This question is about energy changes. Hydrogen peroxide, H2O2\text{H}_2\text{O}_2H2O2, decomposes in the gas phase as shown in Reaction 1:
H2O2(g)→H2O(g)+12O2(g)Reaction 1 \text{H}_2\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \quad \text{\textbf{Reaction 1}} H2O2(g)→H2O(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 |
|---|---|---|
| H2O2(g)\text{H}_2\text{O}_2(\text{g})H2O2(g) | −136.0-136.0−136.0 | 233233233 |
| H2O(g)\text{H}_2\text{O}(\text{g})H2O(g) | −242.0-242.0−242.0 | 189189189 |
| 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 hydrogen peroxide 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.
