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NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice

For the reaction: X 2 O 4 l → 2X O 2 g ΔU = 2 .1 kcal, ΔS = 20 cal K − 1 300 K Hence Δ G is

Options

  1. A9 .3 kcal
  2. B2.7 kcal
  3. C-2 .7 kcal
  4. D– 9.3 kcal

Correct answer

C. -2 .7 kcal

Step-by-step solution

X 2 O 4 l → 2 X O 2 g ; Δ n g = 2 - 0 = 2 Δ H = Δ U + Δ n g R T = 2.1 + 2 × 2 1000 × 300 ΔH = 3 .3 kcal ΔG = ΔH − T ⋅ ΔS = 3 .3 − 300 × 20 1000 ; ΔG = - 2 .7 kcal

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