For a particular reversible reaciton at temperature `T, DeltaH` and `DeltaS` were found to be both `+ve`. If `T_e` is the temperature at equilibrium, the reaciton would be spontaneous when :A. `T_(e)gtT`B. `TgtT_(e)`C. `T_(e)` is `5` times `T`D. `T=T_(e)`

For a particular reversible reaciton at temperature `T, DeltaH` and `D

1.	For a particular reversible reaciton at temperature `T, DeltaH` and `DeltaS` were found to be both `+ve`. If `T_e` is the temperature at equilibrium, the reaciton would be spontaneous when :A. `T_(e)gtT`B. `TgtT_(e)`C. `T_(e)` is `5` times `T`D. `T=T_(e)`
Answer» Correct Answer - b `DeltaG=DeltaH-TDeltaS` at equilibrium, `DeltaG=0` so, `TDeltaS=DeltaH` As `DeltaH` and `DeltaS` are `+ve`, for a reaction to be spontaneous `DeltaG` should be -ve. `TDeltaSgtDeltaH i.e,. TDeltaSgtT_(e)DeltaS` `:. TgtT_(e)`

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For a particular reversible reaciton at temperature `T, DeltaH` and `DeltaS` were found to be both `+ve`. If `T_e` is the temperature at equilibrium, the reaciton would be spontaneous when :A. `T_(e)gtT`B. `TgtT_(e)`C. `T_(e)` is `5` times `T`D. `T=T_(e)`

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