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 temperatures
D. `T gt 298 K`

Answer

B

Explanation

`:' Delta G=DeltaH-TDeltaS`
For a reaction to be spontaneous , `DeltaG=-ve `
i.e., `DeltaH lt T DeltaS`
`:. T gt (DeltaH)/(DeltaS)=(35.5 xx 10^(3)J)/(83.6 JK^(-1))`
i.e., ` T gt 425 K`