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 whenA. `T_(e)` is 5 times TB. `T = T_(e)`C. `T_(e) gt T`D. `T gt T_(e)`

For a particular reverisble reaction at temperature T,`DeltaH` and `De

1.	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 whenA. `T_(e)` is 5 times TB. `T = T_(e)`C. `T_(e) gt T`D. `T gt T_(e)`
Answer» 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)`

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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 whenA. `T_(e)` is 5 times TB. `T = T_(e)`C. `T_(e) gt T`D. `T gt T_(e)`

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