In each part of this problem, we are provided a chemical reaction and we first need to calculate the value of the change in inter P for the overall reaction. And then we need to be able to rationalize why the sign of the changing in standard more entropy is either positive or negative in each case, this is the equation that we used to solve for help s of the reaction we take the total and trippy of the products and subtract the total entropy of the reactant. This is a reaction that were given in part A. So if we use this equation above, we first calculate SP and then s are we can subtract the to to get overall delta s of the reaction at standard conditions. So SP is a total entropy of the products we look at the equation that we have. We know that the products are on the right side of the reactionary Ruth. So we have two moles of H and 03 and one mole of I know So for the products again, that's two moles of H and 03 and one mole of Eno. Then we look up in the appendix, the values for the standard Mueller enter peace for each one of those molecules. For H and 03 this is a value and for n no, this is the value. We add them together to get the total entropy of the products. Since both of those air products, when we want to play the number of moles with its danger more entropy, units of most cancel out and we're left with jewels per Calvin as a total entropy unit. So for the products, this comes out to be the total entropy. Then we do the same for the reactant the reactant, sir. Everything to the left of the reaction hero. That's three moles of CO two and one mole of H 20 So three moles of n 02 and one mole of H 20 as a liquid state of matter. It does make a difference when looking up the standard molar entropy values in the appendix. So for the n 02 as a gas, this is it's dinner, more entropy value. And this is a standard Miller entropy value for water as a liquid. If we look those up in the appendix, we add them together to get the total entropy of the reactant and that comes out to this value. So now that we've calculated S, p and s, are we just attracted to to find Delta s of the reaction. So this is a total entropy of the products we subtract, the total entropy of the reactant. And when we do that, we find that Delta s of the overall reaction comes out to about negative 200 88 jewels her Kelvin. And we see that the change in entropy comes out to be a negative value. In order to rationalize that change, we need to examine the total number of moles of gas on the side of the reaction. Here we see that we have just three moles of gas on the reactive side. Since water is a liquid and we have one mole of gas on the product side since 18 03 is a quickest. So we decrease the total number of moles of gas from three down to one. And since we decrease the number of bowls of gas, that means that we decreased the amount of disorder, which means that we decreased the entropy and when we decrease the entropy is a result of a chemical reaction. That must mean that we had a negative change in the entropy. So that's why we expect that to become a negative value for Part B. This is the reaction that were given, and we walk through the same process. Starting with the product side. We have two moles of solid chromium and three moles of gashes Co two. We can look up each one of the standard molar entropy values for those substances, and we add them together to get the total and trippy of the products. And for the reactant swe have one mole of CR, two of three is a solid and on three moles of carbon monoxide as a gas, we look up their standard molar entropy is in the appendix. Multiply them by the number of molds of each one of those participating on the react. Inside, we add them together to get the total entropy of the reactive. Then we take the total entropy of the products and subtract the total entropy of the reactant that gives us tell Tess of the reaction of standard conditions. And when we calculate that all out that comes out to a positive value of 14 0.7 jewels per Kelvin. This time change in entropy is positive. We left. We see we have three moles of gas on the reactive side and three moles of gas on the product side. So since the total number of bowls of gas doesn't change very much, we should expect the Delta s of the overall reaction to be pretty close to zero. And we see that a small positive value. 14.7 jewels per killed In alliance with that theory in part C, this is the reaction that were given, starting with the products we have. One mole of eso three is a gas. We look up, it's dinner Miller entropy value. And that's the only product. So that ends up being the total entropy of the product side of this reaction. And on the reactant side, we have one mole of eso to gas and a half more of 02 gas. Remember that even though 02 is a naturally occurring die atomic molecule, it still has a non zero value for its dinner Miller entropy. Unlike it's delta h of formacion value and it's Delta G information value. So we do have to include it when we're calculating the total entropy of the reactions. So we look up, it's dinner, more inter fees of both of those and software the total entropy of the reactions in the same way that we have been. And then we take the total entropy of the products and subtract the total entropy of the reactions to get Delta s of the reaction at senior conditions. And for part C, that value should come out to negative 94 Jules Per Kelvin. This time Delta s is negative, as it was in part a And if we look, we see that we have one plus one half so 1.5 moles of gas on the reactant side in one mole of gas on the product side. So as we go from the reactions to the products as a result of this chemical reaction, we decrease the total number of moles of gas. When we decrease the total number of most of gas, we decrease the amount of disorder and therefore, decrease the entropy. If we decrease the entropy, that must mean that the reaction has a negative change in entropy. So that's why that came out to be a negative value. Finally and party. This is the reaction that were given. We start with the products again. We have one month of N two as a gas and four moles of water vapor. You look up their standard molar entropy values and then add the total Enter peas of each one of those together to get the total entropy of the products. Then for the reactant swe, have one mole of into a four gas and formals of H two gas. And again, just like for 02 and part See, we have to include H to an end to because they have non zero values for their standard Mueller enter peas. So again we take the total entropy of the products and subtract the total entropy of the reactant. And that will give us built s of the reaction. And that comes out to be about 120 Jules for Kelvin. And this is a positive value. We can see that on the product side, we have one plus four. So five total moles of gas and one plus four So still, five photo moles of gas on each side. So we should expect this to be a relatively small value for the change in standard Mueller entropy, since the number of moles of gas doesn't change very much and it does come out to be a fairly small positive value, which is what we expected.