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

Answer:

Temperature = 818 K

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:

$\Delta H^{0} = T\Delta S^{0}$

$T=\frac{\Delta H^{0}}{\Delta S^{0}}=\frac{-109}{-0.1332}=818 K$