If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

More rigorous Gibbs free energy / spontaneity relationship

More formal understanding of why a negative change in Gibbs Free Energy implies a spontaneous, irreversible reaction. Created by Sal Khan.

Want to join the conversation?

  • blobby green style avatar for user Jamie Martin
    Wouldn't the irreversible process need more heat than the reversible process, as the heat taken up in the irreversible process will be used not just to move the piston, but also to move the piston against the surface, creating the friction, therefore using more heat?
    (27 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user sindarus
      I think the mass that is left on top of the second piston is less than the one left on top of the first one, that's the only way they can end up at the same height, otherwise, with friction, the right piston would stop before reaching the same height, unless the right reservoir had a higher temperature than in the left one, and then it WOULD need more heat, like you said. But it's not the case since they have the same temperature.

      rigorous proof : the equilibrium is reached when the upper part of the piston is not moving, so, when the sum of the forces it undergoes are zero (i.e. when it is pushed down as much as it's pushed up)
      Let's consider the right piston, with friction : the forces it undergoes are : pressure from atmosphere, weight of the mass, friction and pressure of the gas inside.
      we have mg + Pext/A + F = P1/A (where m is the mass on top on the piston, Pext is the external pressure, A is the top surface of the piston, F is the force caused by friction, if we consider it as a dry friction, P1 is the pressure inside the piston) we then have m = P1/gA - F/g - Pext/gA <=> m = nRT/VgA - Fg - Pext/A (because P1 = nRT/V)
      now, n is constant, because the gaz is closed inside the piston, T is constant because it's isothermic process, V is fixed because we want to reach a certain given height. So, if F is null i.e. there's no friction, there is a certain mass m1. And it F is non null i.e. there's friction, m2 is smaller than m1. I.e : in the left piston, the mass left on top of the piston is smaller !
      If I am not mistaken :p
      (6 votes)
  • blobby green style avatar for user ajkalenkiewicz
    At Sal says that for a spontaneous process, deltaS > 0. But what about when supercooled water freezes?
    (9 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Runpeng Liu
      yes, that's it. spontaneity depends on dS and dH. at temperatures below the freezing point of water, then dH factor contributes more entropy to the surroundings than the entropy lost due to solidifying the liquid-thus, the net entropy change in the universe is positive and process is spontaneous
      (17 votes)
  • leaf green style avatar for user Shred
    change in enthalpy is equal to Q at a constant pressure but the example shows a PV graph where both P and V change. thus, pressure is not constant and enthalpy is not equal to Q. or am I missing something?
    (11 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user sharpless.charlie
      The details get messy, but there is a simple way to think about this. H is a state function, so it does not matter how you get from point A to point B, (delta)H is the same. This process can be carried out in two steps: 1) add heat at constant pressure, the gas expands, you move left to right on the PV diagram, and the temperature inside the piston goes up; 2) carry out an "adiabatic" expansion after lowering the external pressure, no heat is exchanged, the volume increases, the interior gas loses energy via the expansion work, and the internal temperature drops (back to its starting value... ideally). In this two step process, the heat is all exchanged in the first step at constant pressure, so the enthalpy change is clearly well defined as Q at constant P. The second step gets us to the final state without any change in H. Not a rigorous proof, but a simple one, to show that the enthalpy change is (or can be) a well defined quantity with clear relation to Q(P) in this system.
      (11 votes)
  • leaf blue style avatar for user Muhammad Ali
    i keep on feeling that the work done by the irreversible process should be less but when i think about how in the PV diagram the start and end point of both graphs for reversible and irreversible processes are the same they make me wonder that the piston must have moved up to the same height in the irreversible process as well so would that mean both the reversible and irreversible processes did the same amount of work as they displaced the piston by the same change in volume? but just thinking that they would do the same amount of work gives me an uneasy feeling
    can some one help me?
    (7 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user sindarus
      here's how i see it : they did the same amount of work, but in the right piston experiment, some of the work was used to counter the friction. The remaining work was used to move the mass to the top. This mass being lower than the one on the piston on the left, as I explained in my answer to Jamie Martin.
      (2 votes)
  • leaf blue style avatar for user Diep Nguyen
    Sal, at you state that delta SiR is = to QR/T1 = delta SR because they move to the same spot on the PV plot, and that QiR/T1 is less than QR/T1, but, how can this be? If QiR/T1 is less than QR/T1 and Qfriction is generated, this means that the engine would perform less work. They would not be at the spot on the PV diagram.
    (4 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Sidharth Gat
    why is it assumed that every irreversible process is spontaneous?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Anarchasis
    In one of Sal's earlier videos, we've been told that Reversible = quasi-static AND frictionless. OK. This in turn means that Irreversible = not Quasi-static OR with friction

    But then in the "More rigorous Gibbs free energy..." at , Sal explains that BECAUSE the process is irreversible, there is friction, and that this friction generates heat. Does this mean that irreversible = with friction ?
    If not, let us suppose it is frictionless, but still irreversible BECAUSE not quasi-static.
    How can we show that Q will still be less that in the reversible case ?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user liquigital
      Notice that the external pressure is "modelled" by weights on the roof of the canister. When you remove pebbles (in the reversible case) the system lifts the roof just a little, and has to perform work against almost all the remaining pebbles, thereby supplying them with potential energy.

      Since there is equilibrium throughout, the external matches the internal pressure, and that is why you can plot the work (that what you do against the external pressure) in the state diagram of the system (plotting the internal (!) pressure).

      In constrast, if you remove the heavy weight, the system has to "lift up" a less heavy reamining weight until a pressure balance is reached. Think of the removed weights in either case as being removed horizontally ("through the wall") onto a shelf of corresponding height, next to the canister. You see that you had to transfer more internal energy into potential energy alongside this process in the revesible case, since the pebbles' height on these shelfs surmounts in total the one of the two big weights. That is why you have to draw less heat from the reservoir to compensate for that, in the irreversible case. That is true even without friction. The videos are unfortunately not terribly clear about that connection.
      (4 votes)
  • blobby green style avatar for user Peter Xu
    How come at about , Sal doesn't have to take in the Q_f the friction heat when determining the delta S_universe for the irreversible process?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Amruta Bonde
    if enthalpy change is positive and entopy change is negative then what is the condition for the spontanity of a process?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Prabal Jain
    at how can we we lose temperature in absence of reservoir? i did not get it.!
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

I've hopefully given you a bit of a gut feeling behind where the formula of Gibbs Free Energy comes from. In this video I want to do something a little bit more rigorous and actually, I guess you could say derive the formula. So to do that, let's just study two systems that have the exact same change in entropy. To depict that, I'll get out our handy PV diagram. What we're going to do is we're going to compare two systems. One that's this perfect, reversible system and one that's irreversible or a spontaneous system. So they're both going to start here. They're going to go from this point on the PV diagram to this point right here. And before we go into exactly what they're doing, I think it's a good review to kind of talk about what an irreversible process and what a reversible process really is. An irreversible process is this theoretical thing where you have no friction, where you're always so close to equilibrium that you can always go backwards. You're always-- you can kind of view that the reaction is never really proceeding forward or backwards. Although obviously it is moving, so it's a very-- it actually doesn't really exist in nature, but it's a useful theoretical construct. So let me draw a little. So if this is a-- I'll get the good old pistons out. So let's say this is an-- I'll do it twice. So that's my irreversible piston-- That'll be my reversible, this is going to be my irreversible. Let me draw them. OK, and let me label them. So this right here is going to be reversible. And this right here is going to be irreversible. Or it could also be considered spontaneous. Using the key word. That's going to matter for Gibbs Free Energy. But all spontaneous reactions are irreversible. So they're going to go from this state to this state. So in our reversible world, we have this little cap to our piston. And we have our gas in here, exerting some pressure. And what we'd have is a bunch of pebbles. We have a bunch of pebbles out here. And we slowly remove the pebbles one by one. And as we remove the pebbles, our piston-- or this movable ceiling up here-- will move upwards. It'll move upwards. So the pressure will push up, but as we move up, we will have lower and lower pressure because the gases will bounce into the surface less. And we'll have a higher, higher volume. So as we move each infinitesimally small grain of sand, and we're going to do it super slow, that we're always infinitely close to equilibrium. We're going to move from this state to this state. And even better, let's view that this is the first stage of our Carnot cycle. Let's say in either case, we're on top of a reservoir. So in either case, both of these systems are on top of this infinite reservoir that has a temperature T1. And what that does it keeps the temperature constant. So we're going to travel along an isotherm. Because normally if we were to remove these things and if we just allowed it to occur adiabatically, we would actually lose temperature. We would actually lose average kinetic energy as work is done. But in this case we have this reservoir. So heat is just going to be transferred to my system. Heat is going to be transferred. So let me call that QR. Right? If the reservoir wasn't here, our temperature would go down. But since we do have the reservoir, we will constantly be transferring heat. And we've seen this. This is just the first phase of the Carnot cycle. We'll move along an isotherm like this. This is the reversible case. And the only reason why we can even draw the state at every point here is because reversible processes are quasistatic. They're always infinitely close to equilibrium. And when we say reversible, we're also saying there's no friction between this little piston in the cylinder. That if we put a grain of sand back, it'll go exactly to where it was before. And no energy was lost because there was no heat of friction there. So this is what the graph of the reversible process would look like. Now what does the graph of the error irreversible-- Well, actually no. I won't draw the graph, but let's talk about what the irreversible process is going to be like. So it's going to look similar. It's going to look like that. It's going to have its gases there. But for the sake of an argument, to get from that state to that state, instead of moving the pebbles one by one, let's say I have these big blocks. And when I remove one of these big blocks, I go from that state to that state. But obviously all hell breaks loose. So I'm not really defined in this in between state. But I definitely go from that state to that state once I go back to equilibrium. Now the other key thing in the irreversible process-- and every real process in our world is irreversible-- is that you're going to have friction here. As this moves up, it's going to rub against the side of the container and generate heat of friction. So let me call that heat of friction. So let me ask you a question. If, in this case, Q sub R had to be added to the system to maintain its temperature, what's going to be the Q sub irreversible here? How much has to be added to this system in order to keep it at the constant temperature T1? Will it have to be more or less than what was added to the reversible system? Well this guy, as this piston moves up, he's generating some of his own heat. So if this was an adiabatic process, he wouldn't lose as much temperature as this guy would. So he's going to need less heat from the reservoir in order to maintain his temperature. At T1. In order to get to this point on the isotherm. Remember, this irreversible process, we don't know what happens over here. He might be travelling on some crazy path. In fact, we can't even define the path, because it goes out of equilibrium. So it's going to be some crazy thing. But we know it pops back on the PV diagram right there. But because it's generating some of its own heat from the friction, it's going to need less heat from the reservoir. So let me write that down. The heat absorbed by the reversible process is going to be greater than the heat absorbed by the irreversible process. And that's because the irreversible process is generating friction. Fair enough. Now what is the change in entropy for both systems? Well they both started here. And they both ended here. And entropy is a state variable. So the change in entropy for the reversible process is going to be equal to the change in entropy for the irreversible process. They're both going from there to there. And obviously the entropy has changed. We're going from one state to another state. And the entropy, well I won't go too much into it. But let's ask another question. What is the total change in entropy of the universe for the reversible process? So for the universe, that's going to be-- Our universe here is the reservoir in our system. So let me write here. Reversible. I don't want to run out of space. Let me see, I'll do it in a different color. So this is the reversible process. So the change in entropy of the universe is equal to the change in entropy of our reversible process plus the change in entropy of-- oh, I already used R for reversible, so let's call it the reserv-- well, the first three letters are the same, so let me call it of our environment. Right? And then the reversible process, the change in entropy of our reversible process is the heat added for our reversible process. And we can use this definition because it is a reversible process. It's that over T1. And then what's the change in entropy of our environment? Well it's giving away Q sub R. Right? So its heat absorbed is minus Q sub R. And, of course, it's at a constant temperature. It is a heat reservoir. It's at T1. So it equals 0. It equals 0. Interesting. So actually I should take a little side note here. That, the change in entropy of the universe for a reversible process is 0. And actually that should make a lot of intuitive sense because the whole point of a reversal process is you could go in this direction, or you could go in that direction. It's always so close people to equilibrium you can move in either direction. And if the entropy was greater than 0 in one direction, it would have to be less than 0 on the other direction. So it wouldn't be able to go in the other direction by the second law of thermodynamics. So it actually makes sense that the entropy of the universe-- the change in entropy of the universe, not just of the system-- when a reversible process occurs, is 0. Let's see if we can relate that to the irreversible process So if I wanted to figure out the change in entropy of the irreversible process-- What's the change in entropy of the irreversible process? And then let me subtract from that the heat that was taken away from the reservoir, from the irreversible process, Q sub IR, and of course all this over T1. What is this going to be equal to in relation to 0? Remember this is an irreversible, spontaneous process. Well, this value, the irreversible process is starting here and going there. So its change in delta S is the exact same thing as that. Change in delta S is the same as the reversible process. So these two things are equivalent. Now, I just told you that since there's some heat of friction, this guy has to take in less heat from the reservoir than this guy. So if this-- and I wrote it here. I mean, let me clean this thing up a little bit. I wrote it right here. Q sub IR, the heat that the irreversible process has to take from the reservoir, because generating its own heat from friction, is less than the heat from the reversal process. So this number, right here, is less than this number here. Or you could view it this way. This number here was equal to this. So this number here is going to be less than this. So this has to be greater than 0 for the irreversible, spontaneous process. Now let's just do a little bit of mathematics. So this is the heat that was essentially given to the irreversible system. It's a minus here because this is the term. It's kind of taken away from the actual reservoir. So let's just do a little bit of-- let's just multiply all sides of this equation by T1. And we get T1 times delta S of the irreversible process minus Q of the irreversible process is greater than 0. Now, well, let's just-- So what is this? How can we say this? But let's actually just multiply both sides of this by negative 1. And remember, this is true for any irreversible, spontaneous process. If you multiply both sides of this by negative 1, you get this. The heat added to the irreversible process minus T times delta S of the irreversible process is going to be less than 0. This is true for any irreversible, spontaneous process. And at this point, this should look reasonably familiar to you. When we wrote the Gibbs Free Energy formula, we said change in G is equal to delta H minus T times delta S. And we said that if this is less than 0, then we have a spontaneous process. And this all makes sense. Because these two are equivalent to each other. What's the only difference here? Here we wrote heat added to the system. Here we wrote change in enthalpy. And I've done three or four videos right now, where I say that change in enthalpy is equal to the heat added to a system, as long as we're dealing with a constant pressure system. So we get this result just by comparing a irreversible to a reversible system. And this is true for all irreversible, spontaneous processes. And then if we assume that we're dealing with a constant pressure system-- so you can forget a little bit of what I just drew, because we know that this is true, and we assume constant pressure-- then we get to this. And then we know that if something is spontaneous, then this right here must be less than 0. So hopefully you found that a little bit interesting. Actually I'll do one more point here. This kind of gels with the idea that second law of thermodynamics tells us that for any spontaneous process, delta S is going to be greater than or equal-- or well for any spontaneous process, is going to be greater than 0. Because although this right here isn't the formal definition of entropy, because we're not dealing with a reversible process here, you can kind of think of it that way. And so, at least, on an intuitive level it gets you there, that we have delta S greater than 0. And I won't fixate on that too much, because what I did earlier in the video is more rigorous than what I'm heading to right now. So hopefully that gives you a sense of where you can get to the Gibbs Free Energy formula, and how it drives spontaneous reactions from just our basic understanding of what reversible and irreversible processes are, and how they relate to entropy and heat exchange and enthalpy.