Final answer:
To calculate ΔG°rxn at 25 °C for the reaction NO(g) + ½ O₂(g) → NO₂(g), we use ΔG°rxn = ΔH°rxn - TΔS°rxn. Given the values for ΔH° and ΔS°, we can calculate ΔG°rxn to be 19.8 kJ. Since ΔG°rxn is positive, the reaction is non-spontaneous.
Step-by-step explanation:
To calculate the standard free-energy change (ΔG°rxn) at 25 °C for the reaction NO(g) + ½ O₂(g) → NO₂(g), we need to use the equation ΔG°rxn = ΔH°rxn - TΔS°rxn.
Given:
- ΔH° = 91.3 kJ
- ΔS° = 240.1 J/K
- T = 25 °C = 298 K
Converting ΔH° to J (1 kJ = 1000 J), we have ΔH° = 91.3 kJ = 91,300 J. Now, we can calculate ΔG°rxn:
ΔG°rxn = 91,300 J - (298 K)(240.1 J/K) = 91,300 J - 71,459.8 J = 19,840.2 J = 19.8 kJ
Since ΔG°rxn is positive (19.8 kJ), the reaction is non-spontaneous. Therefore, the correct option is c) ΔG°rxn = -18.1 kJ; Non-spontaneous.