Answer:
Temperature = 818 K
Explanation:
The given reaction is:
[tex]2Na2O2 + H2O \rightarrow 4NaOH + O2[/tex]
The change in enthalpy is: ΔH°= -109 kJ
The change in entropy is : ΔS° = -133.2 J/k = -0.1332 kJ/K
Based on the Gibbs-Helmholtz equation the standard free energy change is given as:
[tex]\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}[/tex]
At equilibrium ΔG° = 0. therefore:
[tex]\Delta H^{0} = T\Delta S^{0}[/tex]
[tex]T=\frac{\Delta H^{0}}{\Delta S^{0}}=\frac{-109}{-0.1332}=818 K[/tex]
