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,
$\frac{1}{2}{X}_2+\frac{3}{2}{Y}_2⟶X{Y}_3,△H=-30kJ$, to be at equilibrium, the temperature will be:

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

Answer 1

The correct option is C $750K$

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

$\Delta H-T\Delta S=0$

or $\Delta H=T\Delta S$

For the reaction,
$\frac{1}{2}{X}_2+\frac{3}{2}{Y}_2⟶X{Y}_3;\Delta H=-30kJ$ (given)

Calculating $\Delta S$ for the above reaction, we get:
$\Delta S=50-[\frac{1}{2}\times 60+\frac{3}{2}\times 40]J{K}^{-1}$

$=50-(30+60)J{K}^{-1}=-40J{K}^{-1}$

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

$[\Delta G=0]$

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

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

Hence, option C is correct.