Question

Standard entropy of $X_2, Y_2$ and $XY_3$ are $60, 40$ and $50\, J K^{-1} mol^{-1},$ respectively, for the reaction
$\frac{1}{2}X_{2}+\frac{3}{2}Y_{2}\to XY_{3,}\Delta H=-30 kJ, $ to be at equilibrium, the temperature will be:

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

Answer 1

Answer: (C)

$\frac{1}{2}X_{2}+\frac{3}{2}Y_{2}\to XY_{3}$
$\Delta S_{\text{reaction}}=50-\left(\frac{3}{2}\times40+\frac{1}{2}\times60\right)=-40J mol^{-1}$
$\Delta G=\Delta H-T\Delta S
$ At equilibrium $\Delta G = 0$
$\Delta H=T\Delta S$
$30\times 10^3=T\times 40$
$\therefore T=750$