In learning objective number five, we're going to be be combining work that we did in previous learning objectives. We've learned about the first law of thermodynamics, and we've learned about thermochemical equations. Let's bring them together. From the first law of thermodynamics we know that the change in internal energy, is heat plus work. Heat, is that transfer of thermal energy. And work we learned about as negative p delta v. So if we are undergoing this change of heat, whatever the process is, under constant pressure environment. Then we can replace the q with a delta H, because heat is enthalpy change if it's under constant pressure conditions. So now we have a relationship between delta H and delta E. So, if we are doing a constant pressure reaction, we can determine delta E. Let's determine the delta h for this reaction that we've seen here. Now, let's see if you can pull forward what you learned from a previous section. The thermal chemical equation is provided for the amounts we see here. I want to know the amount of heat released when one gram of sodium reacts. If you got a negative 8.0 kilojoules, you're correct and you might want to skip ahead to the next slide. If you didn't get a negative 8 I'll remind you of how you work this problem. We start with the one gram of sodium. We convert from grams to moles of sodium using the molar mass, which is 23. We use the thermochemical equation to go from moles of sodium to kilojoules. Negative 368.4 for every two moles. Now if you selected number three, you probably forgot to put the two in there. The two is what's in the thermochemical equation. And when you do this work, you're going to get a negative eight. So that's the delta H and that's the heat released for this little quantity. Let's look at this situation here. When you drop that little piece of sodium metal. So I'm going to put a one gram sample, of sodium into a glass of water and allow it to react. And it's in this chamber with a movable piston that can go up and down and it's being pushed on by a constant pressure. What's going to happen as soon as I put that sodium in there is it's going to start generating hydrogen gas. And that hydrogen gas is going to expand in this container, and of course increase the volume. So the delta v is going to be greater than zero. It's going to expand as we do that. So, when you place the water, as the sodium in the water, we know that this reaction is going to take place. We know we going to give off heat, as this happens. We know we going to generate some gas. And so we can utilize the fact that delta E equals delta H minus P delta V to calculate delta E. We've got the delta H for the quantity given here, and we could plug that in. Now, if we wanted to know it for one gram, that would be a little bit different, but if we did it for the balanced equation, that's what we have there. And we could figure out if we know how much it expanded. Okay? So we knew how much it grew there, the delta v, and we knew what pressure was being pushed down, we could get the work term. And that would be all we'd need in order to calculate delta E. If we take that equation. And in the reaction we're looking at there is no change to the number of moles of gas. Then we know we not going to be expanding. If we're not expanding then this term is equal to zero. And so what can we say about delta E? Well we can say delta E equals delta H. Anytime you've got a reaction, and you're not changing the number of moles of gas, and you can not have an expansion or contraction therefore, delta E equals delta H. But the fact of the matter is, that even if you do have an expansion, a gas being produced, or a contraction, and it's getting smaller because you're consuming gas. That second term, it's much smaller than the delta H term. So delta E is always close to delta H. Now you may see as I go through this chapter, and later in thermodynamics in the advanced chemistry class, that I will use interchangeably the change in energy and the change in heat or the change in enthalpy. I'll use those terms, because they are, the change of heat is the change in enthalpy, or heat is enthalpy change. As long as it's under constant pressure conditions. And delta H, and delta E, are either exactly the same, or they're very close. Now we can do a little substitution and sometimes this is convenient for us. We could use PV equals nRT. Up there in that equation and substitute the P delta V for RT delta n. Because why would the volume be changing? It'd be changing because you'd be changing the number of moles and this would be of gas. It wouldn't matter about the solids and liquids because they wouldn't change the volume. Cause you look at the change in the moles of gas you could determine that work term without actually having to calculate volume. So if we examine this reaction here and we're asked to calculate the delta e for one mole of this. And reacts with two moles of oxygen according to this equation. The first thing I would do, is I would look at the equation and say have I changed the number of moles of gas. Well on the left hand side of the equation I have one mole for the methane and two for the oxygen, so I have three moles of gas. And on the righthand side, I have one mole of CO2 and two moles of water gas, so I have three moles. So Delta N is equal to zero, Delta V would be equal to zero. Either way you want to think about it, so work is equal to zero. So if I wanted to know the change in internal energy for this reaction. Delta E would simply be equal to delta H and that is a negative 890.4 kilojoules. Now if it weren't, we would have to add that term, plus RT delta n. I'd have to give you a temperature, which I didn't give you here. But I have to give you a temperature. You watch your units and make sure that this term ends up being in kilojoules as well. Now we see a new reaction, a new delta h given for this reaction so we have a thermal chemical equation provided and we're asked to do four different things for this reaction. We're going to determine the heat released when five grams of hydrogen reacts with 20 grams of chlorine. We're going to determine the work done against a pressure of one atmosphere and 25 degrees celsius. We're going to calculate the Delta E for the reaction for the amounts reacted. These five grams and 20 grams. And we're going to calculate the delta e for the equation as written. Which means for one mole of this reacting with one mole of that. So let's answer this question first. What type of problem gives measure information about both reactants? Well that is a problem, that is a limiting reactant problem. So we have to work the problem a certain way. When you work a limiting reactant problem, you should do which of the following? Well, it'd be number one. This is one way, at least, to do a limiting reactant problem is you could determine which produces a smaller amount of product. And in this case, the product is heat. So, we can work each problem now. We can work this problem. So we can work this problem starting with part one. What is the heat released by the reaction if five grams of hydrogen reacts with twenty grams of chlorine? So the balanced equation tells me when one mole reacts with one mole we product this. But we're not starting with one mole. Let's start with the five grams of H2. And let's figure out the number of moles of, well let's go all the way to the amount of heat produced for those number of moles. So we go from grams to hydrogen to moles of hydrogen. We use the molar mass of hydrogen, which is 2.02, and then we can go from moles hydrogen to kJ using kJ, forget that, using the balanced equation. So a negative 184.6 is released every time one mole reacts. We can then take the 20.0 grams of chlorine and go to moles of chlorine. And 35.45 times 2 would be 70.9 grams. And then we can from moles of chlorine to kilojoules. Okay. So, which one is going to give us the smaller amount? Well here we have 5 divided by 2 which is bigger than 1 and we have 20 divided by 70 which is less than 1. This is going to be the smaller number down here on the bottom. But let S get actual numbers for that. The bottom one is a negative 52.1 kilojoules. I know that's my limit, but to verify it for you, let's plug in the other numbers. The numbers up on the top equation gives me a negative 457. So this is producing if in, all of the hydrogen could react, it would produce negative 457. If all of that chlorine reacts, we're going to give off 52. This is a smaller amount, so this is our answer to number one. Number two. What is the work done against the pressure of one atmosphere. Well, don't go calculating work until you look at your equation and see if your moles of gas have changed. We start with a balanced equation, two moles produce two moles so the gas moles is not changing in the balanced equation, Therefore. Whether you react .1 mole or 4,000 moles of hydrogen, it will only be able to react with the exact same amount of chlorine to produce. You're never going to change the moles of gas, no matter what you're starting with, even though it's a limited reactant problem. So part two, the work done, is 0 joules. You cannot produce work. So what is a delta E for the reaction for the amount that reacted? Well, if there's no work term, then delta E is equal to delta H and we have a negative 52.1 kilojoules as your change in internal energy. What is the delta E for the reaction as balanced? Well, if we look up there at the reaction as balanced,. We know that delta E equals delta H and that's a negative 184.6 kilojoules. Even if a little bit of gas were produced or consumed in the course of this reaction overall. The delta E would be very close to the delta H. Okay, so now we've seen thermochemical equations and we've applied it to the first law of thermodynamics.
