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.

Standard entropies of X_(2) , Y_(2) and XY_(3) are 60,40 and 50 "J/K m

1.	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.
Answer» 750 K 1000 K 1250 K 500 K Solution :`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`