Question

Standard entropy of $X_2,Y_2$ and $X{Y}_3$ are 60,40 and 50 $J{k}^{-1}mo{l}^{-1}$, respectively.

For the reaction, $\frac{1}{2}{X}_2+\frac{3}{2}{Y}_2\to X{Y}_3,\Delta H=-30kJ$, to be at equilibrium, the temperature wil be

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

Answer 1

The correct option is C

750 K

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

For the reaction

$\frac{1}{2}{X}_2+\frac{3}{2}{Y}_2\to X{Y}_3,\Delta H=-30kJ$, (given)

Calculating $\Delta S$ for the above reactions, 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, $T\Delta =\Delta H$ $[\because \Delta G=0]$

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

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