Standard entropies of` X_(2), Y,_(2) and XY_(3)` are 60,40 and 50`JK^(-1) mol^(-1)` respectively. For the reaction:
`(1)/(2) X ^(2) + (3)/(2) Y _(2) to XY _(3)`
`Delta H =- 30 kJ` to be at equilibrium, the temperature will be:

Options

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

Explanation

`(1)/(2) X _(2) + (3)/(2) Y_(2) to XY _(3)`
`Delta S _("Reaction") ^(@) = Delta _(XY_(3)) ^(@) - (1)/(2) Delta S _(X _(2)) ^(@) - (3)/(2) Delta S _(Y _(2)) ^(@) = 50 - (1)/(2) xx 60 - (3)/(2) xx 40 = - 40 JK ^(-1) mol ^(-1)`
`Delta G = Delta H - T Delta S = 0` at equilibrium `therefore T = ( Delta H )/( Delta S ) = ( 30 xx 1000)/(40) = 750 K`