Standard entropies of `x_(2) , y_(2) and xy_(3)` are `60,40 and 50 "J/K 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 should be.

Options

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

Answer

A

Explanation

`Delta S` for the reaction `(1)/(2) X_(2) + (3)/(2) Y_(2) hArr XY_(3)`
`Delta S = 50- (30 + 60)`
`= - 40 J`
For equilibrium `Delta G = 0 = Delta H - T Delta S`
`T= (Delta H )/( Delta S ) = (- 30000)/( - 40)`
`= 750 K`