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Question 23084
NEET Chemistry · Medium

Thermodynamics: Entropy numerics MCQ

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Entropy numericsMediumQuestion 23084

Question

If ΔH=−40\Delta H = -40 kJ and ΔS=−100\Delta S = -100 J K−1^{-1}, the process is spontaneous at 200 K because ΔG\Delta G is:

  1. A
    -20 kJ
    Correct
  2. B
    +20 kJ
  3. C
    -60 kJ
  4. D
    +60 kJ

Correct answer

-20 kJ

Explanation

Convert entropy term: TΔS=200×(−0.1)=−20T\Delta S = 200\times(-0.1) = -20 kJ, so ΔG=−40−(−20)\Delta G = -40 - (-20).

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