JEE Main 202524 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2025
- Date
- 24 January
- Shift
- Morning Shift
- Subject
- Chemistry
- Chapter
- Thermodynamics (C)
JEE Main 2025 · 24 January · Morning Shift · Chemistry · Thermodynamics (C)
Standard entropies of X_2, Y_2 and XY_5 are 70,50 and 110 ~J ~K^ -1...
Standard entropies of and are 70,50 and respectively. The temperature in Kelvin at which the reaction
will be at equilibrium is_______. (Nearest integer)
Correct answer
0
Step-by-step solution
\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}