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` so not vary with temperature)

Options

A. `T lt 425 K`
B. `T gt 425 K`
C. all temperature
D. `T gt 298 K`

Answer

B

Explanation

Accoding to Gibbs-Helmholtz equation,
Gibbs energy `(DeltaG)=DeltaH-TDeltaS`
Where , `DeltaH`=Enthalpy change
`DeltaS` =Entropy change
T=Temperature
For a reaction to be spontaneous
`DeltaG lt 0`
`therefore` Gibbs -Helmholtz equation becomes,
`DeltaG=DeltaH-TDeltaS lt 0`
or , `DeltaH lt TDeltaS`
or, `T gt (DeltaH)/(DeltaS)=(35.5 KJ "mol"^(-1))/(83.6 JK^(-1) "mol"^(-1))=(355xx1000)/(83.6)`
`=425 K`
`T gt 425 K`