Q51 / 200 · NEET 2010 (Screening) · Chemistry
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
- Correct answerA$750 \mathrm{~K}$
- B$1000 \mathrm{~K}$
- C$1250 \mathrm{~K}$
- D$500 \mathrm{~K}$
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,
$$
\begin{aligned}
\Delta \mathrm{G}^{\circ} & =0 \\
\Delta \mathrm{H} & =\mathrm{T} \Delta \mathrm{S}^{\circ} \\
\mathrm{T} & =\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}^{\circ}}=\frac{-30 \times 10^3 \mathrm{~J}}{-40 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}=750 \mathrm{~K}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.