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

Explanation

Let us first calculate the temperature at which the reaction will be in equilibrium, i.e., `DeltaG=0`.
Now, `DeltaG=DeltaH-T DeltaS`
`:. 0=DeltaH-T DeltaS`
or `T=(DeltaH)/(DeltaS)=(400 kJ mol^(-1))/(0.2 kJ K^(-1) mol^(-1))=2000 K`
For reaction to be spontaneous, i.e., for `DeltaG` to be -ve `T` should be greater than `2000 K`.