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Thermodynamics question

2016 · Q105
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Thermodynamics question

2016 · Q105

NEETChemistryThermodynamicsMultiple correct+4 / −1
The correct thermodynamic conditions for the spontaneous reaction at all temperatures is
  1. A
    Δ\DeltaΔH < 0 and Δ\DeltaΔS > 0
  2. B
    Δ\DeltaΔH < 0 and Δ\DeltaΔS < 0
  3. C
    Δ\DeltaΔH < 0 and Δ\DeltaΔS = 0
  4. D
    Δ\DeltaΔH > 0 and Δ\DeltaΔS < 0
View written solutionFree

Correct answer: A, C

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

If Δ\Delta ΔH < 0 and Δ\Delta ΔS > 0
   ΔG=(−ve)−T(+ve)\Delta G = (-ve) - T (+ve)ΔG=(−ve)−T(+ve)

then at all temperatures, ΔG\Delta GΔG = -ve, spontaneous reaction.

If Δ\Delta ΔH < 0 and Δ\Delta ΔS = 0
   ΔG=(−ve)−T(0)\Delta G = (-ve) - T (0)ΔG=(−ve)−T(0)

then at all temperatures, ΔG\Delta GΔG = -ve at all temperatures.

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