For a given reaction, DeltaH = 35.5kJ mol^(-1) and DeltaS = 83.6 JK^(-1) mol^(-1) .The reaction s spontaneous at : ( Assume atDeltaHand DeltaS do not vary with temperature )

For a given reaction, DeltaH = 35.5kJ mol^(-1) and DeltaS = 83.6 JK^(-

1.	For a given reaction, DeltaH = 35.5kJ mol^(-1) and DeltaS = 83.6 JK^(-1) mol^(-1) .The reaction s spontaneous at : ( Assume atDeltaHand DeltaS do not vary with temperature )
Answer» `T lt425 K` `T gt 425 K` All temperatures `T gt 298 K` Solution :`DeltaS= DeltaH - T DeltaS` For the reaction to be spontaneous , `DeltaG = - ve` As`DeltaH` and`DeltaS` both are positive,`DeltaG` can be `-ve` only if`T DeltaS gt DeltaH`or`T gt ( DeltaH)/( DeltaS)` `i.e,T gt ( 35.5xx 1000J)/( 83.6JK^(-1)) ` or `T gt 425 K`