`DeltaH` and `DeltaS` for a reaction are `+30.558 kJ mol^(-1)` and `0.66 kJ mol^(-1)` at `1` atm pressure. The temperature at which free energy is equal to zero and the nature of the reaction below this temperature are

Options

A. `483 K`, spontaneous
B. `443 K`,non- spontaneous
C. `443 K`, spontaneous
D. `463 K`,non- spontaneous

Answer

D

Explanation

`DeltaG=DeltaH-TDeltaS=0`
`:. T=(DeltaH)/(DeltaS)=(+30.558kJmol^(-1))/(0.066 kJ K^(-1) mol^(-1))=463 K`
`DeltaG=DeltaH-TDeltaS`
`DeltaG=(30.558-Txx0.066) kJ mol^(-1)`
At `T lt 463 K`,
`0.066 T lt 463xx0.066`
`0.066 T lt 30.558`
`0 lt 30.558-0.066T`
`0 lt DeltaG`
Thus at `T lt 463 K`, `DeltaG gt ` i.e., process is non spontaneous