Question

Standard entropy of $X_2,Y_2$ and $X{Y}_3$ are $60,40$ and $50J{K}^{-1}{mol}^{-1}$ respectively. For the reaction:
$\frac{1}{2}{X}_2+\frac{3}{2}{Y}_2⟶X{Y}_3,\Delta H=-30kJmo{l}^{-1}$ to be at equilibrium, the temperature will be:

• A. $1250K$
• B. $500K$
• C. $750K$
• D. $1000K$

Answer 1

The correct option is C $750K$

We know at equilibrium,

$\Delta S=\frac{\Delta H}{T}$

$\Delta S=\Delta S_{X{Y}_3}-\frac{3}{2}\times \Delta S_{Y_2}-\frac{1}{2}\times \Delta S_{X_2}=\frac{\Delta H}{T}$

$\Delta S=-40J/K/mol$

$\Delta H=-30000J/mol$

$T=\frac{\Delta H}{\Delta S}=\frac{-30000}{-40}$

$T=750K$

Option C is correct.