Please explain how answer was derived. The species in the reaction, 2 KNO3(s) +

Question

Please explain how answer was derived.

The species in the reaction, 2 KNO3(s) + 2 C(s) rightarrow K2C03(s) + N2(g) + 3/2 CO2(g) have the values for standard enthalpies of formation at 25 degree C: KN03(s), delta H degree f= -492.7 kJ mol-1 C02(g), delta H degree f= -394.0 kJ mol-1 K2C03(s), delta H degree f= -1146 kJ mol-1 Make the assumption that, since the physical states do not change, the values of delta H degree and atm = delta s degree arc constant throughout a broad temperature range, and use this information to determine which of the following conditions may apply. The reaction is spontaneous at all temperatures. The reaction is non-spontaneous at all temperatures. The reaction is spontaneous at low temperatures but non-spontaneous at high temperatures. The reaction is non-spontaneous at low temperatures but spontaneous at high temperatures. It is impossible to make any judgment about spontaneity because insufficient information can be gleaned from the data and equations presented.

Explanation / Answer

The balanced reaction is given by

change in the number of gaseous moles dn = 1 + 3/2 -0 = 5/2

so dn = +ve  

so dS   for the reaction is + ve

consider dH

dH =   dHf K2C03 + dHf N2 + 3/2 dHf   C02   - 5/2 x dHF C - 2 dHf KN03

dH = -1146 +0 -3/2 x 394 -0 + 2 x 492.7

dH = - 751.6 kJ /mol

we know that

dG = dH - TdS

for a reaction to be spontaneous    dG <0

so   dH - TdS <0

from the above we get that    dS = +ve   and   dH = -ve

so       dH - TdS   is always -Ve

so   dG   = -Ve    for very temperatures

so the answer is  

a)   the reaction is spontaneous at all temperatures

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