0:00

In this learning objective we're going to be learning about

Â Thermochemical Equations.

Â It's like a chemical equation only it gives you the information,

Â associated with delta H for the reaction.

Â 0:13

So a thermochemical equation, shows the entropy changes for the reaction, so

Â there's a delta H associated with it.

Â As well as the mass relationship, so

Â if you were stoichiometry that's the balanced equation it is showing you.

Â 0:26

So here's an example of a thermochemical equation.

Â Now this is not a chemical reaction, it's a physical change.

Â We're going from water solid, which is ice, to liquid, so

Â it's the melting process.

Â This is an endothermic process.

Â I see that because I have a positive.

Â Value for delta H,

Â but you have to put in heat to break those attractions between the molecules.

Â 1:27

To work with these thermochemical equations,

Â there's several guidelines that you've got to keep in mind as you work and use them.

Â The first one is that when you see those coefficients of a balanced equation,

Â they're always in terms of moles of the substance.

Â So you would never turn, think of them in terms of individual molecules.

Â Or as we did in gas laws think about em in

Â terms of leaders you must think of em in terms of moles.

Â 1:56

If you reverse the equation and you're, you're going to flip it around and

Â make the products the reactants and

Â reactants the products, then you'll change the sign at the delta H.

Â So if you're going to take a liquid and

Â convert it to a solid, it is going to be an exothermic process.

Â It is going to give off 6.01 kilojules.

Â So you freeze water by sticking it in the freezer, and

Â the freezer is a pump to pull the heat out of the water.

Â 2:29

The next one should make sense for us.

Â If we double a reaction, we would double the amount of heat.

Â So here we're going back from a solid to a liquid, we're melting it.

Â But, we're melting twice as much, and

Â so the delta H would be twice the 6.01, or it's 12.02.

Â 2:54

The other thing that has to happen in a thermal chemical equation,

Â is you would have to always include the states, whether it be solid, liquid,

Â gas, or aqueous, you always put that.

Â In parenthesis after substance.

Â This equation obviously would make no sense if you left those off,

Â you've got H2O going H2O.

Â It is the state that makes it make sense, but

Â it's required for all thermochemical equations.

Â 3:27

A therm, thermochemical equation involves a reaction.

Â This is a reaction to the combustion of methane,

Â which is the gas in what we call natural gas.

Â What does this equation tell us?

Â It tell us, tells us that if we react 2 mols of the methane, the CH4.

Â 3:48

With two, one mole of that.

Â With 2 mols of the oxygen, we would produce 1 mol of carbon monoxide,

Â we would produce 2 mols of liquid water, and we would give off.

Â It's exothermic, so we would give off 890.4 kilojoules of heat.

Â Okay, so what if we had the equation balanced this way?

Â Now it's balanced, it's just not balanced with the lowest possible whole numbers.

Â So it's balanced like this, figure out what the delta H for

Â this reaction would be.

Â 4:30

If you said 1780.8 you'd be correct.

Â You are reversing the reaction, and you are doubling it.

Â So if you chose number two, you changed the sign, but you forgot to double it.

Â If you picked a negative 1780.8 you doubled it, but

Â you forgot to change the sign.

Â 4:53

Okay, let's continue to examine this equation,

Â same one we were looking at before.

Â Combustion of CH4.

Â So it tells us that when one mole of CO2's produced,

Â which is what's in the balanced equation the question is what is entropy change.

Â Well it be a negative 890.4 according to that balanced equation.

Â 5:15

What if we used a number that's not a balanced equation?

Â Balanced equation says that when two moles,

Â react you would get this much heat well what if you don't have two moles.

Â What's the enthalpy change?

Â 5:29

Well anytime you're given a thermal chemical equation,

Â you're giving a relationship between the enthalpy change and

Â the number of moles of that substance in the balanced equation.

Â What we can do is set up a, conversion factor for any substance in there.

Â That is related to the value of the delta H.

Â So this many kilojoules, would be produced in terms of the substance

Â we're interested in, when 2 mols of O2 react.

Â Like I said we can do this for

Â any sets that you can put down here in the denominator 1 mol of methane.

Â We could put in the denominator 1 mol of carbon dioxide.

Â We can put into the denominator 2 mols of water.

Â Since this question is acting for relationship, of heat for

Â the oxygen, we're going to use this first one that I wrote.

Â Don't start with that, start with what's given then we have 2.5 mols of oxygen.

Â And then do your typical dimensional analysis, I don't want moles of oxygen.

Â I want kilojoules of heat.

Â And there are a negative 890.4 kilojoules for every 2 mols of oxygen that react.

Â That gives to us a value of negative

Â 1,113 kilojoules or.

Â To two significant figures.

Â That would be 1.1 times 10 to the 4th kilojoules.

Â Now this is a little bit more, than what 2 mols would provide.

Â And you'd expect that.

Â 2 mols gives off,

Â as you have a minus sign there and there, gives off 890.4 kilojoules.

Â This is more.

Â So, I would expect more heat.

Â And indeed, it is more heat.

Â 7:25

Now here it is written out a little more neatly, on your paper.

Â Just want to re-emphasize that this portion right here,

Â is obtained from the balance equation.

Â It is our conversation factor or our unit factor of conversion,

Â between the amount of a substance and the amount of heat.

Â That comes from our thermochemical equation.

Â 7:52

It's one that you can try using a similar technique.

Â But let's point out something.

Â In this problem I didn't start you out with moles of ammonia, okay?

Â Ammonia is NH3.

Â Okay, so this is a substance.

Â We're going to produce this substance.

Â Then I started you out with a mols.

Â I started you with grams.

Â So if you could, see if you can do dimensional analysis starting with

Â grams of ammonia, and finishing with energy.

Â 8:26

Did you fix number three, pick number 3, then excellent job.

Â If you did, then you can might zoom ahead to the next slide.

Â Otherwise, watch how I've done it.

Â I start with the 1.26 times 10 to the 4th grams of ammonia.

Â 8:45

The first place I want to go is to mols of ammonia,

Â because I know if I can get to mols of ammonia.

Â I can use the thermochemical equation to convert.

Â The molar mass is 17.03 grams per mole,

Â and I can go from moles of ammonia and use my balanced equation to get kilojoules.

Â 9:24

Okay for this problem, we've got some rocket fuel, it's N204 and N2H4 reacting.

Â And it tells me some information.

Â It says that when 10 grams of the N2O4 react, so I'll put a 10 underneath here,

Â 10:01

The next question is, well, if 124 kilojoules is

Â released when 10 grams react, if I wanted to figure out the delta H for

Â this reaction, which I want to go there eventually.

Â Will that delta H, for

Â that chemical equation as balanced be more than 124 kilojoules released?

Â Or less than 124 kilojoules of heat released?

Â 10:50

Well, if you said more, you're correct.

Â N2O4.

Â Has a molar mass of 92.1.

Â So if I have 1 mol sitting right here,

Â I have 92.01 grams of N2O4.

Â This is far less than that.

Â So 10 grams released 124.

Â If I up it to 92.01 grams, it should be way more than 124.

Â 11:46

This information here, gives me that relationship there.

Â I cannot use that relationship up there in the top right-hand corner,

Â unless I convert my moles to grams.

Â So I'll go from mols of N2O4 to grams of N2O4.

Â I get the molar mass of N2O4 and that is, like I said, 92.01.

Â That's how many grams are in a mol?

Â 13:10

or product, and the amount of heat, either given off.

Â Or needed for that reaction.

Â So we can convert between those two using as our tool, a thermochemical equation.

Â 13:35

The work that we're doing?

Â Right now, it's difficult for students to get the handle, get a handle on, so

Â we've got a few more problems just to practice together.

Â In this problem we're starting with 15 grams of methanol,

Â and we want to know how much heat would be produced, in kilojoules.

Â 14:14

So we need the molar mass.

Â Of the methanol.

Â Which is 32.05.

Â Now that we have that, we can go from mols of methanol to kilojoules.

Â And negative 1,452.8 kilojoules is

Â released every time 2 mols reacts.

Â 14:52

And while I'm thinking about it, I want to talk about the way they've

Â written this equation, or the way I've written this equation.

Â Sometimes you will see that the delta H,

Â is written as a negative 1452.8 kilojoules.

Â Some books use it that way, and some use it in kilojoules per mol.

Â When you see it written as kilojoules per mol, what they are saying is.

Â It's kilojoules per mol ratio, of what we saw in the balanced equation here.

Â So it's not per mol of CH3OH.

Â It's per 2 mols of CH3OH.

Â It's not per 1 mol of O2.

Â It's per 3 mols of O2.

Â So sometimes you'll see it written as kilojoules per mol.

Â I more commonly.

Â We'll just write it as kilojoules, knowing that

Â a thermochemical equation is always balanced in terms of mols themselves.

Â 15:45

Okay, now we're going to try to determine the delta H for this reaction.

Â We don't know it, but we do know that we get a releasing.

Â Which is the negative 191 kilojoules, of heat every time 100 grams of NO reacts.

Â 16:03

We want to know it for the balanced equation, and that's for 2 mols of NO.

Â Oops, I didn't write mols, I'll just sneak it in here.

Â Mols of NO.

Â I want to use this relationship which is in terms of grams.

Â So I'm going to go from mols of NO,

Â to grams of NO, and obtain the molar mass of NO.

Â Which is 30.01.

Â Then I can go from grams of NO to kilojoules, which is what's being asked.

Â 16:45

A negative 191 kilojules.

Â So this will give me a negative 115 kilojules and

Â that would be the delta H for the reaction, because it's the amount of

Â heat released when 2 mols see what I have here, when 2 mols reacts.

Â 17:02

Now, I want you to try another one.

Â In this problem, I'm giving you that 55.5 grams of ammonia.

Â Now, that ammonia is NH3, is reacting.

Â It's producing this amount of heat.

Â It's released, 'kay?

Â And then you're going to determine, what is the delta H for this reaction?

Â 17:51

If you got that right, you have, feel, feel comfortable about it,

Â know that this is the end of our learning objective number four,

Â where we're utilizing our thermal chemical equations to do calculations, and

Â seeing what kind of information we can obtain from them.

Â And how they get the values for the delta H,

Â by knowing a relationship between energy and amount.

Â If you did not get this one right and you want to see it again,

Â you'll want to continue watching, because I'll lay it all out for you here.

Â 18:22

So we're trying to determine the delta H for 4 mols.

Â So I have 4 mols of NH3 and it's exactly 4 mols,

Â I'm not going to limit myself to one significant figure in my answer.

Â I'm going to go from mols of NH3 to grams of NH3.

Â I'm going to do this, which was the molar mass of 17.0 for grams per mol.

Â I'm going to do this because in the balanced equation,

Â they are giving me a relationship between the amount of ammonia in grams,

Â so I can put 55.5 grams of NH3 here.

Â