Question

Standard entropies of $X_2 ,Y_2$ and $XY_3$ are $60 , 40$ and $50 \, JK^{-1} mol^{-1}$ respectively. For the reaction ${ \frac{1}{2} X_2 + \frac{3}{2}Y_2 <=> XY_3, \Delta H = -30 kJ}$ to be at equilibrium, the temperature should be

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

Answer 1

Answer: (B)

750 K

Answer 2

${ \frac{1}{2} X_2 + \frac{3}{2} Y_2 <=> XY_3}$

$\Delta S$ reaction = \sum\Delta $$S_{product}-\sum\Delta S_{reactant}

Multipy eq-1 by 2

$\therefore X_2=3Y_2$ → $2XY_3$

Now: $\Delta H = -60KJ$
$\Delta S$ reaction = $X_2+3(Y_2) → 2(XY_3)$
$X_2=60ZJK^{-1}$
$Y_2=40 ZJK^{-1}$
$XY_3=50ZJK^{-1}$

$\Delta S_{reaction}=2\times50-3\times40-1\times60$
=100-120-60
$\Delta S^{\circ} = -80 \, JK^{-1} \, mol^{-1}$

Now $\Delta G=\Delta H-T\Delta S$
$\Delta G=0$, $\therefore \Delta H=T\Delta S$

1000X(-60)= -80XT

T=750K