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

Answer

A

Explanation

`Delta S=sum S_(p)-sum S_(R)`
`Delta S =50 -((1)/(2)xx60+(3)/(2)xx40)`
`DeltaS=-40JK^(-1)mol^(-1)`
`Delta G=DeltaH-TDeltaS`
At equilivrium `DeltaG=0`
`therefore T=(DeltaH)/(DeltaS)=(-30xx10^(3))/(-40)`
`T=750K`