For a particular reverisble reaction at temperature T,DeltaH and DeltaSwere found to be both =ve. If T_(e) is the temperature at equilibrium, the reaction would be spontaneous when

For a particular reverisble reaction at temperature T,DeltaH and Delta

1.	For a particular reverisble reaction at temperature T,DeltaH and DeltaSwere found to be both =ve. If T_(e) is the temperature at equilibrium, the reaction would be spontaneous when
Answer» `T_(e)` is 5times T `T = T_(e)` `T_(e) gt T` `T gt T_(e)` SOLUTION :`DELTAG =DeltaH - T DELTAS`. At equilibrium, `DeltaG = 0`. Hence,`T_(e)DeltaS = DeltaH`. As `DeltaH `and `DeltaS` are`+ve` , for REACTION to be spontaneous `DeltaG` should be `-ve`. This can be so only if `T DELTA S gt DeltaH `, i.e., `T Delta S gt T_(e) DeltaS ` or `T gt T_(e)`