Question

At what temperature is the following reaction at equilibrium? 2Na, O, + 2H,0 + 4NaOH + O, AH° = -109 kJ, ASO = -133.2 J/K

Answer 1

Temperature = 818 K

Step-by-step explanation:

The given reaction is:

`2Na2O2 + H2O \rightarrow 4NaOH + O2`

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:

`\Delta G^(0)=\Delta H^(0)-T\Delta S^(0)`

At equilibrium ΔG° = 0. therefore:

`\Delta H^(0) = T\Delta S^(0)`

`T=(\Delta H^(0))/(\Delta S^(0))=(-109)/(-0.1332)=818 K`