Question 17.7: (a) Calculate the standard free-energy change for the oxidat......
(a) Calculate the standard free-energy change for the oxidation of ammonia to give nitric oxide (NO) and water. Is it worth trying to find a catalyst for this reaction under standard-state conditions at 25 °C?
4 NH_{3}(g) + 5 O_{2}(g)\longrightarrow 4 NO(g) + 6 H_{2}O(l)(b) Is it worth trying to find a catalyst for the synthesis of NO from gaseous N_{2} and O_{2} under standard-state conditions at 25 °C?
STRATEGY
We can calculate ΔG° most easily from standard free energies of formation using the formula:
ΔG° = ΔG°_{f}(products) – ΔG°_{f}(reactants)Remember to multiply the ΔG° value for each substance by its coefficient in the balanced chemical reaction. It’s worth trying to find a catalyst for a reaction only if the reaction has a negative free-energy change.
| IDENTIFY | |
| Known | Unknown |
| Standard free energies of formation ΔG°_{f} (Appendix B) | ΔG° |
(a) ΔG° = [4 ΔG°_{f}(NO) + 6 ΔG°_{f}(H_{2}O)] – [4 ΔG°_{f}(NH_{3}) + 5 ΔG°_{f}(O_{2})]
= [(4 mol)(87.6 kJ/mol) + (6 mol)(-237.2 kJ/mol)] – [(4 mol)(-16.5 kJ/mol) + (5 mol)(0 kJ/mol)]
ΔG° = -1006.8 kJ
It is worth looking for a catalyst because the negative value of ΔG° indicates that the reaction is spontaneous under standard-state conditions. (This reaction is the first step in the Ostwald process for the production of nitric acid. In industry, the reaction is carried out using a platinum–rhodium catalyst.)
(b) It’s not worth looking for a catalyst for the reaction
N_{2}(g) + O_{2}(g)\longrightarrow 2 NO(g) because the standard free energy of formation of NO is positive (ΔG°_{f} = 87.6 kJ/mol). This means that NO is unstable and will decompose to N_{2} and O_{2} under standard-state conditions at 25 °C. A catalyst would only increase the rate of its decomposition. A catalyst can’t affect the composition of the equilibrium mixture (Section 14.11), and so it can’t reverse the direction of the reaction.