Standard entropt 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. `1250K`
B. `500K`
C. `750K`
D. `1000K`

Answer

A

Explanation

`(1)/(2)X_(2)+(3)/(2)Y_(2)rarr XY_(3)`
`DeltaS_("reaction")=50-((3)/(2)xx40+(1)/(2)xx60)=-40Jmol^(-1)`
`DeltaG=DeltaH-TDeltaS`
At equilibrium `DeltaG=0`
`DeltaH=TDeltaS`
`30xx10^(3)=Txx40`
`:. T=750`