Standard entropies of X_(2), Y_(2) and XY_(3) are 60,40 and 50 kJ"mol"^(-1) respectively. For the reaction: 1/2 X_(2) + 3/2Y_(2)iff 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 kJ"mol"

1.	Standard entropies of X_(2), Y_(2) and XY_(3) are 60,40 and 50 kJ"mol"^(-1) respectively. For the reaction: 1/2 X_(2) + 3/2Y_(2)iff XY_(3), Delta H = -30 kJ to be at equilibrium, the temperature should be
Answer» 500 K 750 K 1000 K 1250 K Solution :`DeltaS^(@) = S^(@)(XY_(3)) -1/2S^(@)(X_(2)) - 3/2S^(@)(Y_(2))` ` = 50 -1/2(60)-3/2(40) = -40J K^(-1) "mol"^(-1)` `T = (DeltaH^(@))/(DeltaS^(@)) = (-30 xx 10^(3) J"mol"^(-1))/(-40 J K^(-1)"mol"^(-1))` = 750 K