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Thermodynamics question

2010 · Q125
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Thermodynamics question

2010 · Q125

NEETChemistryThermodynamicsMCQ+4 / −1
Standard entropies of X2, Y2 and XY3 are 60, 40 and 50 J K−-−1 mol−-−1 respectively. For the reaction

1/2X2 + 3/2Y2 ⇌\rightleftharpoons⇌ XY3, Δ\DeltaΔH = −-− 30 kJ,

to be at equilibrium, the temperature should be
  1. A
    750 K
  2. B
    1000 K
  3. C
    1250 K
  4. D
    500 K
View written solutionFree

Correct answer: A

Given reaction is :

12{1 \over 2}21​X2 + 32{3 \over 2}23​Y2 ⇌ XY3

We know,

Δ\Delta ΔSo = ∑Sproductso−∑Sreactan⁡tso\sum {S_{products}^o} - \sum {S_{reac\tan ts}^o} ∑Sproductso​−∑Sreactantso​

= 50 - (30 + 60) = -40 J K-1 mol-1

At equilibrium Δ\Delta ΔGo = 0

Δ\Delta ΔHo = TΔ\Delta ΔSo

∴\therefore∴ T=ΔHoΔSoT = {{\Delta {H^o}} \over {\Delta {S^o}}}T=ΔSoΔHo​ = −30×103−40{{ - 30 \times {{10}^3}} \over { - 40}}−40−30×103​ = 750 K

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