Now we're going to look at another example of calculating Gibbs free energy. The question asks us to calculate the delta G at 25 degrees Celsius based on the given information and decide whether or not the reaction is spontaneous. We're given the balanced chemical equation. We're given delta H of formation values and entropies of substances, but we're not given the enthopy or the entropy of a particular reaction. So the first thing we're going to have to do is calculate both the delta H of formation and the delta S of the system. Remember that when we're looking at our Gibbs free energy, we're using the expression delta G equals delta H minus T delta S. This is the formula we'll use to be able to calculate delta G and determine the spontaneity of the reaction. For our delta H value, we know we want to have delta H of the reaction = delta H of formation of the products- delta H of formation of the reactants. And since all of our coefficients are 1, we don't have to worry about any coefficients in our calculation. So delta H of reaction = delta H of our products, which is our COCl2. So minus 219.1 kilojoules per mol, and there's 1 mol of that,- (-110.5 KJ per mole for the carbon monoxide), again x 1), because there's 1 mole. For the chlorine, I don't need to put anything in there. That's an element in its standard state, so it has enthalpy of formation of 0. Then I find delta H of reaction = -108.6 kilojoules. This is an exothermic process because the delta H value is negative, so we're losing energy from the system. Now I need to calculate the value of delta S of the reaction. Just like I did with my delta H of reaction, I'm going to look at my delta S of reaction = S of the products- S of the reactants. So delta S of the reaction = S of my products, which I have one product, COCl2. So I have 283.5 joules per mol Kelvin. And I'm going to subtract from that the entropy of both of my reactants, so the entropy of CO, which is 197.7 joules per mol Kelvin- the Cl2. Entropies of pure substance or elements in there standard state is not 0, unlike the delta H values. And so, I have 223.1 joules per mol Kelvin. And what I find is that the delta S of my reaction = -137.3 joules per Kelvin. Because each of these is 1 mol, my mols will have cancelled out as well. So now I know my delta S of reaction. I also know the delta H of my reaction. Now, I can use that information with my delta G = delta H- T delta S to solve for delta G. And what I find is delta G = -108.6 kilojoules- T. I'm at 25 degrees Celsius, but I need to convert that to Kelvin. So that's 298 Kelvin x delta S, which is -137.3 joules per Kelvin. But I also need to divide by 1,000 because I want to convert that to kilojoules, so that my entropy value and my enthalpy value are both with respect to kilojoules. Now I have delta G = -67.7 kilojoules. And because delta G is a negative value, this is a spontaneous process at 25 degrees Celsius.