Question

Standard entropies of X${}_2$, Y${}_2$ and XY${}_3$ are 60, 40 and 50 JK${}^{-1}$ mol${}^{-1}$ respectively. For the reaction:
$\frac{1}{2}{X}_2+\frac{3}{2}{Y}_2\rightleftharpoons X{Y}_3,\Delta H=-30$ kJ to be at equilibrium, the temperature should be :

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

Answer 1

The correct option is A 750 K

$\Delta$G= $\Delta$H - T$\Delta$S.

When the chemical reaction is at equilibrium , $\Delta$G = 0

So,

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

$\Delta$H = T$\Delta$S

In the Reaction, $\frac{1}{2}{X}_2$ + $\frac{3}{2}{Y}_2$ ⇌ $X{Y}_3$

$\Delta S_{total}$ = $\Delta S_{product}$ - $\Delta S_{reactants}$

$\Delta S_{total}$ = 50 - ($\frac{1}{2}$(60) + $\frac{3}{2}$(40) ) $J$ $K^{-1}$ $mo{l}^{-1}$

$\Delta S_{total}$ = -40 $J$ $K^{-1}$ $mo{l}^{-1}$

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

$T=\frac{-30\times {10}^3}{-40}$

$T=750K$