This question is about energy changes. Nitrous oxide, N2O\text{N}_2\text{O}N2O, decomposes as shown in Reaction 1:
N2O(g)→N2(g)+12O2(g)Reaction 1 \text{N}_2\text{O}(\text{g}) \rightarrow \text{N}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \quad \text{\textbf{Reaction 1}} N2O(g)→N2(g)+21O2(g)Reaction 1The table shows enthalpy changes of formation and standard entropies: | Substance | ΔHf⊖\Delta H_{\text{f}}^\ominusΔHf⊖ / kJ mol−1\text{kJ mol}^{-1}kJ mol−1 | S⊖S^\ominusS⊖ / J K−1 mol−1\text{J K}^{-1}\text{ mol}^{-1}J K−1 mol−1 | | :--- | :---: | :---: | | N2O(g)\text{N}_2\text{O}(\text{g})N2O(g) | +82.0+82.0+82.0 | 220220220 | | N2(g)\text{N}_2(\text{g})N2(g) | 000 | 192192192 | | O2(g)\text{O}_2(\text{g})O2(g) | 000 | 205205205 | Calculate the free-energy change, ΔG\Delta GΔG, in kJ mol−1\text{kJ mol}^{-1}kJ mol−1, of Reaction 1 at 25 ∘C25\ ^\circ\text{C}25 ∘C. Give your answer to 3 significant figures.
The decomposition of nitrous oxide shown in Reaction 1 is thermodynamically feasible. Suggest why Reaction 1 does not take place at 25 ∘C25\ ^\circ\text{C}25 ∘C despite being feasible.
