JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
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
\(\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 \Delta \mathrm{H}^{\Theta}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
will be at equilibrium is_______. (Nearest integer)
- A 800
- B 900
- C 1000
- D 700
Answer & Solution
Correct Answer
(D) 700
Step-by-step Solution
Detailed 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}…
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