This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.
7.07a Heat capacity of bomb
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来自 University of Kentucky 的课程
Chemistry
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从本节课中
Week 7
In this unit, we will learn about thermochemistry, which is the study of the thermal energy transfer (heat) in chemical reactions. We will learn how these measurements of heat are made via calorimetry. Then, learn to calculate heat transfer using various methods.
与讲师见面
Dr. Allison SoultLecturer
Chemistry
Dr. Kim WoodrumSr. Lecturer
Chemistry
Before you can use a bond calorimeter, it's going to have to be calibrated.
In order to determine what the heat capacity of that calorimeter is.
How much heat does it absorb, how much energy does it absorb for
every degree Celsius the temperature raises.
So you would calibrate it with a reaction that you know the delta H very well for.
Okay, so this problem says that we're going to determine the heat capacity of
a bomb calorimeter, and we're going to put a certain amount of sucrose,
and it tells us how much, here it is.
Sucrose, into the bomb calorimeter.
And we see that the bomb calorimeter is going to have some water in it.
And we see how much sucrose we're placing in there.
And we see how much the temperature changes of this calorimeter.
Okay?
So we know whenever we're doing calorimetry that the q of the reaction,
which is your system, is equal to the negative q of the surroundings.
And the surroundings is made up of two things.
It is the calorimeter, which in this case is a bomb calorimeter,
and the q of the water bath that surrounds that bomb.
So we know those two things are current.
If we further break it out, we know that the heat of the reaction
is equal to the negative of the heat capacity of the bomb,
which is what we're trying to determine, times the change in temperature.
Plus, the mass of the water, the specific heat of the water, and
the change in the temperature of the system of the surround the surroundings,
which is the water and the bomb calorimeter.
Now, we need to note everything in this equation except for
this in order to solve for it, so we need to determine this.
This is what the problems is asking us to do,
so we have to have something to plug in for everything else.
Well, some of them it's quite easy.
We know the mass of the water.
The mass of the water is 2,000 grams, there it is.
Specific heat of water?
Well, we can always look that up, but the specific heat of water
is 4.184 joules per gram degree Celsius.
We use that a lot so you might remember that.
The delta T we know.
The delta T is final minus initial, so
it's 27.913 minus 24.667 degrees Celsius.
Okay, so we know that.
See if I've got a, a number for that written down somewhere.
3.246.
3.246 degrees Celsius, so
we know, we'll put a check mark above everything we know.
Change in temperature, specific heat of water, mass of water,
change in temperature.
We're trying to determine this, so we've gotta know the heat of the reaction.
Well, it didn't give it to us at the problem.
But we do have an ability to get to delta H for
the reaction and let's consider the reaction.
Okay, what is the reaction that's happening here?
It's a combustion of C12H22O11.
This is a sucrose.
This is table sugar.
It's reacting in the presence of oxygen.
That's what you do in a bond calorimeter.
It's a hydrocarbon with some oxygen in there.
And you always produce carbon dioxide and water as your products.
Okay?
At 24 and 27 degrees Celsius, that water is going to be a liquid.
Okay?
And we'll need to balance it.
So there's 12 carbons.
We're going to need 12 carbons.
There's 22 hydrogens, we're going to need 22 hydrogens.
We have got our 11 oxygens taken up for here, and
we've got 24 more oxygens, so I'm going to put a 12 right here, and it's balanced.
So how do we determine the delta H of this reaction,
which we need to know in order to calibrate it?
Well, let us use that the delta H of a reaction.
And let's do it in terms of the delta H formations, okay?
So it would be 12 moles times the delta H of formation of the carbon dioxide,
plus 11 moles, times the delta H of formation of H20 liquid,
minus, so that's products minus the reactants.
We'd have to have the delta H of formation of C12H22O11
and 12 moles of the delta H of formation of O2.
So we can go to the tables of these values here, and
we can determine the delta H.
Okay, so let's go ahead and plug numbers in.
Here the reaction is equal to 12 times.
4 carbon dioxide, it's a negative 393.5 kilajoules per mole.
This was moles, this is kilojoules per mole, so I am left with kilojoules, okay?
And then, 11 times, for H2O liquid
is negative 285.8 kilojoules, okay?
Minus 1 mole, that's how much sucrose we have,
times, and if we look up the value for sucrose, it is 221.7.
Now, 2,221.7.
Okay?
Kilojoules per mole, times the moles.
So, all of these combined will give me the delta H of the reaction.
And I'll get the written number written down here, negative 5644 kilojoules.
Now is that the q, here the reaction?
Well, no.
That's the heat of one mole of this combusting.
But I do not have a whole mole.
I only have 2.1250 grams of the sucrose.
We know that the q of the reaction is going to be determined as the heat for
the 2.1250 grams of sucrose that we put into
the calorimeters we're trying to calibrate it.
We know that we're putting this much sucrose into it, so
we have to figure out how much heat this amount.
We know how much heat a whole mole would produce.
So let's convert the 2.1250
grams of C12H22O11 to moles.
So we'll need the molar mass of the sucrose.
The molar mass of sucrose is 342.
We ought to carry more significant figures.
0.299 grams per mole.
So this is how many grams we have, this will convert it to moles.
And then we'll use the information we got from the previous slide to go
from moles of C12H22O11 to kilojules.
Kay, we know we have a negative 5,644 kilojules for every mole.
So when we multiply and divide these quantities out,
we get a negative 35.039 kilojoules.
So that would be how much heat is given off,
that is the heat of the reaction when the 12.125 grams reacts.
So now we're ready to plug everything into the previous equation,
let's look back up here.
We're going to be plugging everything we know into this equation.
We now have just determined the q of the reaction.
That took a lot of work.
Everything else is pretty much given to us.
So, I'm going to put this in joules, the heat of reaction in joules instead of
kilojoules, and we'll see why that would be the case here in just a minute.
Negative 35039 joules is
equal to a negative, well the heat of the calorimeter that I don't know,
the delta T, we determine that to be 3.246.
3.246 degrees Celsius.
Kay. The mass of the water that was 2000.
4.184 joules per gram degrees Celsius, and again the same delta T, 3.24.
Now I can't remember either.
Was that a six?
Yes it was.
I'll write a little neater here.
6, 246 degrees Celsius.
And there's a decimal point there.
Okay, so there is everything plugged in except for
the heat capacity of the calorimeter.
I'll, divide both sides by negative 1 and get rid of the minus signs.
That'll give me 35039 joules,
is equal to C of the calorimeter times
the 3.246 degrees Celsius plus,
when you multiply all this out you have 27,163 joules.
So we will subtract 27,163 from this side.
So that would be 35031 joules minus
27,163 joules is equal to,
what we have left on the right hand side.
That would give me 7,903 joules.
246 degrees Celsius.
Divide both sides by 3.246 degree Celsius and
that will give me the heat capacity of the calorimeter.
And that is equal to 2,434 joules
per degree Celsius.
So unlike the coffee cup calorimeter or the constant volume calorimeter where we
often see that the heat capacity of the calorimeter is negligible,
certainly this is a very significant heat capacity of this bond calorimeter.
