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What is the value of Δ G kJ / mol at 298 K at some non-equilibrium condition? Given the concentrations of [NH 3 ] is 0.05 M and [NH 4 + ] = [OH - ] = 0.002 M in the presence of excess water. Also Δ G Reaction ∘ = + 2 6 · 8 1 KJ mol . NH 3 (aq) + H 2 O(l) ⇌ NH 4 + (aq) + OH − (aq)

Options

  1. A+ 3.437
  2. B- 9.433
  3. C+ 50.18
  4. D- 50.18

Correct answer

A. + 3.437

Step-by-step solution

Δ G = Δ G ∘ + 2 . 3 0 3 RT log Q = 2 6 . 8 1 kJ + 2 . 3 0 3 × 8 . 3 1 4 × 2 9 8 log 2 × 1 0 - 3 × 2 × 1 0 - 3 5 × 1 0 - 2 = 2 6 . 8 1 kJ + 5 7 0 6 J log 4 5 × 1 0 - 4 = 2 6 . 8 1 + 5 . 7 0 6 - 4 + 0 . 6 0 - 0 . 7 = 2 6 . 8 1 + 5 . 7 0 6 - 4 . 1 = 2 6 . 8 1 - 2 3 . 3 7 = 3.44  kJ mol − 1

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