Thermodynamic favorability and temperature | AP Chemistry | Khan Academy

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, ΔG° is less than zero, the forward reaction is thermodynamically favorable. Therefore, the net reaction goes to the right to make more of the products. When ΔG° 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 ΔG° is to use the following equation: ΔG° = ΔH° - T * ΔS°. Since ΔH° is talking about the standard change in enthalpy for this equation, ΔH° will be called the enthalpy term. Because ΔS° is talking about the standard change in entropy, the absolute temperature times ΔS° will be referred to as the entropy term. So, when we're trying to figure out if ΔG° is negative or if ΔG° is positive, we need to consider the sign for ΔH°, the sign for ΔS°, and sometimes we also need to consider what the temperature is.

There are four possible combinations of signs for ΔH° and ΔS°. As we can see in this table, for each of these four combinations, we're going to think about the temperatures at which ΔG° is less than zero, and we're going to finish filling out this table.

To start with, we're going to think about the situation where ΔH° is negative and ΔS° is positive. For this first possible combination, when ΔH° is negative, we say that's an exothermic reaction. When ΔS° is positive, that's saying there's an increase in entropy as reactants turn into products. So let's think about the sign for ΔG°.

If we plug in some numbers for ΔH°, T, and ΔS°, 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 ΔG°. So for ΔH°, I'm saying that's -100, for T I'm saying that's 100, and for ΔS°, I'm saying that's +0.1. This would be equal to -100 - 10, which is equal to -110.

So ΔG° is negative. Notice how a negative enthalpy term favors getting a negative value for ΔG°, and notice how a positive entropy term, since we're subtracting that in our equation, is also favorable for getting a negative value for ΔG°. Also, notice, no matter what we put in for the temperature, since ΔS° 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 ΔG°.

So whenever ΔH° is negative and when ΔS° is positive, ΔG° will be negative no matter what the temperature is. If we go back to our chart for our first possible combination when ΔH° is negative, that's favorable, and when ΔS° is positive, that's also favorable. Therefore, ΔG° is less than zero at all temperatures.

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 ΔH° is negative. Since we're going from two moles of gas on the reactant side to three moles of gas on the product side, that's an increase in entropy, so ΔS° is positive. Since ΔH° is negative and ΔS° is positive, ΔG° 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 ΔH° is positive and when ΔS° is negative. When ΔH° is positive, that's an endothermic reaction, and when ΔS° is negative, there's a decrease in entropy going from reactants to products. Once again, we're going to plug in some numbers and solve for ΔG°.

So for ΔH°, let's say that's +100, for the temperature term let's say that's 100, and for ΔS°, let's say that's -0.1. When we do the math, we get 100 - (-10), which is equal to +110. So ΔG° is positive. Notice when the enthalpy term is positive, that's not favorable for getting a negative value for ΔG°, 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 ΔG°.

Also, notice, no matter what you put in for the temperature here, you're always going to 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 ΔG°.

So whenever ΔH° is positive and ΔS° is negative, ΔG° will be positive no matter what the temperature is. So going back to our chart, when ΔH° is positive, that was not favorable, and when ΔS° was negative, that was also not favorable. Therefore, it doesn't matter what the temperature is; ΔG° will not be less than zero.

Therefore, on our chart here, we can write in "At no temperature will ΔG° 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 ΔH° is positive.

Since we're going from three moles of gas on the reactant side to two moles of gas on the product side, that's a decrease in entropy, and ΔS° is negative. Since ΔH° is positive and ΔS° is negative, ΔG° 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 ΔH° and ΔS° as being negative. Once again, we're going to plug some numbers into the equation and see what happens to the sign of ΔG°. So let's say ΔH° is -100, the temperature is 100, and ΔS° is -0.1. Doing the math, we get -100 - (-10), so that's -100 + 10, which is equal to -90.

Let's think about why we got a negative value for ΔG°. The enthalpy term is negative, which favors a negative value for ΔG°. The entropy term is also negative, but since we're subtracting the entropy term, it actually gets added for the overall ΔG° calculation. So the entropy term is not favorable. However, in this case, the favorable enthalpy term outweighs the unfavorable entropy term, and that's the reason why we get a negative value for ΔG° at this temperature.

Let's compare the calculation that we just did at a temperature of 100, which gave us a negative value for ΔG°, to a similar calculation, except this time we have a much higher temperature. This time it's 2000. But notice how the values for ΔH° and ΔS° are the same as before. The reason why we can use the same values at a different temperature is that ΔH° and ΔS° don't change much with temperature. However, ΔG° does change a lot with temperature because at this higher temperature, when we do the math, we see that now we get ΔG° 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 ΔG°. Therefore, at a relatively lower temperature, ΔG° is negative; however, at a higher temperature, ΔG° is positive. So when ΔH° is negative, that's favorable, but when ΔS° is negative, that's not favorable.

We just saw from the calculations that ΔG° will be negative at low temperatures, so let's write that in here in our chart. So far, we've been talking about ΔG° relative to a chemical process. However, ΔG° 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 ΔH° will be negative.

Going from a liquid to a solid is a decrease in entropy; therefore, ΔS° is negative. For this combination of signs, ΔG° 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 ΔH° and ΔS° are positive. Let's say ΔH° is equal to +100, ΔS° is equal to +0.1. First, we'll look at the lower temperature of 100. Doing the math, we get 100 - 10, which gives us a positive value for ΔG°. If we do the calculation at a higher temperature of 2000, notice that ΔH° is still +100 and ΔS° is still +0.1. This time we get a negative ΔG°.

Notice how for this example at the higher temperature, the enthalpy term is unfavorable for giving a negative value for ΔG°, however, the entropy term is favorable. A positive value for ΔS° 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 ΔG°.

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. That's the reason why ΔG° ends up being positive at the lower temperature. So for this combination of signs, ΔG° is negative at higher temperatures.

For our fourth combination, when ΔH° is positive, that is unfavorable; however, when ΔS° is positive, that is favorable. We just saw that ΔG° 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. 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 ΔH° is positive. Melting is an endothermic process. Going from a solid to a liquid is an increase in entropy; therefore, ΔS° is positive.

When both signs are positive, our calculations showed us that ΔG° 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 ΔH° and ΔS° are both favorable, the forward process is thermodynamically favorable at all temperatures. When ΔH° and ΔS° are both unfavorable, the forward process is thermodynamically unfavorable at no temperatures.

When ΔH° is favorable and ΔS° is unfavorable, the forward process is only thermodynamically favorable at low temperatures. Finally, if ΔH° is unfavorable but ΔS° is favorable, the forward process is thermodynamically favorable only at high temperatures.