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.
7.04 Thermochemical Equations
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Del curso dictado por University of Kentucky
Chemistry
507 calificaciones
De la lección
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.
Conoce a los instructores
Dr. Allison SoultLecturer
Chemistry
Dr. Kim WoodrumSr. Lecturer
Chemistry
In this learning objective we're going to be learning about
Thermochemical Equations.
It's like a chemical equation only it gives you the information,
associated with delta H for the reaction.
So a thermochemical equation, shows the entropy changes for the reaction, so
there's a delta H associated with it.
As well as the mass relationship, so
if you were stoichiometry that's the balanced equation it is showing you.
So here's an example of a thermochemical equation.
Now this is not a chemical reaction, it's a physical change.
We're going from water solid, which is ice, to liquid, so
it's the melting process.
This is an endothermic process.
I see that because I have a positive.
Value for delta H,
but you have to put in heat to break those attractions between the molecules.
Now, what this equation is stating, is specific amounts.
So when one mole of ice melts to make 1 mole of water.
Then 6.01 kilojoules of heat is absorbed.
I know it's absorbed, because of the positive value of delta H.
We could represent it with this diagram here.
We have water solid you're putting in the 6.01 kilojoules of heat and
it is being converted to H2O liquid.
So a thermochemical equation is giving you,
the information this picture is providing.
To work with these thermochemical equations,
there's several guidelines that you've got to keep in mind as you work and use them.
The first one is that when you see those coefficients of a balanced equation,
they're always in terms of moles of the substance.
So you would never turn, think of them in terms of individual molecules.
Or as we did in gas laws think about em in
terms of leaders you must think of em in terms of moles.
If you reverse the equation and you're, you're going to flip it around and
make the products the reactants and
reactants the products, then you'll change the sign at the delta H.
So if you're going to take a liquid and
convert it to a solid, it is going to be an exothermic process.
It is going to give off 6.01 kilojules.
So you freeze water by sticking it in the freezer, and
the freezer is a pump to pull the heat out of the water.
The next one should make sense for us.
If we double a reaction, we would double the amount of heat.
So here we're going back from a solid to a liquid, we're melting it.
But, we're melting twice as much, and
so the delta H would be twice the 6.01, or it's 12.02.
So whatever you do to the coefficients, you will do to the delta H.
The other thing that has to happen in a thermal chemical equation,
is you would have to always include the states, whether it be solid, liquid,
gas, or aqueous, you always put that.
In parenthesis after substance.
This equation obviously would make no sense if you left those off,
you've got H2O going H2O.
It is the state that makes it make sense, but
it's required for all thermochemical equations.
So let's examine.
A therm, thermochemical equation involves a reaction.
This is a reaction to the combustion of methane,
which is the gas in what we call natural gas.
What does this equation tell us?
It tell us, tells us that if we react 2 mols of the methane, the CH4.
With two, one mole of that.
With 2 mols of the oxygen, we would produce 1 mol of carbon monoxide,
we would produce 2 mols of liquid water, and we would give off.
It's exothermic, so we would give off 890.4 kilojoules of heat.
Okay, so what if we had the equation balanced this way?
Now it's balanced, it's just not balanced with the lowest possible whole numbers.
So it's balanced like this, figure out what the delta H for
this reaction would be.
If you said 1780.8 you'd be correct.
You are reversing the reaction, and you are doubling it.
So if you chose number two, you changed the sign, but you forgot to double it.
If you picked a negative 1780.8 you doubled it, but
you forgot to change the sign.
Okay, let's continue to examine this equation,
same one we were looking at before.
Combustion of CH4.
So it tells us that when one mole of CO2's produced,
which is what's in the balanced equation the question is what is entropy change.
Well it be a negative 890.4 according to that balanced equation.
What if we used a number that's not a balanced equation?
Balanced equation says that when two moles,
react you would get this much heat well what if you don't have two moles.
What's the enthalpy change?
Well anytime you're given a thermal chemical equation,
you're giving a relationship between the enthalpy change and
the number of moles of that substance in the balanced equation.
What we can do is set up a, conversion factor for any substance in there.
That is related to the value of the delta H.
So this many kilojoules, would be produced in terms of the substance
we're interested in, when 2 mols of O2 react.
Like I said we can do this for
any sets that you can put down here in the denominator 1 mol of methane.
We could put in the denominator 1 mol of carbon dioxide.
We can put into the denominator 2 mols of water.
Since this question is acting for relationship, of heat for
the oxygen, we're going to use this first one that I wrote.
Don't start with that, start with what's given then we have 2.5 mols of oxygen.
And then do your typical dimensional analysis, I don't want moles of oxygen.
I want kilojoules of heat.
And there are a negative 890.4 kilojoules for every 2 mols of oxygen that react.
That gives to us a value of negative
1,113 kilojoules or.
To two significant figures.
That would be 1.1 times 10 to the 4th kilojoules.
Now this is a little bit more, than what 2 mols would provide.
And you'd expect that.
2 mols gives off,
as you have a minus sign there and there, gives off 890.4 kilojoules.
This is more.
So, I would expect more heat.
And indeed, it is more heat.
Now here it is written out a little more neatly, on your paper.
Just want to re-emphasize that this portion right here,
is obtained from the balance equation.
It is our conversation factor or our unit factor of conversion,
between the amount of a substance and the amount of heat.
That comes from our thermochemical equation.
So let's see if we can work this problem.
It's one that you can try using a similar technique.
But let's point out something.
In this problem I didn't start you out with moles of ammonia, okay?
Ammonia is NH3.
Okay, so this is a substance.
We're going to produce this substance.
Then I started you out with a mols.
I started you with grams.
So if you could, see if you can do dimensional analysis starting with
grams of ammonia, and finishing with energy.
Did you fix number three, pick number 3, then excellent job.
If you did, then you can might zoom ahead to the next slide.
Otherwise, watch how I've done it.
I start with the 1.26 times 10 to the 4th grams of ammonia.
The first place I want to go is to mols of ammonia,
because I know if I can get to mols of ammonia.
I can use the thermochemical equation to convert.
The molar mass is 17.03 grams per mole,
and I can go from moles of ammonia and use my balanced equation to get kilojoules.
The balanced equation has a 2 with the ammonia, and
we put a negative 92.6 kilojoules and that provides with answer number 3.
Okay, here's the next question we're going to work on.
Okay for this problem, we've got some rocket fuel, it's N204 and N2H4 reacting.
And it tells me some information.
It says that when 10 grams of the N2O4 react, so I'll put a 10 underneath here,
we know that we are going to produce, or release a 124 kilojoules.
So the first question is, what's the sign of delta H?
Well, since it is released, it's exothermic.
That is a negative value.
The next question is, well, if 124 kilojoules is
released when 10 grams react, if I wanted to figure out the delta H for
this reaction, which I want to go there eventually.
Will that delta H, for
that chemical equation as balanced be more than 124 kilojoules released?
Or less than 124 kilojoules of heat released?
Well, to figure this out, think about the balanced equation.
Would this amount, in grams be bigger than 10 or less than 10?
What is this amount?
This amount is 1 mol of N2O4.
So, stop and answer the question.
Well, if you said more, you're correct.
N2O4.
Has a molar mass of 92.1.
So if I have 1 mol sitting right here,
I have 92.01 grams of N2O4.
This is far less than that.
So 10 grams released 124.
If I up it to 92.01 grams, it should be way more than 124.
So let's work this part together.
I want to know it for 1 mol.
So I want to start with that.
1 mol of N2O4.
I know the relationship that.
A negative 124 kilojoules will be released for every 10 grams of N2O4.
This information here, gives me that relationship there.
I cannot use that relationship up there in the top right-hand corner,
unless I convert my moles to grams.
So I'll go from mols of N2O4 to grams of N2O4.
I get the molar mass of N2O4 and that is, like I said, 92.01.
That's how many grams are in a mol?
Now I can use the relationship we saw up in the top right-hand corner.
For every 10 grams of this reactant, it produces or
gives off 124 kilojoules of heat.
That will give me a negative 1.14 times 10 to the 3rd kilojoules of heat.
Over 100, over 1,140 kilojoules of heat would be given off,
when the reaction happens for the balanced amounts we see up there.
So it's exothermic.
It's certainly more than 124 kilojoules of heat released.
Because, I have more than 10 grams.
So we've seen lots of example of using thermochemical equations, and
being able to convert, basically, between the amount, of a reactant
or product, and the amount of heat, either given off.
Or needed for that reaction.
So we can convert between those two using as our tool, a thermochemical equation.
The work that we're doing?
Right now, it's difficult for students to get the handle, get a handle on, so
we've got a few more problems just to practice together.
In this problem we're starting with 15 grams of methanol,
and we want to know how much heat would be produced, in kilojoules.
So we're going to start with the 15 grams, of methanol.
We know that the thermochemical equation tells me every time,
2 mols of this reacts, you're going to produce this much heat.
So let's convert the grams to mols.
So we need the molar mass.
Of the methanol.
Which is 32.05.
Now that we have that, we can go from mols of methanol to kilojoules.
And negative 1,452.8 kilojoules is
released every time 2 mols reacts.
And when you multiply and
divide all of this out, you're going to get negative 340 kilojoules of heat.
And while I'm thinking about it, I want to talk about the way they've
written this equation, or the way I've written this equation.
Sometimes you will see that the delta H,
is written as a negative 1452.8 kilojoules.
Some books use it that way, and some use it in kilojoules per mol.
When you see it written as kilojoules per mol, what they are saying is.
It's kilojoules per mol ratio, of what we saw in the balanced equation here.
So it's not per mol of CH3OH.
It's per 2 mols of CH3OH.
It's not per 1 mol of O2.
It's per 3 mols of O2.
So sometimes you'll see it written as kilojoules per mol.
I more commonly.
We'll just write it as kilojoules, knowing that
a thermochemical equation is always balanced in terms of mols themselves.
Okay, now we're going to try to determine the delta H for this reaction.
We don't know it, but we do know that we get a releasing.
Which is the negative 191 kilojoules, of heat every time 100 grams of NO reacts.
We want to know it for the balanced equation, and that's for 2 mols of NO.
Oops, I didn't write mols, I'll just sneak it in here.
Mols of NO.
I want to use this relationship which is in terms of grams.
So I'm going to go from mols of NO,
to grams of NO, and obtain the molar mass of NO.
Which is 30.01.
Then I can go from grams of NO to kilojoules, which is what's being asked.
And I use the relationship that they give me in the problem, that I'm going to for
every 100 grams of NO that reacts, I'm going to release.
A negative 191 kilojules.
So this will give me a negative 115 kilojules and
that would be the delta H for the reaction, because it's the amount of
heat released when 2 mols see what I have here, when 2 mols reacts.
Now, I want you to try another one.
In this problem, I'm giving you that 55.5 grams of ammonia.
Now, that ammonia is NH3, is reacting.
It's producing this amount of heat.
It's released, 'kay?
And then you're going to determine, what is the delta H for this reaction?
Did you pick number 2?
Then you are correct.
If you picked number,
you didn't keep in mind that the energy is released so it's a negative sign.
And if you picked a negative 225, you've determined it for 1 mol, and
it wants it for 4 mols, so we've got to make sure that we take that in to account.
If you got that right, you have, feel, feel comfortable about it,
know that this is the end of our learning objective number four,
where we're utilizing our thermal chemical equations to do calculations, and
seeing what kind of information we can obtain from them.
And how they get the values for the delta H,
by knowing a relationship between energy and amount.
If you did not get this one right and you want to see it again,
you'll want to continue watching, because I'll lay it all out for you here.
So we're trying to determine the delta H for 4 mols.
So I have 4 mols of NH3 and it's exactly 4 mols,
I'm not going to limit myself to one significant figure in my answer.
I'm going to go from mols of NH3 to grams of NH3.
I'm going to do this, which was the molar mass of 17.0 for grams per mol.
I'm going to do this because in the balanced equation,
they are giving me a relationship between the amount of ammonia in grams,
so I can put 55.5 grams of NH3 here.
And the amount of heat released, which is a negative 734 kilojoules of heat
in this spot, and that is going to give us the value of a negative 901 kilojoules.
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