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)A. `T gt 425 K`B. All temperaturesC. `T gt 298 K`D. `T lt 425 K`

Answer» Correct Answer - A
`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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