Discussion Forum

1)

Standard entropy  of X₂,Y₂  and XY₃  are 60, 40 and 50 JK^-1 mol^-1 , respectively , For the reactions

$\frac{1}{2}X_{2}+\frac{3}{2}Y_{2}\rightarrow XY_{2}, \triangle H=-30kJ, $   , to be at equilibrium , the temperature will be 

Answer:

Option C

Explanation:

For a reaction to be equilibrium   $\triangle$ G=0,   Since ,  $\triangle G=\triangle H-T\triangle S$ sop at equilibrium

 $\triangle H -T\triangle S=0$

 For the reaction

 $\frac{1}{2}X_{2}+\frac{3}{2}Y_{2}\rightarrow XY_{3}$   ;   $\triangle$ H= -30kJ

(given)

 Calculating $\triangle$ S for the above reaction , we get

 $\triangle S= 50-\left[\frac{1}{2}\times 60+\frac{3}{2}\times40\right]JK^{-1}$

 = 50-(30+60)=JK^-1 =-40 JK^-1 

At equilibrium , T $\triangle$ S= $\triangle$ H

           [$\because$   $\triangle$ G=0]

 $\therefore$     $T\times (-40)=-30\times1000[\because 1kJ=1000J]$

 or        $T= \frac{-30\times 1000}{-40} or 750 K$