Standar entropies of x_(2), y_(2) and xy_(3) are 60,40 and 50JK^(-1) mol^(-1) respectively for the reaction to be at equilibrium, the temperature should be (1)/(2) x_(2) +(3)/(2) y_(2) hArr xy_(3) Delta H= - 30kJ

Standar entropies of x_(2), y_(2) and xy_(3) are 60,40 and 50JK^(-1) m

1.	Standar entropies of x_(2), y_(2) and xy_(3) are 60,40 and 50JK^(-1) mol^(-1) respectively for the reaction to be at equilibrium, the temperature should be (1)/(2) x_(2) +(3)/(2) y_(2) hArr xy_(3) Delta H= - 30kJ
Answer» 750K 1000K 1250K 500K Solution :`(1)/(2) y_(2) + (3)/(2) y_(2) hArr xy_(3)` `Delta S = Delta S_(xy) - ((1)/(2) sy_(2) + (3)/(2) sy_(2))` `50 - ((1)/(2) xx 60 + (3)/(2) xx 40) = -40JK^(-1) mol^(-1)` `Delta G = Delta H - T Delta S, Delta G = 0 = T = (Delta H)/(Delta S) = 750k`