Suppose we redefine the standard state as P=5 bar. Find the new standard fG values of the following substances. Part A: HCl(g) Part B: N2O(g) Part C: H(g)

My Answers: Part A: -95.3 kJmol1 Part B: 101.7 and for Part C: 205.3

I need help with parts D-F please.

Part D: Explain the results in terms of the relative entropies of reactants and products of the reaction in part A.

- Since the number of moles of reactants and products are the same, the decrease in volume affects the entropy of both equally, so there is no change in fG

-The entropy of the reactants is decreased more than the entropy of the product. Since the product is relatively more favored, there is no change in fG

-The entropy of the product is decreased more than the entropy of the reactant. Since the product is relatively less favored, there is no change in fG

Part E: Explain the results in terms of the relative entropies of reactants and products of the reaction in part B.

- Since the number of moles of reactants and products are the same, the decrease in volume affects the entropy of both equally, so fG is less positive.

-The entropy of the reactants is decreased more than the entropy of the product. Since the product is relatively more favored, fGis less positive.

-The entropy of the product is decreased more than the entropy of the reactant. Since the product is relatively less favored, fG is less positive.

Part F: Explain the results in terms of the relative entropies of reactants and products of the reaction in part C.

- Since the number of moles of reactants and products are the same, the decrease in volume affects the entropy of both equally, so fG is more positive.

- The entropy of the reactants is decreased more than the entropy of the product. Since the product is relatively more favored, fG is more positive.

- The entropy of the product is decreased more than the entropy of the reactant. Since the product is relatively less favored, fG is more positive.