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N 2 + 3 H 2 ⟶ 2 N H 3 1 mole N 2 and 4 moles H 2 are taken in 15 L flask at 27 o C . After complete conversion of N 2 into N H 3 , 5 L of H 2 O is added. Pressure set up in the flask is (assume water dissolves N H 3 )

Options

  1. A3 × 0.0821 × 300 15 a t m
  2. B2 × 0.0821 × 300 10 a t m
  3. C1 × 0.0821 × 300 15 a t m
  4. D1 × 0.0821 × 300 10 a t m

Correct answer

D. 1 × 0.0821 × 300 10 a t m

Step-by-step solution

N 2 + 3 H 2 ⟶ 2 NH 3 Initial mole 1 4 0 Reacted mole 1 3 Left mole 0 4-3=1 2 H 2 left = 1 mol N H 3 formed = 2 mol But N H 3 g dissolves in water forming N H 4 O H . Residual gas is only H 2 which exerts pressure. Volume = 15 - 5 = 10 L (5L occupied by H 2 O added) ∴ p = n V R T = 1 × 0.0821 × 300 10 a t m

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