Question

• Calculate AH , AS and AG for the reaction below at 115°C. Before you begin calculating, predict signs (+ or -) for enthalpy, entropy, and free energy of reaction. • Is the reaction spontaneous at 115°C? Na 0 (8) + H2O() → NaOH () Na,0 (s) HO (D) NaOH (s) AH°, (kJ/mol) 414.2 -285.8 -427.0 Sº (J/K-mol) 75.1 70.0 64.0

Answer 1

Na2O + H2O ---------> 2NaOH

1)$\Delta$H⁰_rxn = [2$\Delta$H⁰_f(NaOH) ] - [$\Delta$H⁰_f(Na2O) + $\Delta$ H⁰_f(H2O) ]

$\Delta$H⁰_rxn = [2 (-427)] - [-414.2 - 285.8]

$\Delta$H⁰_rxn = [-854] - [-700]

$\Delta$H⁰_rxn = -154 KJ /mol

$\Delta$H⁰_rxn = - ve

2) $\Delta$ S⁰_rxn = [2S⁰_(NaOH)] - [S⁰_Na2O + S⁰_H2O]

$\Delta$S⁰_rxn = [2(64.0)] - [75.1 - 70]

$\Delta$S⁰_rxn = [128] - ] 145.1 ]

$\Delta$S⁰_rxn = -17.1 J/mol K

$\Delta$S⁰_rxn = - ve

3) $\Delta$ G⁰ =  $\Delta$H⁰ - T$\Delta$S⁰

$\Delta$G⁰ = -154 - 388 [-17.1 / 1000]  

$\Delta$G⁰ = -154 + 6.63

$\Delta$G⁰ = - 147.37 kJ/mol

$\Delta$G⁰ = - ve

reaction is spontaneous because  $\Delta$G⁰ = - ve