For a particular reverisble reaction at temperature T,`DeltaH` and `DeltaS` were found to be both `=ve`. If `T_(e)` is the temperature at equilibrium, the reaction would be spontaneous when
A. `T_(e)` is 5 times T
B. `T = T_(e)`
C. `T_(e) gt T`
D. `T gt T_(e)`
A. `T_(e)` is 5 times T
B. `T = T_(e)`
C. `T_(e) gt T`
D. `T gt T_(e)`
Correct Answer – D
`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)`