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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` |
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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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