In this module we're going to look at how we find the entropy of the universe to determine whether or not a process is spontaneous. Our objective is to understand how temperature affects the entropy of the surroundings, and in turn, how that affects the entropy of the universe. When we look at entropy of the universe we know that we must have a positive value in order to have a spontaneous process. We also know that the change in entropy of the universe is equal to the change in entropy of the system plus the change in entropy of the surroundings. these are not necessarily negatives of one another. What we have to look at are the individual components to determine whether or not a process is spontaneous. when we look at the values at the delta S of the system and delta S of the surroundings we see that we can have both values as positive numbers, both fighters negative numbers, or have one value that's positive and one value that's negative. It's the balance and the signs of these numbers that determine whether or not a process is spontaneous. Let's take an example where this process is not always spontaneous. Looking at the freezing of water we know that at lower temperatures this happen spontaneously. At higher temperatures it is a non spontaneous process, so what we have to look at is where's the energy going? We're dispersing energy so we have to look at the temperature of the surroundings to see which one is going to favor the spontaneous process. For both of these we see that the Delta S of the system is the same, but we see that the Delta S of the surroundings is different and therefore the Delta S of the universe ends up being different. It's the Delta S of the surroundings that depends on the temperature because that's going to determine how that energy is dispersed. So remembered that energy tells us about the dispersal of energy, and the qualitative value tells us that entropy is a measure of energy dispersed per unit of temperature. As the temperature increases the amount of entropy for a given amount of energy dispersed decreases, therefore, something at a higher temperature has a lower impact and will not change the Delta S of the system as much. As the temperature decreases or drops the amount of entropy for a given amount of energy dispersed increases. This means that at a lower temperature we have a higher impact, so looking at the relative values of the Delta S of the system versus the delta S of the surroundings the temperature will determine how much the Delta S of the surroundings will impact the value of the delta S of the system and therefore how it will impact the delta S of the universe. So, how do we find the delta S of the surroundings? We know it's dependent on temperature but we have to find the actual value of it. For endothermic reactions, so that will be a reaction where Delta H is less than zero, so a negative value, we see that the delta S of surroundings is greater than zero because an exothermic reaction is releasing energy. We're dispersing that energy and so the Delta S of the surroundings is increasing. An endothermic reaction when Delta H is greater than zero, or a positive value, shows we're absorbing energy from surroundings. We're decreasing the dispersal of energy, and therefore delta S of the surroundings will be less then zero. But this doesn't give us the exact value of delta S of the surroundings to do that we have to combine both the enthalpy and the temperature to understand how they are related. So if I want to look at the relationship between enthalpy and delta S of the surroundings I have to take the negative of the delta H value, the negative enthalpy of the reaction of the system, and divide it by the temperature. Remembered that the temperature must be in units of Kelvin. So now we can find delta S of the universe if we know delta S of the system and delta S of the surroundings. We have our way to find delta S of the surroundings. We looked earlier at how to find delta S of the system by looking at products minus reactants, and from that information we could then find delta S of the universe, and then we can finally determined is a reaction spontaneous or not based on the Delta S of the universe. so as we said before a small T, or a low temperature, we have a large delta S of surroundings. For a large T we have a small Delta S of surroundings. Let's look at an example. Under which of the following conditions will Delta S of the universe always be positive? So for delta S of the universe to be positive we know that our equation for that is delta S of the universe equals delta S of the system plus delta S of the surroundings. So, if delta S of the system is positive and delta S of the surroundings is positive then Delta S of the universe will also be positive making B the correct answer. If we look at when both values are negative, if delta S of the system is negative and delta S of the surrounding is negative then our delta S of the universe will also be negative. however, if we look at the options were one value is positive and one value is negative we can't know whether Delta S of the universe will be positive or negative because it depends on the relative values of those delta S of the system and delta S of the surroundings. So, when will delta S of the surroundings be positive? Delta S of the surroundings will be positive when we have an exothermic reaction Remember, that for an exothermic reaction our delta H value is less than zero. It's negative. so if I put a negative value into my equation I'm going to end up with a positive delta S value, or a value that is greater than zero. Let's look at a specific example of finding delta S of the surroundings. We know that delta S of the surroundings equals minus delta H over T, remember that has to be in Kelvin. So I'm going to take delta S of the surroundings equals negative 34 kilojoules per mole, and I'm actually going to go ahead and multiply that by a thousand because I want to get this in the unit to joules because that's typically how entropy is reported, in joules, and then I'm going to divide by my temperature, which I have to convert to Kelvin, If I add 273. Now, I could take delta S of surroundings equal to -34,000 joules per mole over 298 Kelvin and what I find is that delta S of the surroundings is equal to negative 114 joules per mole Kelvin. now this doesn't necessarily mean that the process will be non spontaneous. What's going to matter is what the change in entropy is of the system so we can see how that compares to our negative 114 joules for the Delta S of the surroundings. Now we've talked about a lot of entropy terms here and we talked about a lot of greater than zero and less than zero and what they exactly mean so we've got a little summary here to go through. Remember, that in order for reaction to be spontaneous delta S of the universe must be greater than 0. That's the second law of thermodynamics. When I look at delta D of the system I see that it's greater than zero, when I have more disorder, more dispersal of energy. It's less than zero when I have less disorder and less dispersal of energy. for delta S of the surroundings if I have an exothermic or negative value for delta H I'm going to end up with a positive Delta S of surroundings, and for an endothermic process that has a positive delta H I'm gonna end up with a negative Delta S of surroundings. Next we're gonna look at Gibbs free energy.
