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Thermodynamics question

2017 · Q108
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Thermodynamics question

2017 · Q108

NEETChemistryThermodynamicsMCQ+4 / −1
For a given reaction, Δ\DeltaΔH = 35.5 kJ mol−-−1 and Δ\DeltaΔS = 83.6 J K−-−1 mol−-−1. The reaction is spontaneous at (Assume that Δ\DeltaΔH and Δ\DeltaΔS do not vary with temperature.)
  1. A
    T > 425 K
  2. B
    all temperatures
  3. C
    T > 298 K
  4. D
    T < 425 K
View written solutionFree

Correct answer: A

For a spontaneous reaction,

Δ\Delta ΔG < 0 i.e. Δ\Delta ΔH - TΔ\Delta ΔS < 0

T>ΔHΔST \gt {{\Delta H} \over {\Delta S}}T>ΔSΔH​

T>(35.5×100083.6=424.6≈425K)T \gt \left( {{{35.5 \times 1000} \over {83.6}} = 424.6 \approx 425K} \right)T>(83.635.5×1000​=424.6≈425K)

∴T>425 K \therefore T \gt 425\,K∴T>425K

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