For a given reaction, DeltaH = 35.5 kJ"mol"^(-1) and DeltaS = 83.6 J K^(-1) "mol"^(-1). The reaction is spontaneous at (assume that DeltaH and DeltaS do not vary with temperature.

For a given reaction, DeltaH = 35.5 kJ"mol"^(-1) and DeltaS = 83.6 J K

1.	For a given reaction, DeltaH = 35.5 kJ"mol"^(-1) and DeltaS = 83.6 J K^(-1) "mol"^(-1). The reaction is spontaneous at (assume that DeltaH and DeltaS do not vary with temperature.
Answer» `T gt425K` all temperatures `T lt298K` `T lt425K` Solution :`DeltaG = DeltaH -TDeltaS` For a reaction to be at EQUILIBRIUM, `DeltaG =0` `DeltaH = TDeltaS` or `T = (DeltaH)/(DeltaS) = (35.5 xx 10^(3))/(83.6)` ` = 425 K` Since reaction is endothermic, it will be sponatneous at `T gt 425 K`