1.

For a given reaction, DeltaH=35.5 kJ mol^(-1) and DeltaS=83.6 JK^(-1) mol^(-1). The reaction is spontaneous at : (Assume that DeltaH and DeltaS do not very with temperature)

Answer»

`T gt 425 K`
All TEMPERATURES
`T gt 298 K`
`T LT 425 K`

SOLUTION :`DeltaG=DeltaH-TDeltaS`
for equilibrium `DeltaG=0`
`DeltaH-TDeltaS`
`T_(eq).=(DeltaH)/(DELTAS)=(35.5xx1000)/(83.6)=425 K`
Since the reaction is endothermic it will be spontaneous at `T gt 425 K`.


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