Let's look at some problems where we're calculating Gibbs free energy. For the values of delta H and delta S, predict if the reaction would be spontaneous at 25 degrees Celsius. If the reaction is nonspontaneous, at what temperature would it become spontaneous? So the first thing we need to note is that we are looking for a delta G value less than zero for a reaction to be spontaneous. And we also need to know that we're going to be using our temperature in Kelvin. So, we want to say 25 + 273 = 298 Kelvin. So that's the temperature we'll use in our calculations. Now we want to look at our first problem here, we're given a delta H value of 15 kilojoules per mol and delta S of 29.5 joules per mol Kelvin. We're going to use the equation delta G = delta H- T delta S to solve for our Gibbs free energy. We do need to be careful because our delta H is in terms of kilojoules and our delta S in terms of joules. So we want to say delta G = 15.0 kilojoules per mol and I'm going to go ahead and put the conversion in to take this one to joules. So 1000 joules in 1 kilojoules- my 298 Kelvin for my temperature times my delta S value, which is 29.5 joules per mol Kelvin. And when I solve this, what I see is I get a delta G value = 6209 joules per Kelvin. Now it doesn't really matter whether I convert to joules or kilojoules, as long as we get both values with respect to the same unit. My delta G value here is greater than zero. So that lets me know my reaction is nonspontaneous at this particular temperature. But I want to see if there's a temperature at which it will become spontaneous. And so what I want to do is I want to find that temperature. So I'm going to set up the same equation but solve it for the temperature this time instead of the delta G value. So what I want to do is for my delta G value this time, I'm going to put in a value of 0. Because I know that I have a positive delta G value. I want to get to a negative delta G value. In order to get there, I'm going to have to go through 0, and that's also where I'll change from being a nonspontaneous process to a spontaneous process. So I say 0 = and I convert this. I have my 15,000 joules per mol because I've already converted my delta H to joules. Minus T, I don't know that this time, times my 29.5 joules per mol Kelvin and when I can rearrange that and I end up with 15,000 joules per mol = T x 29.5 joules per mol Kelvin. And when I solve then I end up with a temperature equal to 508 Kelvin. And if I subtract 273 from that, I get 235 degrees Celsius. So at temperatures greater than 235 degrees Celsius I will have a spontaneous process. Note that at 25 degrees, we were a non-spontaneous process. As we go up to 235 we get a delta G greater than zero. If I continue on past 235, I'm going to see that my delta G value will now become negative. Now let's look at another example. We're going to start this one the same way. We're going to have delta G = delta H- T delta S. My delta G value is -250 kilojoules per mol. And just so I can show you can do it either way, I'm going to leave it in kilojoules. I'm going to convert my S value to joules, so I have -298K(17.0Jper mol Kelvin. And I'm going to convert that joules to kilojoules so 1,000 joules is one kilojoule. And when I solve for my delta G value, I end up with -255 kilojoules per mol. Since this value is less than zero, I'm dealing with a spontaneous process. So delta G less than zero, spontaneous, delta G greater than zero non spontaneous.
