Main content

## Chemistry library

### Unit 15: Lesson 4

Gibbs free energy- Gibbs free energy and spontaneity
- 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
- Changes in free energy and the reaction quotient
- Standard change in free energy and the equilibrium constant
- 2015 AP Chemistry free response 2c

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 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?

- 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)
- 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)

- At13:07Sal says that for a spontaneous process, deltaS > 0. But what about when supercooled water freezes?(9 votes)
- 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)

- 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)
- 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)

- 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)- 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)

- Sal, at9:40you 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)
- It would mean that the engine would have to draw more Q from the heat source to perform the same amount of work due to the inefficiency.(3 votes)

- why is it assumed that every irreversible process is spontaneous?(4 votes)
- This is not the case, as per I understand it.

On the other hand, we know that only irreversible processes are spontaneous.(1 vote)

- 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..." at4:50, 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)- 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)

- How come at about9:15, 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)
- because that's part of the Q_IR, which he said was < Q_R, because of the heat of friction(4 votes)

- if enthalpy change is positive and entopy change is negative then what is the condition for the spontanity of a process?(1 vote)
- Then the process can't be spontaneous under any conditions.

Because, delG=delH -T*delS

delH>0, delS<0. And T can't be less than 0 (Kelvin Scale)

Hence both the terms are positive. Thus, delG>0.

This implies process non spontaneous. :)(4 votes)

- at3:19how can we we lose temperature in absence of reservoir? i did not get it.!(1 vote)
- because the piston has moved up and kinetic energy is transformed to mechanical work, so the average kinetic energy inside the cylinder becomes less..(4 votes)

## 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.