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» 500 K 750 K 1000 K 1250 K Solution :`Delta S = 50 - ((60)/(2) + (3)/(2) XX 40)` `= 50 (30+ 60) = - 40 "J/K mol"` `Delta G = 0` `therefore Delta H = T Delta S therefore T = (-30 xx 10^(3) )/( - 40) = 750 K`