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

Answer

B

Explanation

`(1)/(2) X_(2) + (3)/(2) Y_(2) rarr XY_(3)`
`Delta S= 50- ((60)/(2) + (3)/(2) xx 40)= 50- (30+60)= -40`J/kmol
at equilibrium `Delta G=0`
`Delta H = T Delta S, T= (Delta H)/(Delta S) = (-30 xx 10^(3))/(-40)= 750K`