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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.
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.
