In Part A of this problem, we are asked to calculate the standard free energy for the reaction. So the standard free energy for this reaction is what we're tying to determine. Now we know the following, that the delta G standard of any reaction can be determined if you're given the delta G as a formation by the following equation. You sum up the delta G of formation of the products. And I'm going to put an n here to represent the number of moles of that, minus the sum of and once again the number of moles and the delta g of formation of reactants. So it's products- reactants or final- initial. Now if we look up there at the balanced equation, we have 1 mole of Cu2O. So we'll take 1 mole that times the delta G of formation of Cu2O which is copper one oxide. The other products, we're going to add the summation of it. We're going to add one half a mole of the delta G of formation of 02. So that's the sum of the products, and then we'll subtract that. We only have one reactant, so it'll be 2 moles. Times the delta G, a formation of CuO which is copper two oxide. Now, let's plug our numbers in. We have 1 mole times, if you look up there to given numbers, we have 166.69 kilojoule per mole plus one half a mol times a negative, oops, no, there's no number given, I was about ready to put that number in there but there's no number given for oxygen. So what do we do here? Well, we have simply got to put in the value that we know it is even though it's not given, and that would be 0. Okay, elemental oxygen, any element in its most stable form, has a value of 0. And now we do -2 moles times a delta G formation of CuO, and that is a -155.2 kilojoules per mole. So that's going to give me a -166.69 kilo joules minus a negative or we'll add to it 310.4 kilo joules. This will give us a value of 143.7 kilo joules for this reaction. So what does a 143.7 or a positive value for the delta G of the reaction mean? Since it's in standard state conditions, that's what this circle means. We know that if we were to start with 1.0 atmosphere of oxygen. And some of the other solid in that reaction chamber, in this reaction, would have to proceed to the left to get to equilibrium. In other words, it's spontaneous in reverse direction.