 Okay guys, let's start the day today with calculating the standard entropy of reaction. So the actual problem says calculate the delta S in standard conditions of the reaction for the combustion of one mole of propane at 25 degrees Celsius. So in order to do these types of problems, if you see an unbalanced equation, you've got a balanced first, okay? So let's just balance it, so 5, what do we have, 3 and 4 that are correct? So now it's balanced to one mole of propane here, okay, so that's propane. So how do I do this? Well, you're going to get a table of standard entropy values or they may be given to you in the particular problem, okay? And what you're going to do is just like how you did with enthalpy, okay? So the change in standard entropy of the reaction is going to equal the sum of the moles times the entropy of the products minus the sum times the moles of the entropy of the reactants, okay? So let's write that down. So the products, the first one is carbon dioxide there, so we go over to our table, okay, and we see standard entropy of carbon dioxide there, so bracket, okay, so there's 3 moles times standard entropy, we'll start mole kiln, then we're going to add that to the 4 moles of water down there, remember you've got to get these in the right state of matter. So if you write down the entropy for moles of water or the entropy of water in gaseous state it's going to be different than the entropy of water in the liquid state, okay? So those are all the products there, I like to bracket them, and then we're going to subtract those from the reactants here, so 1 mole of propane up there, 269, and that to 5 moles of oxygen, so hopefully everybody could have said that one, and then now it's just a plug-in shed, so I'm sorry, joules, also say joules per mole kelvin and not use these moles in here, okay, either way would be fun, the answer usually will be presented in joules per mole kelvin for these entropy values, okay? Are there any questions about this one? Pretty straightforward.