MCQ
Standard entropy of $X_2, Y_2$ and $XY_3$ are $60, 40$ and $50 \,JK^{-1} \,mol^{-1}$, respectively. For the reaction, $\frac{1}{2} X_2\,+\,\frac{3}{2} Y_2 \to XY_3, \Delta H = -30\,kJ$, to be at equilibrium, the temperature will be .....$K$
  • A
    $500$
  • $750$
  • C
    $1000$
  • D
    $1250$

Answer

Correct option: B.
$750$
b
$\frac{1}{2} \mathrm{X}_{2}+\frac{3}{2} \mathrm{Y}_{2} \rightarrow \mathrm{XY}_{3}$

$\Delta S =50-\left(\frac{60}{2}+\frac{3}{2} \times 40\right)=50-(30+60) $

$=-40 \,\mathrm{J} / \mathrm{Kmol} \text { at equilibrium } \Delta \mathrm{G}=0$

$\Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} ; \mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{-30 \times 10^{3}}{-40}=750\, \mathrm{K}$

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