JEE MAIN - Chemistry (2025 - 24th January Morning Shift - No. 25)
Standard entropies of $\mathrm{X}_2, \mathrm{Y}_2$ and $\mathrm{XY}_5$ are 70, 50 and $110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature in Kelvin at which the reaction
$$\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 \Delta \mathrm{H}^{\ominus}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
will be at equilibrium is __________ (Nearest integer)
Answer
700
Explanation
$$\begin{aligned}
& \frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 \\
& \Delta \mathrm{~S}_{\mathrm{Rxn}}^0=110-\left[\left(\frac{1}{2} \times 70\right)+\left(\frac{5}{2} \times 50\right)\right] \\
& =110-160=-50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}^0=0 \text { at eqb } \\
& \Delta \mathrm{G}^0=\Delta \mathrm{H}^0-\mathrm{T} \Delta \mathrm{~S}^0 \\
& 0=-35000-\mathrm{T}(-50) \\
& \mathrm{T}=700 ~\mathrm{Kelvin}
\end{aligned}$$
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