,
it's constant volume.
That'll be inside our calorimeter.
It's insulated, and there's still a thermometer, so we can
measure the temperature.
So this is still adiabatic.
It's insulated, but now it's constant volume, OK.
Now, you know with constant volume, now it's not going to
be delta H that's straightforward to measure,
it's going to be dealt u, all right.
But it's going to be almost the same, right?
So let's just think about what happens here.
Now it's the cycle that we've got will still
basically look the same.
That is its reactants at T1 plus calorimeter at T1, going
to products at T2 plus calorimeter at T2, right?
And it's still adiabatic, but now it's constant volume.
And it's also reversal right.
So this is our process one.
Now, process two, we're going to end up here with our
products, again at T1 plus our calorimeter at T1, right.
So now we have a constant volume
reversible temperature change.
And so this part now isn't exactly what we want to
determine delta H, because it's a isothermal constant
volume, reversible products or reversible process that takes
a reactant to T1 to a product of T1.
What these are going to give us are delta u values.
This'll give us delta u of reaction, right, T1, right.
It's a constant volume.
All right, so in the end, we're going to determine delta
u here, and then in the end, we're going to have to relate
that to delta H, but that's straight forward enough to do.
So let's look at how we'll analyze what happens.
So one adiabatic, constant volume process, right.
What's zero in that case?
In this case delta H was zero in the
constant pressure example.
Now we've got a constant volume process.
What's zero?
It's u, because u is to q plus w right, heat and work, but
it's adiabatic.
So there's no heat, exchange with the environment, and it's
constant volume, so there's no p dV work, right.
So q is zero adiabatic.
Work is zero.
Delta u1 is zero.
And again just in the constant pressure case, what happens
that we're going to measure in step one, which is to say in
actually running the reaction is the
temperature's going to change.
Right?
The whole thing's going to come to some new equilibrium
temperature between the products and the oil or
whatever's around it, and we're going to measure that.
OK, two, now it's a temperature change, right?
We know how to calculate delta u for a temperature change.
It's very similar here, but what's going to be different
from the case with constant pressure?
There's a real important detail that's different if you
want to calculate what happens at constant pressure when you
change the temperature?
What happens at constant volume?
Looking at this expression, what's got to be different?
Cv, right.
We're not going to have the constant pressure heat
capacity, we're going to have the constant volume heat
capacity, right.
So delta u for step two, that's our q under constant
volume conditions, integral from T1 to T2 Cv for our
whole system dT.
Integral from T2 to T2, and now I can separate again the
calorimeter from the product, and at least approximately,
again, it'll be Cv for the calorimeter dT, which is to
say then it's Cv of the calorimeter
times delta T. Great.
Three, well delta u1 was zero.
Delta u2 was there Cv delta T, so now all we need since the
whole thing is going around in a cycle, Since we know delta
u3 must be negative delta u2, and that is our delta u of
reaction, right?
So delta u of reaction is approximately equal to
negative Cv for the calorimeter times delta T. So,
this is what we're going to measure, and this is what
we're going to determine.
And now we're almost done, except what we really want is
delta H and not delta V, right.
So now we're going to use the fact that H is u plus pV.
Delta H is delta u plus delta pV, and now this is all at
constant temperature in the end, right?
We've determined delta u for some temperature T1, right.
So what happens then we're going to use
the ideal gas law.
So it's approximately delta u plus delta nRT.
That's a constant.
So it's delta u plus RT, we can say T1 is the temperature
we've used here, delta n of the gas.
In other words, what matters here in changing the pressure
volume product?
What matters is we turned some reactants into some products.
How many moles of gas are there in each case, in
reactants and products?
If that changes, of course you know that the pressure in
there is going to change at constant volume if the amount
of gas in there is changing.
And nothing else is going to make a significant
contribution to it, OK?
So finally, then delta H of reaction for our temperature
T1 is approximately minus Cv of the calorimeter times delta
T plus R T1 delta n of gas.
Change of the number of moles of gas, right.
We're going to measure that, and now we're going to
determine this.
In practice, we'll already know the heat capacity of our
calorimeter, when we buy it, right?
So we don't really need to put in a certain amount of heat
and change the temperature of the products and the
calorimeter and so on.
What we need to do is just measure how much the
temperature changed, OK.
It's worth getting just sort of roughly calibrated how big
is this compared to the rest of this stuff.
That is, how different is delta u from
delta H, all right.
And of course it's straightforward to do this,
and I've written this out in the notes, so I won't re-write
the numbers here.
But I gave an example looking at combining
HCl and oxygen right.
So 4 HCl plus oxygen gas.
This two in the gas going to water and the liquid, plus
chlorine gas at room temperature.
Well, what you find out is delta u of reaction is minus
195 kilojoules for the reaction as written.
And it turns out, as written, if you say OK, how much, what
changed for the pV product?
Well here we've got four moles of gas, five moles of gas.
Here just two, so we changed the number of
moles of gas by three.
All right, how much did it matter, right?
Well it matters.
It's measurable.
So now delta H of reaction turns out to be minus 202
kilojoules.
So, you know, seven out of a couple of a hundred, right, a
few percent, kind of typical.
In other words, if you look at energetics of ordinary
reactions where you're, you know, you're making and
breaking covalent bonds, there's a fair amount of
energy stored in those, right?
The additional change due to changing pressure volume is
certainly measurable.
You don't want to just ignore it, but in cases like that,
it's usually a small fraction of the total.
A few percent is typical, okay.
All right, let me just go through one numerical example
of a calorimetry calculation, OK.
I won't put all the numbers up on the board because our time
is running short, but I just want to outline it.
All I really want to do is calibrate you a little bit for
what happens with ordinary calorimeter heat capacities
that makes the calculation turn out to be relatively
easy, right.
So let's just write it out.
Let's take iron sulfide as a solid, plus 11 halves oxygen
gas to make iron oxide, also a solid, plus
sulphur dioxide gas.
All right.
We'll start at T1 is 298 Kelvin.
T2, maybe we don't know it yet, right.
OK, so we know how to calculate what's going to
happen, delta H and delta u, because we can look up the
heats of formation and so forth of all
the compounds, right.
So if we do that, what we discover is that delta H of
formation whoops, something's wrong here.
This is sulphur -- what am I doing, boy, S and it's a two,
sorry, and the heat of formation is minus 180
kilojoules per mole, that's oxygen, it's zero.
Boy, I'm, I don't know what is about this reaction that's
vexing me but it's not that complicated.
Here it's minus 824 kilojoules per mole minus 297 kilojoules
per mole, right.
So that's our input thermodynamic data.
So first of all, let's just do a heat of reaction
calculation, right.
Taking the product minus the reaction, right.
So it's minus 824 plus 4 times minus 297 minus
2 times minus 180.
Right?
In other words I've got the stoichiometric coefficients in
there and the values, and I'm subtracting the reactants from
products wind up with minus 1652 kilojoules per mole.
Well, it depends on what we write, what we consider mole,
right, maybe I should just write kilojoules as written.
OK, I'm going to skip the delta and get the change in
moles of gas calculation.
It's straightforward to do.
What I really want to do is just give an example of what
happens when you throw the thing, the material into a
calorimeter and see how much the temperature changes.
So let's imagine we start with 0.1 moles of our iron sulfide,
and then we have a stoichiometric amount of
oxygen, and the whole thing is done in a constant volume
calorimeter, and we see what happens.
Now the crucial element is what is the heat capacity of
the calorimeter, right?
And it's, again it's a macroscopic pretty big thing,
so typical might be 10 kilojoules per Kelvin, and
that's pretty big, right?
Noticed that's not per mole, right.
I mean the calorimeter is a big thing filled the little
oil or whatever is inside it, right?
And it's for that whole unit that you've
got some heat capacity.
How much heat does it take the warm the entire thing up or
the insides of the thing up by a degree?
It's that number right.
Ordinary heat capacities are in Joule's per Kelvin mole,
not kilojoules, right.
And what that's telling you is probably your reactants and
products, so the amount of heat that's involved in
changing the air temperature is going to be negligible
compared to what happens to the whole calorimeter.
OK?
So now if we say, OK, we we've done this calculation starting
with 2 moles of this, but now we're going to be at 0.1 mole,
so we're going to need to divide by 20, right.
So instead of minus 1652, it's going to turn out delta u is
minus 1648 kilojoules, so, and then divide by twenty, we end
up with minus 82.4 kilojoules, that is, that's the delta u of
reaction for what happens if you'd put in not two moles of
this, but 0.1 mole of this, right.
Practical amounts is the reason I'm using numbers like
this, right.
So now, what's delta T?
Well here's Cv, right, so delta T is just our minus 10
kilojoules per degree oh sorry, it's our minus 82.4
kilojoules.
That's the heat released, divided by minus 10 or 10
kilojoules per Kelvin, right.
It's 8.2 Kelvin.
In other words, how much does the temperature of the whole
thing change when you put an ordinary amount of material in
there and run a reaction, right.
Well, what do you do?
You calculate how much heat is released in the reaction.
And then what's going to matter is what's the heat
capacity of the whole, of the calorimeter?
I didn't even need to know that heat capacity of the
product, right.
Because it's effect the thermal mass of the product is
negligible compared to the thermal mass of the
calorimeter.
Now in real practice, I'll do the calculation in reverse.
I'll measure how much the temperature changed in the
calorimeter, right.
I'll know the heat capacity, and what I'll really be
calculating is OK, how much heat must have been released
in the reaction to make that temperature change happen?
And that in the case of constant volume, in this case
that's my delta u, and then I'll add my little delta n
term to get delta H.
Any questions on calorimetry?
OK.
See you Friday.
We'll finish on calorimetry and thermochemistry and then
we'll start in on one of the really most difficult topics
that we'll deal with all semester, which is a second
law and our special function that we've seen just a little
bit so far.