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For a particular reversible reaction at temperature T, `DeltaHandDeltaS` 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)`

Text Solution

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The correct Answer is:
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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