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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 9

Lesson 2: Gibbs free energy and thermodynamic favorability# Thermodynamic favorability and temperature

The temperature conditions under which a process is thermodynamically favored (Δ

*G*° < 0) can be predicted from the signs of Δ*H*° and Δ*S*°. When Δ*H*° < 0 and Δ*S*° > 0, the process is favored at all temperatures. When Δ*H*° > 0 and Δ*S*° < 0, the process is favored at no temperature. When Δ*H*° < 0 and Δ*S*° < 0, the process is favored only at low temperatures. When Δ*H*° > 0 and Δ*S*° > 0, the process is favored only at high temperatures. Created by Jay.## Want to join the conversation?

- 8:36he mentions delta entropy doesn’t change much with temperature, but wouldn’t an increase of over 1000 kelvin increase the number of micro states in a gas by a considerable amount?(7 votes)
- I believe what you're referring to is ΔS (change in entropy), which is equal to:

k * ln(W).

In other words ΔS is directly proportional to the number of microstates, which would increase with temperature. However, as I understand it,

What's being talked about is**absolute or standard entropy**- ΔS⁰ (delta S nought), which is ΔS of 1 mole of a pure solid or liquid,**at a pressure of 1atm and a specified temperature**. In the previous video, Jay mentions that often that temperature is**25°C**.

Moreover, the units of ΔS⁰ are J/(mol * K), whereas the units of ΔH are kJ/mol. (Ignore the difference in kJ and J)

So, ΔS⁰ stays constant regardless of temperature, as Kelvin is in the denominator.

That's also why we multiply it with the temperature while calculating the Gibbs Free Energy (ΔG⁰).

Hope this helps!(2 votes)

- What is the relationship between thermodynamic favorability and the equilibrium constant (K)?(1 vote)
- We can relate Gibbs free energy and the equilibrium constant of a reaction via this equation: ΔG° = -RTln(K). This tells us that a larger equilibrium constant value yields a more negative free energy value which means the forward reaction is favored. If we had a very small equilibrium constant value then that yields a more positive free energy value which means that the reverse reaction is favored.

Hope that helps.(4 votes)

- As Jay mentioned in the video and also in the description: "When ΔH° < 0 and ΔS° > 0, the process is favored at all temperatures". The example in the video plugged in a temperature of 100 degrees. However if I instead plugged in a temperature of -2000 degrees, wouldn't the ΔG°>0?(2 votes)
- The unit for entropy, ΔS, is joules per kelvin, J/K. This means the temperature has to be in Kelvins. And the kelvin scale has no negative values since the lowest temperature is absolute zero, 0 K.

Hope that helps.(2 votes)

- What is the relationship between osmosis and thermodynamics?(1 vote)
- Osmosis is the spontaneous movement of solvent molecules through a semipermeable membrane (like a cell’s membrane) in order to create equal solute concentrations on either side of the membrane. If want to think about this using Gibbs free energy, then this process has a negative free energy value since it’s spontaneous; it happens naturally without the need of external work to make it happen. And if we want to think about it even more fundamentally in terms of entropy, the entropy of the universe increases as the solvent molecules disperse over a greater region.

Hope that helps.(1 vote)

## Video transcript

- [Instructor] The
thermodynamic favorability of a chemical reaction can be affected by the temperature. Let's say we have a generic reaction where the reactants
turn into the products. As a quick review, when the
standard change in free energy, delta G naught, is less than zero, the forward reaction is
thermodynamically favorable. Therefore, the net
reaction goes to the right to make more of the products. And when delta G naught
is greater than zero, the forward reaction is
thermodynamically unfavorable. That means that the reverse
reaction is favorable and the net reaction would go to the left to make more of the reactants. One way to calculate delta G naught is to use the following equation. Delta G naught is equal to delta H naught minus the absolute temperature
times delta S naught. And since delta H naught is talking about the
standard change enthalpy, for this equation delta H naught will be
called the enthalpy term. And because delta S naught is talking about the
standard change in entropy, the absolute temperature
times delta S naught will be referred to as the entropy term. So when we're trying to figure out if delta G naught is negative or if delta G naught is positive, we need to consider the
sign for delta H naught, the sign for delta S naught, and sometimes we also need to consider what the temperature is. There are four possible
combinations of signs for delta H naught and delta S naught as we can see in this table. For each of these four combinations, we're gonna think about the temperatures at which delta G naught is less than zero and we're gonna finish
filling out this table. To start with, we're gonna
think about the situation where delta H naught is negative and delta S naught is positive. So for this first possible combination, when delta H naught is negative, we say that's an exothermic reaction. And when delta S naught is positive, that's saying there's
an increase in entropy as reactants turn into products. So let's think about the
sign for delta G naught if we plug in some numbers
for delta H naught, temperature, and delta S naught. And when I plug in some numbers, I'm not really concerned
about a particular reaction. I'm not concerned about units. I'm not concerned about
significant figures. I'm just interested in
looking at the numbers and seeing how the math affects
the sign for delta G naught. So for delta H naught,
I'm saying that's -100. For T, I'm saying that's 100, and for delta S naught,
I'm saying that's +0.1. So this would be equal to -100 - 10, which is equal to -110. So delta G naught is negative. Notice how a negative entropy term favors getting a negative
value for delta G naught. And notice how a positive entropy term, since we're subtracting
that in our equation is also favorable for
getting a negative value for delta G naught. Also notice, no matter what
we put in for the temperature since delta S naught is positive, the entropy term will always be positive. And since we're subtracting
it in our equation, we would always get a negative
value for delta G naught. So whenever delta H naught is negative and when delta S naught is positive, delta G naught will be negative no matter what the temperature. So if we go back to our chart for our first possible combination, when delta H naught is
negative, that's favorable. And when delta S naught as
positive, that's also favorable. Therefore delta G naught is
less than 0 at all temperatures. So I'm gonna go ahead and write in here, 'at all temperatures' in our chart. As an example of this, let's consider the conversion
of ozone gas into oxygen gas. Since energy is released, this is an exothermic reaction, and delta H naught is negative. And since we're going from 2 moles of gas on the reactant side to 3 moles of gas on the product side, that's an increase in entropy. So delta S naught is positive. Since delta H naught is negative and delta S naught is positive, delta G naught for this reaction is less than zero at all temperatures. That means the forward reaction is thermodynamically favorable, and ozone gas would turn into oxygen gas, and this reaction would
occur at all temperatures. The second possible
combination for the signs is when delta H naught is positive and when delta S naught is negative. When delta H naught is positive, that's an endothermic reaction. And when delta S naught is negative, there's a decrease in entropy going from reactants to products. Once again we're gonna
plug in some numbers and solve for delta G naught. So for delta H naught, let's say that's +100, for the temperature term,
let's say that's 100, and delta S naught, let's say that's -0.1. When we do the math, we get positive 100 - negative 10, which is equal to +110. So delta G naught is positive. Notice when the entropy term is positive, that's not favorable for
getting a negative value for delta G naught. And when the entropy term is negative, since we're subtracting this negative, we would actually be
adding the entropy term. And that would not favor a negative value for delta G naught. Also notice, no matter what you put in
for the temperature here, you're always gonna get a negative value for the entropy term. And if the enthalpy term is positive, since we would mathematically
be adding the entropy term, we would always get a positive
value for delta G naught. So whenever delta H naught is positive and delta S naught is negative, delta G naught will be positive no matter what the temperature. So going back to our charge
when delta H naught is positive, that was not favorable, and when delta S naught was negative, that was also not favorable. So it doesn't matter
what the temperature is. Delta G naught will not be less than zero. Therefore, on our chart
here we can write in, at no temperature will delta
G naught be less than zero. As an example of this, let's look at the reverse
of the reaction we saw for the first possible combination. Before we had ozone
turning into oxygen gas, and now we have oxygen
gas turning into ozone. Because heat is on the reactant side, we know this reaction is endothermic and delta H naught is positive. And since we're going from 3 moles of gas on the reactant side to 2 moles of gas on the product side, that's a decrease in entropy, and delta S naught is negative. Since delta H naught is positive and delta S naught is negative, delta G naught will
never be less than zero. It doesn't matter what the temperature is. And so the forward reaction is always thermodynamically unfavorable. The third possible combination of signs has both delta H naught and delta S naught as being negative. Once again we're gonna plug
some numbers into the equation and see what happens to
the sign of delta G naught. So let's say delta H naught is -100. The temperature is 100 and
delta S naught is -0.1. Doing the math, we get
negative 100 minus negative 10. So that's -100 + 10, which is equal to -90. Let's think about why
we got a negative value for delta G naught. The enthalpy term is negative, which favors a negative
value for delta G naught. The entropy term is also negative, but since we're subtracting
the entropy term, it actually gets added for the overall delta
G naught calculation. So the entropy term is not favorable. However, in this case, the favorable entropy term outweighs the unfavorable entropy term. And that's the reason why
we get a negative value for delta G naught at this temperature. Let's compare the
calculation that we just did at a temperature of 100, which gave us a negative
value for delta G naught, to a similar calculation, except this time we have
a much higher temperature. This time it's 2000. But notice how the
values for delta H naught and delta S naught are the same as before. The reason why we can use the same values at a different temperature is because delta H and delta S naught don't change much with temperature. However, delta G naught does
change a lot with temperature. Because at this higher
temperature, when we do the math, we see that now we get delta
G naught is equal to +100. So at this higher temperature, even though the enthalpy
term is favorable, now the unfavorable entropy term outweighs the favorable enthalpy term and gives us a positive
value for delta G naught. Therefore, at a relatively
lower temperature, delta G naught is negative. However, at a higher temperature, delta G naught is positive. So when delta H naught is
negative, that's favorable, but when delta S naught is
negative, that's not favorable. And we just saw from the calculations that delta G naught will be
negative at low temperatures. So let's write that in here in our chart. So far we've been talking
about delta G naught relative to a chemical process. However, delta G naught also
applies to a physical process. In this case we're looking at liquid water
turning into solid water. So this is the process of freezing. The process of freezing gives off energy. So this process is exothermic, and delta H naught will be negative. Going from a liquid to a solid
is a decrease in entropy, therefore, delta S naught is negative. For this combination of signs,
delta G naught is negative only at relatively low temperatures. Therefore, the forward process of freezing is thermodynamically
favorable at low temperatures. And this matches with what
we already know about water, water freezes at a
relatively low temperature. Finally, let's look at our
fourth combination of signs, and that's when both delta H naught and delta S naught are positive. So let's say delta H
naught is equal to +100, delta S naught is equal able to +0.1, and first we'll look at the
lower temperature of 100. Doing the math, we get positive 100 - 10, which gives us a positive
value for delta G naught. And if we do the calculation at a higher temperature of 2000, notice that delta H naught is still +100 and delta S not is still +0.1. This time we get -100 for delta G naught. Notice how for this example
at the higher temperature, the entropy term is unfavorable for giving a negative
value for delta G naught. However, the entropy term is favorable. A positive value for delta S naught gives a positive value
for the entropy term. And since we are subtracting it, the entropy term outweighs
the enthalpy term, giving a negative value
for delta G naught. However, for the first example
at the lower temperature, since the temperature is lower, the entropy term is a smaller value, and is unable to overcome the
unfavorable enthalpy term. And that's the reason why delta G naught ends up being positive
at the lower temperature. So for this combination of signs, delta G naught is negative
at higher temperatures. So for our fourth combination, when delta H naught is positive, that is unfavorable. However, when delta S naught
is positive, that is favorable. And we just saw that delta
G naught is less than zero at high temperatures. So I'll go ahead and write
in on our table here, 'at high temperatures.' As an example of this, let's consider solid water
turning into liquid water. So the forward process is
the process of melting. It takes energy to disrupt the intermolecular
forces of the solid, therefore, heat is on the reactant side and delta H naught is positive. Melting is an endothermic process. Going from a solid to a liquid
is an increase in entropy. Therefore, delta S naught is positive. And when both signs are positive, our calculations showed us that delta G naught is less
than zero at high temperatures. Therefore, the forward process of melting is thermodynamically favorable at relatively high temperatures. And this matches what we
already know from experience. At relatively low temperatures
ice does not melt. However, at relatively high
temperatures ice does melt. Now that we have our chart filled out, let's summarize what we've learned about these four possible combinations. When delta H naught and delta
S naught are both favorable, the forward process is
thermodynamically favorable at all temperatures. When delta H naught and delta
S naught are both unfavorable, the forward process is
thermodynamically favorable at no temperatures, or you could say the forward process is thermodynamically
unfavorable at all temperatures. When delta H naught is favorable and delta S naught is unfavorable, the forward process is only
thermodynamically favorable at low temperatures. And finally, if delta H
naught is unfavorable, but delta S naught is favorable, the forward process is
thermodynamically favorable only at high temperatures.