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

Answer

D

Explanation

`(1)/(2)X_(2)+(3)/(2)Y_(2)toXY_(3)`
`DeltaS_("reaction")^(@)=DeltaS_(XY_(3))^(@)-(1)/(2)DeltaS_(X_(2))^(@)-(3)/(2)DeltaS_(Y_(2))^(@)`
`=50-(1)/(2)xx60-(3)/(2)xx40=40JK^(-1)" "mol^(-1)`
`DeltaG=DeltaH-TDeltaS=0` at equilibrium
`thereforeT=(DeltaH)/(DeltaS)=(-30xx1000)/(40)=750K`