The free energy for a reaction having DeltaH=31400cal,DeltaS=32" cal " – InterviewSolution

1.	The free energy for a reaction having DeltaH=31400cal,DeltaS=32" cal "K^(-1)mol^(-1) at 1000^(@)C is
Answer» `-9336cal` `-7386cal` `-1936cal` `+9336cal` SOLUTION :From Gibbs-Helmholtz EQUATION, we KNOW `DeltaG=DeltaH-T*DeltaS=31400-1273xx32` `=31400-40736=-9336cal`

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