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• Calculate AH , AS and AG for the reaction below at 115°C. Before you begin calculating, predict signs (+ or -) for enthalpy
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Answer #1

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

1)\DeltaH0rxn = [2\DeltaH0f(NaOH) ] - [\DeltaH0f(Na2O) + \Delta H0f(H2O) ]

\DeltaH0rxn = [2 (-427)] - [-414.2 - 285.8]

\DeltaH0rxn = [-854] - [-700]

\DeltaH0rxn = -154 KJ /mol

\DeltaH0rxn = - ve

2) \Delta S0rxn = [2S0(NaOH)] - [S0Na2O + S0H2O]

\DeltaS0rxn = [2(64.0)] - [75.1 - 70]

\DeltaS0rxn = [128] - ] 145.1 ]

\DeltaS0rxn = -17.1 J/mol K

\DeltaS0rxn = - ve

3) \Delta G0 =  \DeltaH0 - T\DeltaS0

\DeltaG0 = -154 - 388 [-17.1 / 1000]  

\DeltaG0 = -154 + 6.63

\DeltaG0 = - 147.37 kJ/mol

\DeltaG0 = - ve

reaction is spontaneous because  \DeltaG0 = - ve

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