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## MCAT

### Course: MCAT > Unit 9

Lesson 20: Bioenergetics- Bioenergetics questions
- Bioenergetics questions (2)
- An analogy for Gibbs free energy
- Bioenergetics: The transformation of free energy in living systems
- Why we need metabolism?
- Insulin and glucagon
- Tissue specific metabolism and the metabolic states

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# An analogy for Gibbs free energy

Created by Jasmine Rana.

## Want to join the conversation?

- I don't understand how to apply this analogy. Is it saying that at higher temperatures, delta S will tend to be positive, and that the forward reaction will occur spontaneously? and that at lower temperatures, delta H will tend to be positive, meaning the forward rxn isn't spontaneous?(20 votes)
- Ditch the analogy and just focus on the values of H and S in the chart. The analogy is a bit convoluted for the very simple terms outlined in the equation for G.(80 votes)

- For those who didnt get the analogy. When ΔS is positive and ΔH is negative, a process is always spontaneous

When ΔS is positive and ΔH is positive, the relative magnitudes of ΔS and ΔH determine if the reaction is spontaneous. High temperatures make the reaction more favorable, because exothermicity plays a small role in the balance.

When ΔS is negative and ΔH is negative, the relative magnitudes of ΔS and ΔH determine if the reaction is spontaneous. Low temperatures make the reaction more favorable, because exothermicity is important.

When ΔS is negative and ΔH is positive, a process is not spontaneous at any temperature, but the reverse process is spontaneous.(28 votes) - Am I the only one who didn't hate the analogy? I understand free energy pretty well, but sometimes get confused with the temperature variable and found her reasoning helpful. Yes, sometimes it's good to have a chemistry-based example, but I thought this was relatable and explains it correctly.(19 votes)
- I really liked it too, it's good to have a backup way of remembering something just in case you blank during the exam because of nerves.(5 votes)

- What is the state of the reaction when delta G =0?(4 votes)
- The system is then in a state of equilibrium.(11 votes)

- This video was quite convoluted and oversimplified. I had to forward through it in order to NOT confuse myself.(8 votes)
- The analogy she gave makes the entire thing much more confusing than it needs to be. I understand the intent, but it's a very basic, simple concept, and the mathematic illustration is a much more accurate, helpful, and understandable explanation than such a broad, unclear, and inapplicable analogy.(5 votes)
- very confusing and convoluted analogy(4 votes)
- Thanks for the great video. The analogy helped me very much!(4 votes)
- At3:58, what if the temperature was negative? Or is the temperature measured in kelvin?(2 votes)
- Temperature is in Kelvin. If H is + and S is -, delta G will be positive for all temperatures. The only time temperature can help figure out if delta G will be + or - is when both H and S have the same sign.(4 votes)

- So the idea is to try and achieve a spontaneous reaction where ever possible?(2 votes)

## Video transcript

So in this video I want
to go ahead and talk more about the mathematical
expression for delta G, or the change in Gibbs free
energy, which I've went ahead and already written out here. So let's remind ourselves that
this change in free energy is a quantity that is equal
to the change in enthalpy, or heat content for a reaction
minus the temperature at which the reaction is run
times the change in entropy or, broadly
speaking, disorder in going from
reactants to products for a particular
chemical reaction. OK so that was kind
of a mouthful, right? So my ultimate goal in making
this video is to ultimately present to you an
analogy that I hope will give you a
better intuitive feel for how the terms of
enthalpy and entropy contribute to the
overall sign of delta G. Because remember it is
the sign of delta G, that is whether it's
negative or positive, that tells us whether
or not a reaction is spontaneous or nonspontaneous. But before talking about
analogies let's go ahead and get a better handle
on enthalpy and entropy from more of a chemistry
and mathematical lens. So recall that
enthalpy is really a good proxy for the
change in bond energy that occurs during a reaction,
that is to say whether or not energy was released or absorbed
during a bond rearrangement. And because all systems
want to achieve their lowest possible energy, if we're
talking about enthalpy in isolation from entropy,
let's say, generally speaking we might
intuitively say that having a negative
delta H, which describes a release of energy--
remember-- going from reactants to products, would be more
favorable than a positive delta H value. On the other hand, the
second law of thermodynamics says that all systems
tend toward disorder and since entropy, or
the chain in entropy, is a measure of whether
something is getting more disordered or
less disordered. In isolation, if we're
just talking about entropy, we might intuitively say that
having a positive delta S value, that is to say describing
an increase in entropy from going to products
from reactants, would be more favorable than
a negative delta S value. Now mathematically our equation
supports our hypothesis, right? A negative number
minus a positive number will always be negative. That is to say regardless
of what the temperature is, the delta G value will
always be negative, the reaction will
always be spontaneous. But since delta H is
not always negative, and since delta S is not always
positive for all reactions, what will the delta G be
for other combinations of delta H and delta S? So to explore this
further I thought we could go ahead and
create a table like such to write out essentially
the different signs that delta H and
delta S could take on, and determine for each
of these combinations whether the delta G will
be positive or negative. So we've already
gone through one. We've said that when
delta H is negative, and delta S is positive,
delta G will be negative. But of course there
are other combinations too, delta H could be positive,
and delta S could be negative, or both delta H and delta
S could be positive, or delta H and delta S
could both be negative. So let's go ahead
and take these signs and plug and chug into
our equation above. So if we have a positive
number and we're subtracting a
negative number, we are always going to
get a positive number. So for all situations in
which delta H is positive and delta S is negative we will
have a positive delta G value. This makes sense, right? Because it's
essentially the opposite of the previous situation. But what about the
case in which you have a positive number minus
another positive number? Well, you might say
it really depends on the relative magnitude
of these two numbers, right? Because if you have a big
positive number minus a smaller positive number you would
get a positive number. But if you had a small positive
number minus a bigger positive number you would get
a negative number. So ultimately it really
depends, delta G can either be positive or negative. And this value of temperature
plays a pretty big role in kind of tipping
the scale towards one direction or another. Because if it's really high,
for example, our second term might be more positive, and
therefore this overall value might be more negative. But if it's not very high
and delta H is really, really large, well, perhaps then
delta G would be positive. So finally let's
look at the situation where we have a
negative number-- in our last row-- minus
another negative number. Well, you might say again
this really depends, right? If you have a small
negative number and you're adding a very
large positive quantity to it, it could be positive. But if it's a smaller
number that you're adding on to a negative number,
it could still be negative. So again this really depends,
and again temperature plays a big role in
tipping the scale one direction or the other. Now for the analogy
that I want to present, I really want to focus
on these last two rows here in which temperature
plays a big role in determining whether or not a reaction
is spontaneous or not. So how can we understand
this trade-off between delta H and delta S
at high or low temperatures? So I'm going to go
ahead and scroll down and I'm going to
present an analogy that I hope isn't too corny. But hopefully it's
useful, and you might even consider thinking
of your own analogy to help you understand this
trade-off between entropy and enthalpy. So imagine you are
a chemistry student, and for spring break you
decide to go with your friends somewhere really
tropical, where there are a bunch of palm
trees and great beaches. And one day you and your
friends go ahead out to one of the beaches and
you and a couple of friends set out a large beach
towel and decide to sunbathe on the beach, while
some of your other friends decide to go out into
the ocean and swim. Now of course being
a chemistry nerd you describe this
phenomenon in terms of a chemical reaction, that
is to say a reaction in going from lying down on a beach
towel to swimming in the water, or the reverse that is going
from the water to lying on a beach towel. Now at this point you
begin thinking back to your equation
for delta G, which remember is equal to delta
H minus the temperature times delta S. And
you begin to realize that the forward reaction that
is going from a beach towel to the ocean is more
favorable, that is to say if more of your
friends are in the water when the sun is really at its
peak, and everyone really wants to go to the
water to cool off. On the other hand,
the reverse reaction that is going from the
water to the beach towel is more favorable when
the sun is less intense, when it's less
scorching hot outside and it's easier to
relax on the beach. In other words, you conclude
the sun is figuratively, or perhaps even literally here,
analogous to the temperature variable in our equation. Depending on how hot
the sun is ultimately determines whether
or not it is more spontaneous to go into the
water or come out of the water. Now since you happen to also
be an English major and love metaphors, you go
ahead and think up a metaphor for both the enthalpy
and entropy contributions in your hypothetical
chemical reaction. Enthalpy, you decide,
you can think about in terms of the amount
of energy that you're expending whether you're on a
beach towel or in the water. So we all know that
lying on a beach towel is super relaxing
and very peaceful, and requires little to
no energy and you're in a very, very stable place. But on the other hand, when
you're swimming in the water it requires a lot of energy
be expended because you're of course swimming and
throwing around a beach ball. On the other hand--
I'm going to go ahead and scroll down here--
lying on a beach towel can kind of get boring
after a while, right? But being in the
water is lots of fun, and there's really
never a boring moment. And this interplay
between fun and boring is something that
you can kind of think of as your
entropy variable. Now in an ideal world you'd
want to be both relaxed and having fun. And in chemistry
language that's really another way of saying that you'd
want to have a negative delta H value and a positive
delta S value. But of course in this
particular setup, these things are
mutually exclusive. So what wins? Ultimately, in the
end, you conclude that when the sun is
out this fun term, or what we're kind of
calling entropy wins, and the forward
reaction is spontaneous. But when the sun isn't as
hot outside this relaxation, or enthalpy term, wins
and the reverse reaction is spontaneous. And with this mighty conclusion
that you have come up with on this relaxing
day at the beach you go home and reminisce about
how chemistry and life are far more similar than
you once thought.