0:02

In this module we're going to look at how we find the entropy of the universe

Â to determine whether or not a process is spontaneous.

Â Our objective is to understand how temperature affects the entropy of the

Â surroundings,

Â and in turn, how that affects the entropy of the universe.

Â When we look at entropy of the universe we know that we must have a positive value

Â in order to have a spontaneous process.

Â We also know that the change in entropy of the universe is equal to the change in

Â entropy of the system

Â plus the change in entropy of the surroundings. these are not necessarily

Â negatives of one another. What we have to look at

Â are the individual components to determine whether or not a process is

Â spontaneous.

Â when we look at the values at the delta S of the system and delta S of the

Â surroundings

Â we see that we can have both values as positive numbers, both fighters negative numbers,

Â or have one value that's positive and one value that's negative.

Â It's the balance and the signs of these numbers that determine whether or not a

Â process is spontaneous. Let's take an example where

Â this process is not always spontaneous. Looking at the freezing of water we know

Â that at lower temperatures this happen spontaneously. At higher temperatures

Â it is a non spontaneous process, so what we have to look at is

Â where's the energy going? We're dispersing energy

Â so we have to look at the temperature of the surroundings to see

Â which one is going to favor

Â the spontaneous process. For both of these we see that the Delta S of the system

Â is the same, but we see that the Delta S of the surroundings

Â is different and therefore the Delta S of the universe ends up being

Â different. It's the Delta S of the surroundings that depends on the temperature

Â because that's going to determine how that energy is dispersed.

Â So remembered that energy tells us about the dispersal of energy,

Â and the qualitative value tells us that entropy is a measure of energy dispersed

Â per unit of temperature.

Â As the temperature increases the amount of entropy for a given amount of

Â energy dispersed decreases, therefore, something at a higher temperature

Â has a lower impact and will not change the Delta S of the system as much.

Â As the temperature decreases or drops the amount of entropy for a given amount of

Â energy dispersed increases.

Â This means that at a lower temperature

Â we have a higher impact, so looking at the relative values of the Delta S

Â of the system

Â versus the delta S of the surroundings the temperature will determine

Â how much the Delta S of the surroundings will impact

Â the value of the delta S of the system and therefore how it will impact the

Â delta S of the universe.

Â So, how do we find the delta S of the surroundings? We know it's dependent on

Â temperature but we have to find the actual value of it.

Â For endothermic reactions, so that will be a reaction

Â where Delta H is less than zero,

Â so a negative value, we see that the delta S of surroundings is greater than zero

Â because an exothermic reaction is releasing energy. We're dispersing that energy

Â and so the Delta S of the surroundings is increasing.

Â An endothermic reaction when Delta H

Â is greater than zero, or a positive value,

Â shows we're absorbing energy from surroundings. We're

Â decreasing the dispersal of energy, and therefore delta S of the surroundings

Â will be

Â less then zero. But this doesn't give us the exact value of delta S of the

Â surroundings

Â to do that we have to combine both the enthalpy and the temperature

Â to understand how they are related.

Â So if I want to look at the relationship between

Â enthalpy and delta S of the surroundings I have to take the negative

Â of the delta H value, the negative enthalpy of the reaction of the system,

Â and divide it by the temperature. Remembered that the temperature must be

Â in units of Kelvin.

Â So now

Â we can find delta S of the universe if we know delta S of

Â the system and delta S of the surroundings. We have our way to find

Â delta S of the surroundings.

Â We looked earlier at how to find delta S of the system

Â by looking at products minus reactants, and from that information we could then

Â find delta S

Â of the universe, and then we can finally determined is a reaction

Â spontaneous or not

Â based on the Delta S of the universe.

Â so as we said before a small T, or a low temperature,

Â we have a large delta S of surroundings. For a large T

Â we have a small Delta S of surroundings.

Â Let's look at an example. Under which of the following conditions will Delta S

Â of the universe

Â always be positive?

Â 4:57

So for delta S of the universe to be positive we know that our equation for that is

Â delta S of the universe

Â equals delta S of

Â the system plus delta S

Â of the surroundings. So, if delta S of the system is positive

Â and delta S of the surroundings is positive then Delta S of the universe

Â will also be positive

Â making B the correct answer. If we look at when both values are negative, if delta S

Â of the system is negative and delta S of the surrounding is negative

Â then our delta S of the universe will also be negative.

Â however, if we look at the options were one value is positive

Â and one value is negative we can't know

Â whether Delta S of the universe will be positive or negative

Â because it depends on the relative values of

Â those delta S of the system and delta S of the surroundings.

Â 6:02

Delta S of the surroundings will be positive when we have an exothermic reaction

Â Remember, that for an exothermic reaction our delta H value

Â is less than zero. It's negative. so if I put a negative value

Â into my equation I'm going to end up with

Â a positive delta S value, or a value that is greater

Â than zero.

Â Let's look at a specific example of finding delta S of the surroundings.

Â We know that delta S of the surroundings

Â equals minus delta H

Â over T, remember that has to be in Kelvin.

Â So I'm going to take delta S of the surroundings

Â equals negative 34 kilojoules per mole,

Â and I'm actually going to go ahead and multiply that by a thousand because I

Â want to get this in the unit to joules because that's typically how

Â entropy is reported, in joules, and then I'm going to divide

Â by my temperature, which I have to convert to Kelvin,

Â If I add 273. Now, I could take delta S of surroundings

Â equal to -34,000 joules per mole

Â over 298 Kelvin

Â and what I find is that delta S of the surroundings

Â is equal to negative 114

Â joules per mole Kelvin.

Â now this doesn't necessarily mean that the process will be non spontaneous.

Â What's going to matter is what the change in entropy is of the system

Â so we can see how that compares to our negative 114 joules

Â for the Delta S of the surroundings.

Â Now we've talked about a lot of entropy terms here and we talked about a lot of

Â greater than zero and less than zero and what they exactly mean so we've got a little summary here

Â to go through.

Â Remember, that in order for reaction to be spontaneous delta S of the universe

Â must be greater than 0.

Â That's the second law of thermodynamics.

Â When I look at delta D of the system I see that it's greater than zero,

Â when I have more disorder, more dispersal of energy.

Â It's less than zero when I have less disorder and less dispersal of energy.

Â for delta S of the surroundings if I have an exothermic or negative value for

Â delta H

Â I'm going to end up with a positive Delta S of surroundings,

Â and for an endothermic process that has a positive delta H

Â I'm gonna end up with a negative Delta S of surroundings.

Â Next we're gonna look at Gibbs free energy.

Â