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Thermodynamics question

2025 · 24 Jan · Shift 1 · Q25
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Thermodynamics question

2025 · 24 Jan · Shift 1 · Q25

JEE MainChemistryThermodynamicsNumerical+4 / −1
Standard entropies of X2,Y2\mathrm{X}_2, \mathrm{Y}_2X2​,Y2​ and XY5\mathrm{XY}_5XY5​ are 70, 50 and 110 J K−1 mol−1110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}110 J K−1 mol−1 respectively. The temperature in Kelvin at which the reaction 12X2+52Y2⇌XY5ΔH⊖=−35 kJ mol−1\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}21​X2​+25​Y2​⇌XY5​ΔH⊖=−35 kJ mol−1 will be at equilibrium is ‾\underline{\hspace{2cm}}​ (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 700

  1. Condition for equilibrium

For a reaction to be at equilibrium under standard conditions,

ΔG∘=ΔH∘−TΔS∘=0\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} = 0ΔG∘=ΔH∘−TΔS∘=0

So,

T=ΔH∘ΔS∘T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}T=ΔS∘ΔH∘​
  1. Calculate standard entropy change

Reaction:

12X2+52Y2→XY5\frac{1}{2}X_2 + \frac{5}{2}Y_2 \rightarrow XY_521​X2​+25​Y2​→XY5​

Given standard entropies:

S∘(X2)=70 J K−1mol−1S^{\circ}(X_2)=70\,\text{J K}^{-1}\text{mol}^{-1}S∘(X2​)=70J K−1mol−1 S∘(Y2)=50 J K−1mol−1S^{\circ}(Y_2)=50\,\text{J K}^{-1}\text{mol}^{-1}S∘(Y2​)=50J K−1mol−1 S∘(XY5)=110 J K−1mol−1S^{\circ}(XY_5)=110\,\text{J K}^{-1}\text{mol}^{-1}S∘(XY5​)=110J K−1mol−1

Now,

ΔS∘=S∘(products)−S∘(reactants)\Delta S^{\circ} = S^{\circ}(\text{products}) - S^{\circ}(\text{reactants})ΔS∘=S∘(products)−S∘(reactants) ΔS∘=110−(12×70+52×50)\Delta S^{\circ} = 110 - \left(\frac{1}{2}\times 70 + \frac{5}{2}\times 50\right)ΔS∘=110−(21​×70+25​×50) =110−(35+125)= 110 - (35 + 125)=110−(35+125) =110−160=−50 J K−1mol−1= 110 - 160 = -50\,\text{J K}^{-1}\text{mol}^{-1}=110−160=−50J K−1mol−1
  1. Use enthalpy change

Given:

ΔH∘=−35 kJ mol−1=−35000 J mol−1\Delta H^{\circ} = -35\,\text{kJ mol}^{-1} = -35000\,\text{J mol}^{-1}ΔH∘=−35kJ mol−1=−35000J mol−1
  1. Compute equilibrium temperature

At equilibrium,

T=ΔH∘ΔS∘=−35000−50T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{-35000}{-50}T=ΔS∘ΔH∘​=−50−35000​ T=700 KT = 700\,\text{K}T=700K
  1. Final answer

Nearest integer:

700\boxed{700}700​
  1. Comparison with stored answer

Stored correct answer = 700700700, which matches the derived result.

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