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 temperaturesC. `Tgt298K`D. `T lt 425K`

For a given reaction`DeltaH=35.5kJmol^(-1)`and`DeltaS=83.6JK^(-1)mol^(

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 temperaturesC. `Tgt298K`D. `T lt 425K`
Answer» Correct Answer - 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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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 temperaturesC. `Tgt298K`D. `T lt 425K`

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