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

Answer

c

Explanation

`1/2X_(2)+3/2Y_(2)toXY_(3)`
`DeltaS_("reaction")=50-(3/2xx40+1/2xx60)`
`=-40 J mol^(-1), DeltaG=DeltaH-TDeltaS`
at equilibrium , `DeltaG=0`
`DeltaH=TDeltaS`
`30xx10^(3)=Txx40, :. T=750 K`