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For a given reactionDeltaH=35.5kJmol^(-1...

For a given reaction`DeltaH=35.5kJmol^(-1)`and`DeltaS=83.6JK^(-1)mol^(-1)`.The reaction is spontaneous at :(Assume that `DeltaH`and`DeltaS` do no vary with temperature)

A

`Tgt425K`

B

All temperatures

C

`Tgt298K`

D

`T lt 425K`

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaG=DeltaH-TDeltaS`
For equilibrium `DeltaG=0`
`DeltaH-TDeltaS`
`T_(eq).=(DeltaH)/(DeltaS)=(35.5xx1000)/(83.6)=425K`
Since the reaction is endothemic it will be spontaneous at `Tgt 425K` . Option (1)
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