Standard entropies of $\mathrm{X}_2, \mathrm{Y}_2$ and $\mathrm{XY}_3$ are 60,40 and $50 \mathrm{~J}…

Standard entropies of $\mathrm{X}_2, \mathrm{Y}_2$ and $\mathrm{XY}_3$ are 60,40 and $50 \mathrm{~J} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. For the reaction $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature should be
  1. $750 \mathrm{~K}$
  2. $1000 \mathrm{~K}$
  3. $1250 \mathrm{~K}$
  4. $500 \mathrm{~K}$

Solution

For the reaction, $\begin{aligned} & \frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ} \\ & \Delta \mathrm{S}^{\circ}=\mathrm{S}_{\left(\mathrm{XY}_3\right)}^{\circ}-\left[\frac{1}{2} \mathrm{~S}_{\mathrm{X}_2}^{\circ}+\frac{3}{2} \mathrm{~S}_{\mathrm{Y}_2}^{\circ}\right] \\ & =50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \\ & =50-[30+60]=50-90 \\ & =-40 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned}$ We know that, $\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ At equilibrium, $\begin{aligned} \Delta \mathrm{G}^{\circ} & =0 \\ \Delta \mathrm{H} & =\mathrm{T} \Delta \mathrm{S}^{\circ} \\ \mathrm{T} & =\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}^{\circ}}=\frac{-30 \times 10^3 \mathrm{~J}}{-40 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}=750 \mathrm{~K} \end{aligned}$

Asked in: NEET 2010 (Screening)

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