Question

Standard entropies of X2, Y2 and XY3 are 60,40 and 50JK–1mol–1 respectively. For the reaction 12X2+32Y2⇋XY3,ΔH=−30kJ to be at 2 equilibrium, the temperature should be:

Answer 1

Answer:

The correct option is : (a) 750 K

Explanation:

Given reaction is :

1/2 X₂ + 3/2 Y₂⇄XY₃

we know,

ΔS°=∑S°products - ∑S°reactant

     =50 - (30+60)

     =-40 J K^-1 mol^-1

at equilibrium ΔG° = 0

ΔH° = TΔS°

∴T = ΔH°/ΔS°

 T = -30 × 10³ J mol^-1 / -40 J K^-1 mol^-1

∴T = 750 K

Answer 2

Answer:

It concept good..... it will help you.... yuvraj_sharan