Standard entropy of `X_(2)` , `Y_(2)` and `XY_(3)` are `60, 40 ` and `50JK^(-1)mol^(-1)` , respectively. For the reaction, `(1)/(2)X_(2)+(3)/(2)Y_(2)rarrXY_(3),DeltaH=-30KJ` , to be at equilibrium, the temperature will be:

Options

A. `1000 K`
B. `1250 K`
C. `500 K`
D. `750 K`

Answer

D

Explanation

In the reaction
`1//2X_(2)+3//2Y_(3)hArrXY_(3)`, `DeltaH=-30kJ`
`DeltaS^(@)=S_((XY_(3)))^(@)[1//2S_(X_(2))^(@)+3//2S_(Y_(2))^(@)]`
`=50[1//2xx60+3//2xx40]=50[30+60]`
`=(50-90)=-40 J K^(-1) mol^(-1)`
`DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
At equilibrium, `DeltaG^(@)=0`
`:. DeltaH^(@)=TDeltaS^(@)` or `T=(DeltaH^(@))/(DeltaS^(@))`
`T=(-30xx10^(3)J)/(-40JK^(-1)mol^(-1))=(3000)/(40)K=750 K`