For the reaction at `298 K`
`2A+B rarr C`
`DeltaH=400 kJ mol^(-1)` and `DeltaS=0.2 kJ K^(-1) mol^(-1)`
At what temperature will the reaction becomes spontaneous considering `DeltaH` and `DeltaS` to be contant over the temperature range.

Answer

2000K

Explanation

From the expression,
`DeltaG=DeltaH-TDeltaS`
Assumingb the reaction at equilibrium , `DeltaT` for the reaction would be :
`T=(DeltaH-DeltaG)(1)/(DeltaS)`
`(DeltaH)/(DeltaS)(DeltaG=0` at equilibrium )
`=(400 kJ "mol"^(-1))/(0.2 kJ^(-1)"mol"^(-1))`
T=2000 K
For the reaction to be spontaneous, `DeltaG` must be negative . Hence , for the given reaction to be spontaneous, T should be greater than 2000 K.