- Thermochemistry questions
- Phase diagrams
- Heat of formation
- Hess's law and reaction enthalpy change
- Gibbs free energy and spontaneity
- Gibbs free energy example
- More rigorous Gibbs free energy / spontaneity relationship
- A look at a seductive but wrong Gibbs spontaneity proof
- Endothermic vs. exothermic reactions
Standard heat of formation or standard enthalpy change of formation. How heats of formation are calculated. Created by Sal Khan.
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- what does 'enthalpy' mean in the bigger picture? What does it do for the system?(22 votes)
- Enthalpy is similar to energy. It can be thought of as energy in most cases. In all honesty, enthalpy is a broader term than energy because it accounts for pressure and volume in addition to all that energy accounts for. Its a state function of the system. Thus, it doesn't "do" anything. Its just a quantity in a system.(41 votes)
- how is energy released? i mean in exothermic.. heat is given out ? how?(14 votes)
- is the heat of formation the same thing as the heat of reaction?(8 votes)
- sorry but i would like to quote that heat of formation is only equal to heat of reaction in cases where compounds are formed directly from its elements.but when comounds are formed from compounds there is heat of reaction but is it heat of formation i guess no it is equal to differnce in hp-hr(9 votes)
- Does anyone else have suggestions for other enthalpy/calorimetry videos and practice problems? I found this to be even more confusing, and I still don't understand it.(8 votes)
- What is difference between heat and temperature?(3 votes)
- Temperature is the average kinetic energy of the particles of the system whereas
Heat is the quantity of energy that flows across the boundary of a system during a change in its state due to the difference in temperature between the system and the surroundings.(8 votes)
- I would like to ask why the heat when released is 74kJ and not -74kJ, I thought when heat is released, it would be an exothermic process, so shouldn't it be negative?(5 votes)
- deltaH is -74kJ but when you write that in a equation in the left side it is written as +74kJ as it is a product of the reaction "heat is evolved "(2 votes)
- How do you tell whether an equation is endothermic or exothermic? For example what would C2H4 yields 2C t 2H2 + 52.3 kJ(2 votes)
- In the equation q=mc∆T , is q equal to ∆H?(2 votes)
- Yes, if the experiment is done under conditions of constant pressure. If the pressure is not held constant, then q is not equal to ∆H.(3 votes)
- Sal says 1/2 mole of O2. Is that possible? Can you break a n atom or a molecule into two and consider it as two equal parts?(2 votes)
- Molecular oxygen is 2 oxygen atoms combined into 1 molecule. In chemical reactions molecules can be split into parts and recombined, this is basically what chemical reactions are.
Also keep in mind that a 1 mole of O2 is 6 * 10^23 molecules of O2 so there is no difficulty of getting a 1/2 a mole which would be 3 * 10^23 molecules of O2.(3 votes)
- at2:20, he mentions that 74 kJ of heat are being produced. How do we know that? Is this just a random number he picked or is there an actual formula?(1 vote)
We saw in the last video that if we defined enthalpy, H, as being equal to the internal energy of a system plus the pressure of the system times the volume of the system-- and this is an almost arbitrary definition. But we know that this is a valid state variable. That no matter what you do in terms of how you get there, you're always going to have the same value. Because it's the sum and product of other valid state variables. But this by itself isn't that useful or that intuitive. But we saw in the last video that if you assume constant pressure-- and that's a big assumption, but it's not an unreasonable assumption for most chemical reactions. Because most chemical reactions, you know, we're sitting at the beach with our beakers, and they're exposed to just standard temperature and pressure, or at least some pressure, that's not changing as the reaction occurs. If we assume constant pressure, we saw that the change in enthalpy becomes the heat added to the system at that constant pressure. That P there is just to show you that hey, this is assuming that we're dealing with heat being added at a constant pressure. Fair enough. So how can we use these concepts in any useful way? Let's say that I had some carbon in its elemental form as graphite, and I add to that-- say I have a mole of carbon. And I add to that 2 moles of hydrogen in its elemental form. It's going to be a gas, and it's going to be as a molecule, right? If I just have a bunch of hydrogen in its gaseous state, let's say, in a balloon, I'm not going to have individual atoms of hydrogen. They're going to bond and form these diatomic molecules. And if I react them, I'm going to produce a mole of methane. CH4. But that's not all I'm going to produce. I'm also going to produce some heat. I'm going to produce 74 kilojoules of heat-- plus 74 kilojoules of heat-- when I produce that one mole. I'll do a lower case k for the kilo. When I produce that one mole of methane. So what's happening here? So first of all, how much heat is being added to the system? And let's assume that this heat just gets released from the system. That this isn't an adiabatic process. I haven't insulated the system from anything. That this just gets released. It just goes away. It gets released. So my question is, how much-- you know, I started off with this container, I guess we could call it, that's at a standard, fixed pressure. And maybe I had a bunch of-- well, I wanted to do the carbon. I'll do it in gray. I have a bunch of solid carbon lying around. Maybe some type of dust. And then I have some molecular hydrogen gas. Each of those dots is actually two hydrogen atoms. And then, I don't know, maybe I shake it up or something to make them react, and then I get a bunch of methane gas. I'll do that in green. So now I just have a bunch of methane gas, and I released 74 kilojoules. So how much heat was added to the system? Well, we released heat from the system. We released 74 kilojoules. So the heat added to the system was minus 74 kilojoules. Right? If I asked you the heat released, then I would have said 74. But remember, we care about the heat added to the system, is 74 kilojoules. And I just showed you that that's the exact same thing as the change in enthalpy. So how can we think about this? What is the enthalpy of this system relative to this system? Well, it's going to be lower, right? Because if you take enthalpy-- so the change in enthalpy is the enthalpy of your final system, minus the enthalpy of your initial system. And we got a negative number. We got minus 74 kilojoules. So this has to be lower than this by 74 kilojoules. So this enthalpy right here is less than this enthalpy right here. So if we were to actually draw it on a diagram, if I were to draw the reaction-- let's say that this is just time or something, as the reaction proceeds, that axis. And on the y-axis, I'll draw enthalpy. So the reaction starts off at your initial enthalpy, Hi and that's this state right here. So you start there. I'll do it in the yellow of that container. You start there. And then, I don't know, you shake it up, or I'm not going to go into the activation energies, so it might have little hump and all of that, but who knows. But then we end up at our final enthalpy. We have this final enthalpy, right here, after the reaction has occured. That's this state right here. This is H final. So you can see, you've had this drop off in enthalpy. And what's interesting here is, is not so much what the absolute value of this enthalpy is here, or what the absolute value of this enthalpy here is. But now that we have enthalpy, we can kind of have a framework for thinking about how much heat energy is in this system relative to this system. And given that there's less heat energy in this system than that energy system, we must have released energy. And you know, to some degree, I told you that from the beginning, right? I told you that energy is released. And the word for this we use is exothermic. Now, if you want to go the other way-- let's say we wanted to go from methane and go back, you'd have to add heat into the reaction. If you wanted to go backwards through this reaction, go upwards, you would have to add that heat content to get that positive delta H, and then you would have an endothermic reaction. So if a reaction releases energy, exothermic. If a reaction needs energy to occur, it's endothermic. Now you might be asking, Sal, where'd that energy come from? So I started at this enthalpy here. And enthalpy has this weird definition right here, and then'll ended up as that other enthalpy here. And as you see, enthalpy, the pressure we're assuming is constant. Let's say the volume isn't changing much in this situation, or maybe doesn't change at all. So most of the change is going to come from the change in internal energy, right? There's some higher internal energy here, and some lower internal energy here that's causing the main drop in enthalpy. And that change in internal energy is really a conversion from some potential energy, up here, into the heat that's released. So there was some heat that was released, 74 kilojoules, and so our internal energy dropped. And what all of this does is, it gives us a framework so that if we know how much heat it takes to form or not form certain products, then we can kind of predict how much heat will either be released, or how much heat will be absorbed by different reactions. And so here I'm going to touch on another notion. The notion of heat of formation, or sometimes it's change in enthalpy of formation. So the way they talk about it is, the change in enthalpy of formation. And it's normally given at some standard temperature and pressure. So you put a little, usually it's a naught, sometimes it's just a circle in there. And what that is is, what is the change in enthalpy to get to some molecule from its elemental form? So for example, if we want it for methane-- if we have methane there, and we want to figure out its heat of formation, we say, look, if we form methane from its elemental forms, what is the delta H of that reaction? Well, we just learned what the delta H of that reaction was. It was minus 74 kilojoules. Which means that if you form methane from its elemental, I guess, building blocks, you're going to release 74 kilojoules of energy. That this is an exothermic reaction. Because you've released heat, you can say that the methane is in a lower energy state, or it has a lower potential energy, than these guys did. And because it has lower potential energy, it's more stable. I mean, one way to think of it is, if you have a guy, you have a mountain here, and down here, you have a ball. And this isn't, you know, a complete, direct analogy. But the analogy to potential energy is, look. When you're down at a lower potential energy state, you tend to be more stable. And so, in the everyday world, if you have a bunch of methane sitting around, the fact that it has a negative heat of formation, or a standard heat of formation, because I have that naught here, or a negative standard change in enthalpy of formation-- those are all the same things-- tells me that methane is stable relative to its constituent compounds. And actually, you can look these things up. You don't have to memorize them, but it's good to know what they are. And I copied all of this stuff-- actually, let me get the actual tables from Wikipedia down here. I copied all of these directly from Wikipedia. These give you the standard heat of formation of a bunch of things. And if you look down here, for-- let's see if they have methane-- right there, this is what we were dealing with. They're telling us essentially the delta H of the reaction that forms methane. This point table right there tells us that if we start off with some carbon in a solid state, plus two moles of hydrogen in a gaseous state, and we form one mole of methane, that if you take the enthalpy here minus the enthalpy here-- so the change in enthalpy for this reaction-- at standard temperature and pressure, is going to be equal to minus 74 kilojoules. Per mole. And this is all given per mole. So if you have a mole of this, two moles of this, to form one mole of methane, you're going to release 74 kilojoules of heat. So this is a stable reaction. Now there's a couple of interesting things here, and we'll keep using this table over the next few videos. You see here, monoatomic oxygen has a positive standard heat of formation. Which means it takes energy to form it. Right? That if you have a reaction, let me just say, the reaction, I'll write it this way. One half of molecular oxygen as a gas to go to one mole of oxygen in its gaseous state. This tells us that this state has more potential than this state. And in order for this reaction to occur, you have to add energy to it. You have to put the energy on the other side. So you'd have to say, plus 249 joules. So you might say, hey, Sal, that doesn't make sense. Oxygen is just oxygen. Why is there a heat of formation of oxygen? And that's because you always use the elemental form as your reference point. So oxygen, if you just had a bunch of oxygen sitting around, it's going to be in the O2 form. If you have a bunch of hydrogen, it's going to be H2. If you have a bunch of nitrogen, it's going to be N2. Carbon, on the other hand, is just C, and it tends to be in its solid form as graphite. So all heats of formation are relative to the form that you find that element when you have a pure version of it. Not necessarily its atomic form, although sometimes it is its atomic form. Now, in the next video, we're going to use this table, which is a very handy table-- I cut and pasted parts of it-- to actually solve problems. In this last video, I gave you the heat of formation, and we just thought about it a little bit. In the next few videos, we're going to use this table that gives us standard heats of formation to actually figure out whether reactions are endothermic-- meaning they absorb energy-- or exothermic-- meaning they release energy-- and we'll figure out how much.