For a certain process, DeltaH=280 kJ and DeltaS=140 J K^(-1) "mol"^(-1 – InterviewSolution

1.	For a certain process, DeltaH=280 kJ and DeltaS=140 J K^(-1) "mol"^(-1) .What is the minimum temperature at which the process will be spontaneous?
Answer» 2000 K 1200 K 1400 K 1420 K SOLUTION :Temperature has MINIMUM VALUE at equilibrium i.e., at `DeltaG=0` . SINCE `DeltaG=DeltaH-TDeltaS, therefore T=(DeltaH)/(DeltaS)` `T=(280xx10^3)/140` = 2000 K

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